Regime-Switching Interest Rate Models With Randomized Regimes · Regime-Switching Interest Rate...
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Regime-Switching Interest Rate Models WithRandomized Regimes
James G. Bridgeman, FSA
University of Connecticut
Actuarial Science Seminar Jan. 29, 2008
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 1
/ 36
Introduction
Work in progress on cash �ow testing interest rate models
Empirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):
Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):
Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tail
Mean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):
Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulder
Randomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):
Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes it
Trial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):
Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration
2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):
Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):
Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):
Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormal
A surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormal
Couldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targets
ARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):
Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):Asymptotic closed form calibration with randomized targets
Interesting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):Asymptotic closed form calibration with randomized targetsInteresting probability results/techniques
ARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensions
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 2
/ 36
Introduction
Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):
Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings
Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1
More Recent Results (2007):Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1
This year? Numerical examples and extensionsBridgeman (University of Connecticut) Random Regimes
Actuarial Science Seminar Jan. 29, 2008 2/ 36
Example: 55 Years of the 10-year Treasury Rate
10 YEAR TREASURY RATE 19532007 (monthly data)
0
2
4
6
8
10
12
14
16
18
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 3
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The Distribution of those Interest Rates
FREQUENCY OF 10 YEAR RATES
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
DATA LOGNORMAL
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 4
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Lognormal 4th Moment Is Just Too High (6th too)
FREQUENCY OF 10 YEAR RATES
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
DATA LOGNORMAL
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 5
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55 Years of Changes in the 10 Year Treasury Rate
MONTHLY LOGCHANGE IN 10 YEAR RATE
0.2
0.15
0.1
0.05
0
0.05
0.1
0.15
0.2
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 6
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What is the Distribution of Those Changes?
FREQUENCY OF MONTHLY LOGCHANGE IN 10 YEAR RATES
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
DATA GAUSSIAN
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 7
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For Rate Changes, Lognormal 4th Moment Too Low
FREQUENCY OF MONTHLY LOGCHANGE IN 10 YEAR RATE
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
DATA GAUSSIAN
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 8
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The Fix: Randomize the Reversion Target
50 YEAR SAMPLE PATH (A DANGEROUS ONE)
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1PATH TARGET
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 9
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Lognormal Models
Unconstrained:
d ln (r t )=Dtdt + σpdtNt
d ln(rt ) = Dtdt + σdWt
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 10
/ 36
Lognormal Models
Unconstrained:
d ln (r t )=Dtdt + σpdtNt
d ln(rt ) = Dtdt + σdWt
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 10
/ 36
Lognormal Models
Unconstrained:
d ln (r t )=Dtdt + σpdtNt
d ln(rt ) = Dtdt + σdWt
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 10
/ 36
Lognormal Models
Unconstrained:
d ln (r t )=Dtdt + σpdtNt
d ln(rt ) = Dtdt + σdWt
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 10
/ 36
Lognormal Models
Unconstrained:
d ln (r t )=Dtdt + σpdtNt
d ln(rt ) = Dtdt + σdWt
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)
d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 10
/ 36
Lognormal Models
Unconstrained:
d ln (r t )=Dtdt + σpdtNt
d ln(rt ) = Dtdt + σdWt
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 10
/ 36
Lognormal Models
Unconstrained:
d ln (r t )=Dtdt + σpdtNt
d ln(rt ) = Dtdt + σdWt
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 10
/ 36
Lognormal Models
Unconstrained:
d ln (r t )=Dtdt + σpdtNt
d ln(rt ) = Dtdt + σdWt
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�Bridgeman (University of Connecticut) Random Regimes
Actuarial Science Seminar Jan. 29, 2008 10/ 36
Lognormal Models
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�d ln(rt ) = � ln (1� F )
"∞
∑j=0
1[j ,j+1)(t) ln(Tj )� ln(rt )#dt
+Dtdt + σdWt
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 11
/ 36
Lognormal Models
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)
d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�d ln(rt ) = � ln (1� F )
"∞
∑j=0
1[j ,j+1)(t) ln(Tj )� ln(rt )#dt
+Dtdt + σdWt
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 11
/ 36
Lognormal Models
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�d ln(rt ) = � ln (1� F )
"∞
∑j=0
1[j ,j+1)(t) ln(Tj )� ln(rt )#dt
+Dtdt + σdWt
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 11
/ 36
Lognormal Models
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�d ln(rt ) = � ln (1� F )
"∞
∑j=0
1[j ,j+1)(t) ln(Tj )� ln(rt )#dt
+Dtdt + σdWt
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 11
/ 36
Lognormal Models
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�
d ln(rt ) = � ln (1� F )"
∞
∑j=0
1[j ,j+1)(t) ln(Tj )� ln(rt )#dt
+Dtdt + σdWt
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 11
/ 36
Lognormal Models
Mean-reverting:
d ln (r t )=h1� (1� F )dt
i[ln(T0)� ln(rt�dt )]
+ (1� F )dt Dtdt + (1� F )dt σpdtNt
actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)
With Randomized Reversion Target
d ln (r t )=h1� (1� F )dt
i " ∞
∑j=0
1[j,j+1)(t) ln(Tj )� ln(rt�dt )#
+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is
the indicator for t to be in a random interval�tj, tj+1
�d ln(rt ) = � ln (1� F )
"∞
∑j=0
1[j ,j+1)(t) ln(Tj )� ln(rt )#dt
+Dtdt + σdWt
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 11
/ 36
Drift Compensation and Calibration: plain mean-reversion
It would be intuitive to have:
E [rt ] = r(1�F )t0 T
[1�(1�F )t ]0
To �nd out what drift Dt will ensure it, you can integrate d ln(rt ) :
ln(rt ) = ln(r0) (1� F )tdt dt + σ
pdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h1� (1� F )dt
i tdt
∑s=1(1� F )(s�1)dt (= notice geom. series
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt which simpli�es to:
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)�1� (1� F )t
�+ dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt , which is
Gaussian.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 12
/ 36
Drift Compensation and Calibration: plain mean-reversion
It would be intuitive to have:
E [rt ] = r(1�F )t0 T
[1�(1�F )t ]0
To �nd out what drift Dt will ensure it, you can integrate d ln(rt ) :
ln(rt ) = ln(r0) (1� F )tdt dt + σ
pdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h1� (1� F )dt
i tdt
∑s=1(1� F )(s�1)dt (= notice geom. series
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt which simpli�es to:
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)�1� (1� F )t
�+ dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt , which is
Gaussian.Bridgeman (University of Connecticut) Random Regimes
Actuarial Science Seminar Jan. 29, 2008 12/ 36
Drift Compensation and Calibration: plain mean-reversion
Since ln(rt ) is Gaussian, E [rt ] = eµ+ 12 σ2where the µ and σ2 are
some mess determined by the constants in the expression for ln(rt ).
If you work that mess out and set it equal to r (1�F )t
0 T[1�(1�F )t ]0 ,
and require that it be true for all t, you can arrive at what the driftcompensation function Dt must be to deliver the intuitive E [rt ] :
Dt = � 12σ2(1�F )dt
1+(1�F )dth1+ (1� F )2t�dt
i, or
Dt = � 14σ2h1+ (1� F )2t
iin the continuous case
There is a similar closed form for the variance of rt based onE�r2t�= e2µ+ 1
2 (2σ)2which can help calibrate the model to historical
variance F = 1�n1� σ2obsdt
ln(Vobs+T 2)�ln(T 2)
o 12dt
Practical work with the randomized reversion target model all butrequires you to know similar closed forms for drift compensation andvariance, but now when you integrate no geometric series appears.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 13
/ 36
Drift Compensation and Calibration: plain mean-reversion
Since ln(rt ) is Gaussian, E [rt ] = eµ+ 12 σ2where the µ and σ2 are
some mess determined by the constants in the expression for ln(rt ).
If you work that mess out and set it equal to r (1�F )t
0 T[1�(1�F )t ]0 ,
and require that it be true for all t, you can arrive at what the driftcompensation function Dt must be to deliver the intuitive E [rt ] :
Dt = � 12σ2(1�F )dt
1+(1�F )dth1+ (1� F )2t�dt
i, or
Dt = � 14σ2h1+ (1� F )2t
iin the continuous case
There is a similar closed form for the variance of rt based onE�r2t�= e2µ+ 1
2 (2σ)2which can help calibrate the model to historical
variance F = 1�n1� σ2obsdt
ln(Vobs+T 2)�ln(T 2)
o 12dt
Practical work with the randomized reversion target model all butrequires you to know similar closed forms for drift compensation andvariance, but now when you integrate no geometric series appears.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 13
/ 36
Drift Compensation and Calibration: plain mean-reversion
Since ln(rt ) is Gaussian, E [rt ] = eµ+ 12 σ2where the µ and σ2 are
some mess determined by the constants in the expression for ln(rt ).
If you work that mess out and set it equal to r (1�F )t
0 T[1�(1�F )t ]0 ,
and require that it be true for all t, you can arrive at what the driftcompensation function Dt must be to deliver the intuitive E [rt ] :
Dt = � 12σ2(1�F )dt
1+(1�F )dth1+ (1� F )2t�dt
i, or
Dt = � 14σ2h1+ (1� F )2t
iin the continuous case
There is a similar closed form for the variance of rt based onE�r2t�= e2µ+ 1
2 (2σ)2which can help calibrate the model to historical
variance F = 1�n1� σ2obsdt
ln(Vobs+T 2)�ln(T 2)
o 12dt
Practical work with the randomized reversion target model all butrequires you to know similar closed forms for drift compensation andvariance, but now when you integrate no geometric series appears.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 13
/ 36
Drift Compensation and Calibration: plain mean-reversion
Since ln(rt ) is Gaussian, E [rt ] = eµ+ 12 σ2where the µ and σ2 are
some mess determined by the constants in the expression for ln(rt ).
If you work that mess out and set it equal to r (1�F )t
0 T[1�(1�F )t ]0 ,
and require that it be true for all t, you can arrive at what the driftcompensation function Dt must be to deliver the intuitive E [rt ] :
Dt = � 12σ2(1�F )dt
1+(1�F )dth1+ (1� F )2t�dt
i, or
Dt = � 14σ2h1+ (1� F )2t
iin the continuous case
There is a similar closed form for the variance of rt based onE�r2t�= e2µ+ 1
2 (2σ)2which can help calibrate the model to historical
variance F = 1�n1� σ2obsdt
ln(Vobs+T 2)�ln(T 2)
o 12dt
Practical work with the randomized reversion target model all butrequires you to know similar closed forms for drift compensation andvariance, but now when you integrate no geometric series appears.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 13
/ 36
Drift Compensation and Calibration: plain mean-reversion
Since ln(rt ) is Gaussian, E [rt ] = eµ+ 12 σ2where the µ and σ2 are
some mess determined by the constants in the expression for ln(rt ).
If you work that mess out and set it equal to r (1�F )t
0 T[1�(1�F )t ]0 ,
and require that it be true for all t, you can arrive at what the driftcompensation function Dt must be to deliver the intuitive E [rt ] :
Dt = � 12σ2(1�F )dt
1+(1�F )dth1+ (1� F )2t�dt
i, or
Dt = � 14σ2h1+ (1� F )2t
iin the continuous case
There is a similar closed form for the variance of rt based onE�r2t�= e2µ+ 1
2 (2σ)2which can help calibrate the model to historical
variance F = 1�n1� σ2obsdt
ln(Vobs+T 2)�ln(T 2)
o 12dt
Practical work with the randomized reversion target model all butrequires you to know similar closed forms for drift compensation andvariance, but now when you integrate no geometric series appears.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 13
/ 36
Drift Compensation and Calibration: plain mean-reversion
Since ln(rt ) is Gaussian, E [rt ] = eµ+ 12 σ2where the µ and σ2 are
some mess determined by the constants in the expression for ln(rt ).
If you work that mess out and set it equal to r (1�F )t
0 T[1�(1�F )t ]0 ,
and require that it be true for all t, you can arrive at what the driftcompensation function Dt must be to deliver the intuitive E [rt ] :
Dt = � 12σ2(1�F )dt
1+(1�F )dth1+ (1� F )2t�dt
i, or
Dt = � 14σ2h1+ (1� F )2t
iin the continuous case
There is a similar closed form for the variance of rt based onE�r2t�= e2µ+ 1
2 (2σ)2which can help calibrate the model to historical
variance F = 1�n1� σ2obsdt
ln(Vobs+T 2)�ln(T 2)
o 12dt
Practical work with the randomized reversion target model all butrequires you to know similar closed forms for drift compensation andvariance, but now when you integrate no geometric series appears.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 13
/ 36
Drift Compensation & Calibration: random mean-reversion
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+h1� (1� F )dt
i tdt
∑s=1
∞
∑j=01[j,j+1)(sdt) ln(Tj ) (1� F )t�sdt (ugly
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 14
/ 36
Drift Compensation & Calibration: random mean-reversion
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+h1� (1� F )dt
i tdt
∑s=1
∞
∑j=01[j,j+1)(sdt) ln(Tj ) (1� F )t�sdt (ugly
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt , but switch the order of summation:
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h(1� F )(t�t1)+ � (1� F )t
i+
∞
∑j=1ln(Tj )
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i(=after telescoping
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt , so rt 6= lognormal, = log-log-gamma?
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 15
/ 36
Drift Compensation & Calibration: random mean-reversion
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+h1� (1� F )dt
i tdt
∑s=1
∞
∑j=01[j,j+1)(sdt) ln(Tj ) (1� F )t�sdt (ugly
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt , but switch the order of summation:
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h(1� F )(t�t1)+ � (1� F )t
i+
∞
∑j=1ln(Tj )
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i(=after telescoping
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt , so rt 6= lognormal, = log-log-gamma?Bridgeman (University of Connecticut) Random Regimes
Actuarial Science Seminar Jan. 29, 2008 15/ 36
Condition on the Times When Regimes Switch
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h(1� F )(t�t1)+ � (1� F )t
i+
∞
∑j=1ln(Tj )
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt
Conditioning on the tj random variables and using a lognormal model(reasonable) for the random targets Tj (so each ln(Tj ) is Gaussian)we again have a (messy) Gaussian for the conditional ln(rt ). Canthat help in calculating an unconditioned E [rt ] and variance?
The answer is "Yes" ... up to an approximate expansion.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 16
/ 36
Condition on the Times When Regimes Switch
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h(1� F )(t�t1)+ � (1� F )t
i+
∞
∑j=1ln(Tj )
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt
Conditioning on the tj random variables and using a lognormal model(reasonable) for the random targets Tj (so each ln(Tj ) is Gaussian)we again have a (messy) Gaussian for the conditional ln(rt ). Canthat help in calculating an unconditioned E [rt ] and variance?
The answer is "Yes" ... up to an approximate expansion.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 16
/ 36
Condition on the Times When Regimes Switch
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h(1� F )(t�t1)+ � (1� F )t
i+
∞
∑j=1ln(Tj )
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt
Conditioning on the tj random variables and using a lognormal model(reasonable) for the random targets Tj (so each ln(Tj ) is Gaussian)we again have a (messy) Gaussian for the conditional ln(rt ). Canthat help in calculating an unconditioned E [rt ] and variance?
The answer is "Yes" ... up to an approximate expansion.
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 16
/ 36
Edgeworth Expansion for the Unconditioned Moments
We expect the tails of ln(rt ) to be supressed in favor of the shoulders.That suggests that E [rt ], and higher moments as well, might beapproximated e¢ ciently by an Edgeworth expansion for ln(rt ). Itworks out to be surprisingly simple:
Eh(rt )
li� e lµ+ 1
2 (lσ)2n1+ l4
4!
�µ4 � 3σ4
�oHere σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.
� e lµ+ 12 (lσ)
2n1+ l4
4!
�µ4 � 3σ4
� �1� 3
4! (lσ)2�+ l6
6!
�µ6 � 15σ6
�o=
e lµ+12 (lσ)
2
(1+ limN!∞
N∑j=2
l2j(2j)!
hµ2j � (2j)?σ2j
i N�j∑n=0
(�1)n(2n)?(2n)! (lσ)2n
)where (2n)? = (2n� 1) (2n� 3) � � � (1)Conditional Gaussian ensures that odd higher moments vanish.
The problem now is to calculate µ, σ2, and the µ2j
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 17
/ 36
Edgeworth Expansion for the Unconditioned Moments
We expect the tails of ln(rt ) to be supressed in favor of the shoulders.That suggests that E [rt ], and higher moments as well, might beapproximated e¢ ciently by an Edgeworth expansion for ln(rt ). Itworks out to be surprisingly simple:
Eh(rt )
li� e lµ+ 1
2 (lσ)2n1+ l4
4!
�µ4 � 3σ4
�o
Here σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.
� e lµ+ 12 (lσ)
2n1+ l4
4!
�µ4 � 3σ4
� �1� 3
4! (lσ)2�+ l6
6!
�µ6 � 15σ6
�o=
e lµ+12 (lσ)
2
(1+ limN!∞
N∑j=2
l2j(2j)!
hµ2j � (2j)?σ2j
i N�j∑n=0
(�1)n(2n)?(2n)! (lσ)2n
)where (2n)? = (2n� 1) (2n� 3) � � � (1)Conditional Gaussian ensures that odd higher moments vanish.
The problem now is to calculate µ, σ2, and the µ2j
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 17
/ 36
Edgeworth Expansion for the Unconditioned Moments
We expect the tails of ln(rt ) to be supressed in favor of the shoulders.That suggests that E [rt ], and higher moments as well, might beapproximated e¢ ciently by an Edgeworth expansion for ln(rt ). Itworks out to be surprisingly simple:
Eh(rt )
li� e lµ+ 1
2 (lσ)2n1+ l4
4!
�µ4 � 3σ4
�oHere σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.
� e lµ+ 12 (lσ)
2n1+ l4
4!
�µ4 � 3σ4
� �1� 3
4! (lσ)2�+ l6
6!
�µ6 � 15σ6
�o=
e lµ+12 (lσ)
2
(1+ limN!∞
N∑j=2
l2j(2j)!
hµ2j � (2j)?σ2j
i N�j∑n=0
(�1)n(2n)?(2n)! (lσ)2n
)where (2n)? = (2n� 1) (2n� 3) � � � (1)Conditional Gaussian ensures that odd higher moments vanish.
The problem now is to calculate µ, σ2, and the µ2j
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 17
/ 36
Edgeworth Expansion for the Unconditioned Moments
We expect the tails of ln(rt ) to be supressed in favor of the shoulders.That suggests that E [rt ], and higher moments as well, might beapproximated e¢ ciently by an Edgeworth expansion for ln(rt ). Itworks out to be surprisingly simple:
Eh(rt )
li� e lµ+ 1
2 (lσ)2n1+ l4
4!
�µ4 � 3σ4
�oHere σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.
� e lµ+ 12 (lσ)
2n1+ l4
4!
�µ4 � 3σ4
� �1� 3
4! (lσ)2�+ l6
6!
�µ6 � 15σ6
�o
=
e lµ+12 (lσ)
2
(1+ limN!∞
N∑j=2
l2j(2j)!
hµ2j � (2j)?σ2j
i N�j∑n=0
(�1)n(2n)?(2n)! (lσ)2n
)where (2n)? = (2n� 1) (2n� 3) � � � (1)Conditional Gaussian ensures that odd higher moments vanish.
The problem now is to calculate µ, σ2, and the µ2j
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 17
/ 36
Edgeworth Expansion for the Unconditioned Moments
We expect the tails of ln(rt ) to be supressed in favor of the shoulders.That suggests that E [rt ], and higher moments as well, might beapproximated e¢ ciently by an Edgeworth expansion for ln(rt ). Itworks out to be surprisingly simple:
Eh(rt )
li� e lµ+ 1
2 (lσ)2n1+ l4
4!
�µ4 � 3σ4
�oHere σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.
� e lµ+ 12 (lσ)
2n1+ l4
4!
�µ4 � 3σ4
� �1� 3
4! (lσ)2�+ l6
6!
�µ6 � 15σ6
�o=
e lµ+12 (lσ)
2
(1+ limN!∞
N∑j=2
l2j(2j)!
hµ2j � (2j)?σ2j
i N�j∑n=0
(�1)n(2n)?(2n)! (lσ)2n
)where (2n)? = (2n� 1) (2n� 3) � � � (1)
Conditional Gaussian ensures that odd higher moments vanish.
The problem now is to calculate µ, σ2, and the µ2j
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 17
/ 36
Edgeworth Expansion for the Unconditioned Moments
We expect the tails of ln(rt ) to be supressed in favor of the shoulders.That suggests that E [rt ], and higher moments as well, might beapproximated e¢ ciently by an Edgeworth expansion for ln(rt ). Itworks out to be surprisingly simple:
Eh(rt )
li� e lµ+ 1
2 (lσ)2n1+ l4
4!
�µ4 � 3σ4
�oHere σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.
� e lµ+ 12 (lσ)
2n1+ l4
4!
�µ4 � 3σ4
� �1� 3
4! (lσ)2�+ l6
6!
�µ6 � 15σ6
�o=
e lµ+12 (lσ)
2
(1+ limN!∞
N∑j=2
l2j(2j)!
hµ2j � (2j)?σ2j
i N�j∑n=0
(�1)n(2n)?(2n)! (lσ)2n
)where (2n)? = (2n� 1) (2n� 3) � � � (1)Conditional Gaussian ensures that odd higher moments vanish.
The problem now is to calculate µ, σ2, and the µ2j
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 17
/ 36
Edgeworth Expansion for the Unconditioned Moments
We expect the tails of ln(rt ) to be supressed in favor of the shoulders.That suggests that E [rt ], and higher moments as well, might beapproximated e¢ ciently by an Edgeworth expansion for ln(rt ). Itworks out to be surprisingly simple:
Eh(rt )
li� e lµ+ 1
2 (lσ)2n1+ l4
4!
�µ4 � 3σ4
�oHere σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.
� e lµ+ 12 (lσ)
2n1+ l4
4!
�µ4 � 3σ4
� �1� 3
4! (lσ)2�+ l6
6!
�µ6 � 15σ6
�o=
e lµ+12 (lσ)
2
(1+ limN!∞
N∑j=2
l2j(2j)!
hµ2j � (2j)?σ2j
i N�j∑n=0
(�1)n(2n)?(2n)! (lσ)2n
)where (2n)? = (2n� 1) (2n� 3) � � � (1)Conditional Gaussian ensures that odd higher moments vanish.
The problem now is to calculate µ, σ2, and the µ2j
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 17
/ 36
Expected Value Easy
Remember,
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h(1� F )(t�t1)+ � (1� F )t
i+
∞
∑j=1ln(Tj )
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt and condition on the tj
So µ = E[ ln(rt )] is given by
ln(r0) (1� F )t + ln(T0)n
Eh(1� F )(t�t1)+
i� (1� F )t
o+µTE
"∞
∑j=1
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i#
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt where µT = E [ln(Tj )]
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 18
/ 36
Expected Value Easy
Remember,
ln(rt ) = ln(r0) (1� F )t + σpdt
tdt
∑s=1
Nt�(s�1)dt (1� F )sdt
+ ln(T0)h(1� F )(t�t1)+ � (1� F )t
i+
∞
∑j=1ln(Tj )
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt and condition on the tj
So µ = E[ ln(rt )] is given by
ln(r0) (1� F )t + ln(T0)n
Eh(1� F )(t�t1)+
i� (1� F )t
o+µTE
"∞
∑j=1
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i#
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt where µT = E [ln(Tj )]Bridgeman (University of Connecticut) Random Regimes
Actuarial Science Seminar Jan. 29, 2008 18/ 36
Expected Value Easy
So µ = E[ ln(rt )] is given by
ln(r0) (1� F )t + ln(T0)n
Eh(1� F )(t�t1)+
i� (1� F )t
o+µTE
"∞
∑j=1
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i#
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt where µT = E [ln(Tj )]
Telescoping,= ln(r0) (1� F )t + ln(T0)
nEh(1� F )(t�t1)+
i� (1� F )t
o+µT
n1�E
h(1� F )(t�t1)+
io+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt
Eh(1� F )(t�t1)+
iturns out to be a Laplace transform that we can
calculuate (later).
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 19
/ 36
Expected Value Easy
So µ = E[ ln(rt )] is given by
ln(r0) (1� F )t + ln(T0)n
Eh(1� F )(t�t1)+
i� (1� F )t
o+µTE
"∞
∑j=1
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i#
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt where µT = E [ln(Tj )]
Telescoping,= ln(r0) (1� F )t + ln(T0)
nEh(1� F )(t�t1)+
i� (1� F )t
o+µT
n1�E
h(1� F )(t�t1)+
io+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt
Eh(1� F )(t�t1)+
iturns out to be a Laplace transform that we can
calculuate (later).
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 19
/ 36
Expected Value Easy
So µ = E[ ln(rt )] is given by
ln(r0) (1� F )t + ln(T0)n
Eh(1� F )(t�t1)+
i� (1� F )t
o+µTE
"∞
∑j=1
h(1� F )(t�tj+1)+ � (1� F )(t�tj )+
i#
+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt where µT = E [ln(Tj )]
Telescoping,= ln(r0) (1� F )t + ln(T0)
nEh(1� F )(t�t1)+
i� (1� F )t
o+µT
n1�E
h(1� F )(t�t1)+
io+dt
tdt
∑s=1
Dt�(s�1)dt (1� F )sdt
Eh(1� F )(t�t1)+
iturns out to be a Laplace transform that we can
calculuate (later).
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 19
/ 36
Higher Moments Hard
Remembering that the even central moments of std normal are(2n)? = (2n� 1) (2n� 3) � � � (1), the even central moments ofln(rt ) are E
hfln(rt )�E [ln(rt )]g2n
i= (2n)?E
248<:σ2dt
tdt
∑s=1(1� F )2sdt + σ2T
∞
∑j=1e2j
9=;n35
= (2n)?E
"(σ2dt (1� F )2dt 1�(1�F )
2t
1�(1�F )2dt+ σ2T
∞
∑j=1e2j
)n#
σ2T is the common variance of the ln(Tj ) Gaussians
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofor each j
The fgn part can be expanded binomially, but that still leaves termslike...
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 20
/ 36
Higher Moments Hard
Remembering that the even central moments of std normal are(2n)? = (2n� 1) (2n� 3) � � � (1), the even central moments ofln(rt ) are E
hfln(rt )�E [ln(rt )]g2n
i= (2n)?E
248<:σ2dt
tdt
∑s=1(1� F )2sdt + σ2T
∞
∑j=1e2j
9=;n35
= (2n)?E
"(σ2dt (1� F )2dt 1�(1�F )
2t
1�(1�F )2dt+ σ2T
∞
∑j=1e2j
)n#σ2T is the common variance of the ln(Tj ) Gaussians
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofor each j
The fgn part can be expanded binomially, but that still leaves termslike...
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 20
/ 36
Higher Moments Hard
Remembering that the even central moments of std normal are(2n)? = (2n� 1) (2n� 3) � � � (1), the even central moments ofln(rt ) are E
hfln(rt )�E [ln(rt )]g2n
i= (2n)?E
248<:σ2dt
tdt
∑s=1(1� F )2sdt + σ2T
∞
∑j=1e2j
9=;n35
= (2n)?E
"(σ2dt (1� F )2dt 1�(1�F )
2t
1�(1�F )2dt+ σ2T
∞
∑j=1e2j
)n#σ2T is the common variance of the ln(Tj ) Gaussians
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofor each j
The fgn part can be expanded binomially, but that still leaves termslike...
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 20
/ 36
Higher Moments Hard
Remembering that the even central moments of std normal are(2n)? = (2n� 1) (2n� 3) � � � (1), the even central moments ofln(rt ) are E
hfln(rt )�E [ln(rt )]g2n
i= (2n)?E
248<:σ2dt
tdt
∑s=1(1� F )2sdt + σ2T
∞
∑j=1e2j
9=;n35
= (2n)?E
"(σ2dt (1� F )2dt 1�(1�F )
2t
1�(1�F )2dt+ σ2T
∞
∑j=1e2j
)n#σ2T is the common variance of the ln(Tj ) Gaussians
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofor each j
The fgn part can be expanded binomially, but that still leaves termslike...
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 20
/ 36
Still Need To Evaluate Terms Like
... E
" ∞
∑j=1e2j
!m#where the
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofail to be independent and
are each complicated in their own right.
But they do have a uniform correlation property
Lemma: Ehe2a1j1 � � � e2akjk
i= ρa1,...,akE
he2a1j1
i� � �E
he2akjk
iindependent of fj1, ..., jkg for distinct fj1, ..., jkgρa1,...,ak can be computed using Laplace transforms and there�s even arecursive relationship ρa1,...,ak = ρa1,a2+...+ak ρa2,...,akHow does that help?
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 21
/ 36
Still Need To Evaluate Terms Like
... E
" ∞
∑j=1e2j
!m#where the
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofail to be independent and
are each complicated in their own right.
But they do have a uniform correlation property
Lemma: Ehe2a1j1 � � � e2akjk
i= ρa1,...,akE
he2a1j1
i� � �E
he2akjk
iindependent of fj1, ..., jkg for distinct fj1, ..., jkgρa1,...,ak can be computed using Laplace transforms and there�s even arecursive relationship ρa1,...,ak = ρa1,a2+...+ak ρa2,...,akHow does that help?
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 21
/ 36
Still Need To Evaluate Terms Like
... E
" ∞
∑j=1e2j
!m#where the
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofail to be independent and
are each complicated in their own right.
But they do have a uniform correlation property
Lemma: Ehe2a1j1 � � � e2akjk
i= ρa1,...,akE
he2a1j1
i� � �E
he2akjk
iindependent of fj1, ..., jkg for distinct fj1, ..., jkg
ρa1,...,ak can be computed using Laplace transforms and there�s even arecursive relationship ρa1,...,ak = ρa1,a2+...+ak ρa2,...,akHow does that help?
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 21
/ 36
Still Need To Evaluate Terms Like
... E
" ∞
∑j=1e2j
!m#where the
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofail to be independent and
are each complicated in their own right.
But they do have a uniform correlation property
Lemma: Ehe2a1j1 � � � e2akjk
i= ρa1,...,akE
he2a1j1
i� � �E
he2akjk
iindependent of fj1, ..., jkg for distinct fj1, ..., jkgρa1,...,ak can be computed using Laplace transforms and there�s even arecursive relationship ρa1,...,ak = ρa1,a2+...+ak ρa2,...,ak
How does that help?
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 21
/ 36
Still Need To Evaluate Terms Like
... E
" ∞
∑j=1e2j
!m#where the
ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+
ofail to be independent and
are each complicated in their own right.
But they do have a uniform correlation property
Lemma: Ehe2a1j1 � � � e2akjk
i= ρa1,...,akE
he2a1j1
i� � �E
he2akjk
iindependent of fj1, ..., jkg for distinct fj1, ..., jkgρa1,...,ak can be computed using Laplace transforms and there�s even arecursive relationship ρa1,...,ak = ρa1,a2+...+ak ρa2,...,akHow does that help?
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 21
/ 36
For Example
E
24 ∞
∑j=1e2j
!235 = E
"∞
∑j=1e4j +
∞
∑j=1e2j
( ∞
∑i=1e2i
!� e2j
)#
So E
24 ∞
∑j=1e2j
!235 = E
"∞
∑j=1e4j
#+
ρ1,1
8<:
E
"∞
∑j=1e2j
#!2� E
"∞
∑j=1e2j E
he2ji#9=; using monotone
convergence to run expectations across ∞ sums
It gets complicated fast
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 22
/ 36
For Example
E
24 ∞
∑j=1e2j
!235 = E
"∞
∑j=1e4j +
∞
∑j=1e2j
( ∞
∑i=1e2i
!� e2j
)#
So E
24 ∞
∑j=1e2j
!235 = E
"∞
∑j=1e4j
#+
ρ1,1
8<:
E
"∞
∑j=1e2j
#!2� E
"∞
∑j=1e2j E
he2ji#9=; using monotone
convergence to run expectations across ∞ sums
It gets complicated fast
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 22
/ 36
For Example
E
24 ∞
∑j=1e2j
!235 = E
"∞
∑j=1e4j +
∞
∑j=1e2j
( ∞
∑i=1e2i
!� e2j
)#
So E
24 ∞
∑j=1e2j
!235 = E
"∞
∑j=1e4j
#+
ρ1,1
8<:
E
"∞
∑j=1e2j
#!2� E
"∞
∑j=1e2j E
he2ji#9=; using monotone
convergence to run expectations across ∞ sums
It gets complicated fast
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 22
/ 36
For m=3
E
24 ∞
∑j=1e2j
!335=E
26666666664
∞
∑j=1e6j + 3
∞
∑j=1e4j
( ∞
∑i=1e2i
!� e2j
)
+∞
∑j=1e2j
8>>>><>>>>:
∞
∑i=1e2i
" ∞
∑k=1
e2k
!� e2i � e2j
#!
�e2j
" ∞
∑k=1
e2k
!� e2j
#+ e4j
9>>>>=>>>>;
37777777775=E
"∞
∑j=1e6j
#+ 3ρ2,1E
"∞
∑j=1e4j
#E
"∞
∑j=1e2j
#
��3ρ2,1 � ρ1,1,1
�E
"∞
∑j=1e4j E
he2ji#+ρ1,1,1
8<:
E
"∞
∑j=1e2j
#!3
�3E"
∞
∑j=1e2j
#E
"∞
∑j=1e2j E
he2ji#
+E
"∞
∑j=1e2j�
Ehe2ji�2#)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 23
/ 36
Complicated, but each piece is simpler
Now all you need to be able to evaluate are terms like
E
"∞
∑j=1e2nj
n∏k=1
�Ehe2kji�nk#
, where ∑nk=1 knk � m� n
In fact, we will develop a calculation that includes the odd powers
too, E
"∞
∑j=1enj
n∏k=1
�Ehekji�nk#
Some notation: to save ink later let ν(x) stand for xnn∏k=1
E�xk�nk so
our expression abbreviates to E
"∞
∑j=1
ν (ej )
#
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 24
/ 36
Complicated, but each piece is simpler
Now all you need to be able to evaluate are terms like
E
"∞
∑j=1e2nj
n∏k=1
�Ehe2kji�nk#
, where ∑nk=1 knk � m� n
In fact, we will develop a calculation that includes the odd powers
too, E
"∞
∑j=1enj
n∏k=1
�Ehekji�nk#
Some notation: to save ink later let ν(x) stand for xnn∏k=1
E�xk�nk so
our expression abbreviates to E
"∞
∑j=1
ν (ej )
#
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 24
/ 36
Complicated, but each piece is simpler
Now all you need to be able to evaluate are terms like
E
"∞
∑j=1e2nj
n∏k=1
�Ehe2kji�nk#
, where ∑nk=1 knk � m� n
In fact, we will develop a calculation that includes the odd powers
too, E
"∞
∑j=1enj
n∏k=1
�Ehekji�nk#
Some notation: to save ink later let ν(x) stand for xnn∏k=1
E�xk�nk so
our expression abbreviates to E
"∞
∑j=1
ν (ej )
#
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 24
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law d
De�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)
Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of d
The density f�d (x) =P[d�x ]
E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ dj
Let J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")
De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < J
Set�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � t
So t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Set-Up
Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =
P[d�x ]E[d]
De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval
Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 25
/ 36
The Result
E
"∞
∑j=1
ν (ej )
#=
Kn
Ehν�(1� F )�d^t
�i�P [�d �t] ν
�(1� F )t
�o E[ν(1�(1�F )d)]1�E[ν((1�F )d)]
+Ehν�1� (1� F )�d^t
�i�P [�d �t] ν
�1� (1� F )t
�
Where K = 1�E
��Ehν�(1� F )d
�i�J�2j J > 1
�= 1� (1� G )t E[(1�G )�
�d^t ]�P[�d�t ](1�G )�t
E[(1�G )�d^t ]�P[�d�t ](1�G )t
And G is de�ned by (1� G ) = expn�L�1d (E
hν�(1� F )d
�i)o,
Ld being the Laplace transformMeaningEh(1� G )d
i= Ld
nL�1d (E
hν�(1� F )d
�io= E
hν�(1� F )d
�i
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 26
/ 36
The Result
E
"∞
∑j=1
ν (ej )
#=
Kn
Ehν�(1� F )�d^t
�i�P [�d �t] ν
�(1� F )t
�o E[ν(1�(1�F )d)]1�E[ν((1�F )d)]
+Ehν�1� (1� F )�d^t
�i�P [�d �t] ν
�1� (1� F )t
�Where K = 1�E
��Ehν�(1� F )d
�i�J�2j J > 1
�= 1� (1� G )t E[(1�G )�
�d^t ]�P[�d�t ](1�G )�t
E[(1�G )�d^t ]�P[�d�t ](1�G )t
And G is de�ned by (1� G ) = expn�L�1d (E
hν�(1� F )d
�i)o,
Ld being the Laplace transformMeaningEh(1� G )d
i= Ld
nL�1d (E
hν�(1� F )d
�io= E
hν�(1� F )d
�i
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 26
/ 36
The Result
E
"∞
∑j=1
ν (ej )
#=
Kn
Ehν�(1� F )�d^t
�i�P [�d �t] ν
�(1� F )t
�o E[ν(1�(1�F )d)]1�E[ν((1�F )d)]
+Ehν�1� (1� F )�d^t
�i�P [�d �t] ν
�1� (1� F )t
�Where K = 1�E
��Ehν�(1� F )d
�i�J�2j J > 1
�= 1� (1� G )t E[(1�G )�
�d^t ]�P[�d�t ](1�G )�t
E[(1�G )�d^t ]�P[�d�t ](1�G )t
And G is de�ned by (1� G ) = expn�L�1d (E
hν�(1� F )d
�i)o,
Ld being the Laplace transform
MeaningEh(1� G )d
i= Ld
nL�1d (E
hν�(1� F )d
�io= E
hν�(1� F )d
�i
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 26
/ 36
The Result
E
"∞
∑j=1
ν (ej )
#=
Kn
Ehν�(1� F )�d^t
�i�P [�d �t] ν
�(1� F )t
�o E[ν(1�(1�F )d)]1�E[ν((1�F )d)]
+Ehν�1� (1� F )�d^t
�i�P [�d �t] ν
�1� (1� F )t
�Where K = 1�E
��Ehν�(1� F )d
�i�J�2j J > 1
�= 1� (1� G )t E[(1�G )�
�d^t ]�P[�d�t ](1�G )�t
E[(1�G )�d^t ]�P[�d�t ](1�G )t
And G is de�ned by (1� G ) = expn�L�1d (E
hν�(1� F )d
�i)o,
Ld being the Laplace transformMeaningEh(1� G )d
i= Ld
nL�1d (E
hν�(1� F )d
�io= E
hν�(1� F )d
�iBridgeman (University of Connecticut) Random Regimes
Actuarial Science Seminar Jan. 29, 2008 26/ 36
Asymptotically
limt!∞ E
"∞
∑j=1
ν (ej )
#=
E[ν((1�F )�d)]E[ν(1�(1�F )d)]
1�E[ν((1�F )d)]+E
hν�1� (1� F )�d
�i
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 27
/ 36
Everything Is Closed Form
Everything is now of the form P [�d � t] and E [xv] for v one of therandom variables d, �d and�d^ t
E [xv] = Lv [� ln (x)], where Lv is the Laplace transform of vIf, for example, we take the interarrival distribution d forregime-switches to be gamma(α, β), then
Ld(x) = (1+ βx)�α, L�1d (y) = 1β
�y�
1α � 1
�,
L�d (x) = 1αβx
h1� (1+ βx)�α
i, L�d^t (x) =
1αβx
n1� e�xt
h1� Γ
�α; tβ
�i� (1+ βx)�α Γ
�α; (
1+βx )tβ
�o+e�xt
n1� Γ
�α+ 1; tβ
�� t
αβ
h1� Γ
�α; tβ
�io,
P [�d � t] = 1� Γ�
α+ 1; tβ
�� t
αβ
h1� Γ
�α; tβ
�i
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 28
/ 36
Everything Is Closed Form
Everything is now of the form P [�d � t] and E [xv] for v one of therandom variables d, �d and�d^ tE [xv] = Lv [� ln (x)], where Lv is the Laplace transform of v
If, for example, we take the interarrival distribution d forregime-switches to be gamma(α, β), then
Ld(x) = (1+ βx)�α, L�1d (y) = 1β
�y�
1α � 1
�,
L�d (x) = 1αβx
h1� (1+ βx)�α
i, L�d^t (x) =
1αβx
n1� e�xt
h1� Γ
�α; tβ
�i� (1+ βx)�α Γ
�α; (
1+βx )tβ
�o+e�xt
n1� Γ
�α+ 1; tβ
�� t
αβ
h1� Γ
�α; tβ
�io,
P [�d � t] = 1� Γ�
α+ 1; tβ
�� t
αβ
h1� Γ
�α; tβ
�i
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 28
/ 36
Everything Is Closed Form
Everything is now of the form P [�d � t] and E [xv] for v one of therandom variables d, �d and�d^ tE [xv] = Lv [� ln (x)], where Lv is the Laplace transform of vIf, for example, we take the interarrival distribution d forregime-switches to be gamma(α, β), then
Ld(x) = (1+ βx)�α, L�1d (y) = 1β
�y�
1α � 1
�,
L�d (x) = 1αβx
h1� (1+ βx)�α
i, L�d^t (x) =
1αβx
n1� e�xt
h1� Γ
�α; tβ
�i� (1+ βx)�α Γ
�α; (
1+βx )tβ
�o+e�xt
n1� Γ
�α+ 1; tβ
�� t
αβ
h1� Γ
�α; tβ
�io,
P [�d � t] = 1� Γ�
α+ 1; tβ
�� t
αβ
h1� Γ
�α; tβ
�i
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 28
/ 36
Everything Is Closed Form
Everything is now of the form P [�d � t] and E [xv] for v one of therandom variables d, �d and�d^ tE [xv] = Lv [� ln (x)], where Lv is the Laplace transform of vIf, for example, we take the interarrival distribution d forregime-switches to be gamma(α, β), then
Ld(x) = (1+ βx)�α, L�1d (y) = 1β
�y�
1α � 1
�,
L�d (x) = 1αβx
h1� (1+ βx)�α
i, L�d^t (x) =
1αβx
n1� e�xt
h1� Γ
�α; tβ
�i� (1+ βx)�α Γ
�α; (
1+βx )tβ
�o+e�xt
n1� Γ
�α+ 1; tβ
�� t
αβ
h1� Γ
�α; tβ
�io,
P [�d � t] = 1� Γ�
α+ 1; tβ
�� t
αβ
h1� Γ
�α; tβ
�i
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 28
/ 36
Even Those Uniform Correlation Coe¢ cients
ρa,b =1D
nEh(1� F )2(a�b)�d^t
i�P [�d � t] (1� F )2(a�b)t
o
where D =n
Eh(1� F )2a�d^t
i�P [�d � t] (1� F )2at
o�n
Eh(1� F )�2b�d^t
i�P [�d �t] (1� F )�2bt
oand ρa1,...,ak = ρa1,a2+...+ak ρa2,...,ak recursively
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 29
/ 36
Even Those Uniform Correlation Coe¢ cients
ρa,b =1D
nEh(1� F )2(a�b)�d^t
i�P [�d � t] (1� F )2(a�b)t
owhere D =
nEh(1� F )2a�d^t
i�P [�d � t] (1� F )2at
o�n
Eh(1� F )�2b�d^t
i�P [�d �t] (1� F )�2bt
o
and ρa1,...,ak = ρa1,a2+...+ak ρa2,...,ak recursively
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 29
/ 36
Even Those Uniform Correlation Coe¢ cients
ρa,b =1D
nEh(1� F )2(a�b)�d^t
i�P [�d � t] (1� F )2(a�b)t
owhere D =
nEh(1� F )2a�d^t
i�P [�d � t] (1� F )2at
o�n
Eh(1� F )�2b�d^t
i�P [�d �t] (1� F )�2bt
oand ρa1,...,ak = ρa1,a2+...+ak ρa2,...,ak recursively
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 29
/ 36
Where Does the Edgeworth Come From?
De�ne the Fourier Transform f̂ (t) =Z ∞
�∞e�itx f (x) dx
Let W have mean 0 and variance 1 and let φ be std normal density
Write cfW (t) = hcfW (t) � 1bφ(t)�i bφ (t) and Taylor expand the bracket
cfW (t) =(
∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0tn) bφ (t)
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 30
/ 36
Where Does the Edgeworth Come From?
De�ne the Fourier Transform f̂ (t) =Z ∞
�∞e�itx f (x) dx
Let W have mean 0 and variance 1 and let φ be std normal density
Write cfW (t) = hcfW (t) � 1bφ(t)�i bφ (t) and Taylor expand the bracket
cfW (t) =(
∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0tn) bφ (t)
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 30
/ 36
Where Does the Edgeworth Come From?
De�ne the Fourier Transform f̂ (t) =Z ∞
�∞e�itx f (x) dx
Let W have mean 0 and variance 1 and let φ be std normal density
Write cfW (t) = hcfW (t) � 1bφ(t)�i bφ (t) and Taylor expand the bracket
cfW (t) =(
∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0tn) bφ (t)
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 30
/ 36
Where Does the Edgeworth Come From?
De�ne the Fourier Transform f̂ (t) =Z ∞
�∞e�itx f (x) dx
Let W have mean 0 and variance 1 and let φ be std normal density
Write cfW (t) = hcfW (t) � 1bφ(t)�i bφ (t) and Taylor expand the bracket
cfW (t) =(
∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0tn) bφ (t)
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 30
/ 36
Where Does the Edgeworth Come From?
De�ne the Fourier Transform f̂ (t) =Z ∞
�∞e�itx f (x) dx
Let W have mean 0 and variance 1 and let φ be std normal density
Write cfW (t) = hcfW (t) � 1bφ(t)�i bφ (t) and Taylor expand the bracket
cfW (t) =(
∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0tn) bφ (t)
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rule
hcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 30
/ 36
Where Does the Edgeworth Come From?
De�ne the Fourier Transform f̂ (t) =Z ∞
�∞e�itx f (x) dx
Let W have mean 0 and variance 1 and let φ be std normal density
Write cfW (t) = hcfW (t) � 1bφ(t)�i bφ (t) and Taylor expand the bracket
cfW (t) =(
∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0tn) bφ (t)
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 30
/ 36
Where Does the Edgeworth Come From?
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rule
hcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
0 =hbφ (t) � 1bφ(t)
�i(n)t=0
=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
bφ(j) (0) � 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=1
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=3
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0
because W is mean 0 variance 1
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 31
/ 36
Where Does the Edgeworth Come From?
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
0 =hbφ (t) � 1bφ(t)
�i(n)t=0
=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
bφ(j) (0) � 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=1
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=3
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0
because W is mean 0 variance 1
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 31
/ 36
Where Does the Edgeworth Come From?
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
0 =hbφ (t) � 1bφ(t)
�i(n)t=0
=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
bφ(j) (0) � 1bφ(t)�(n�j)t=0
hcfW (t) � 1bφ(t)�i(n)
t=0=
n
∑j=1
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=3
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0
because W is mean 0 variance 1
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 31
/ 36
Where Does the Edgeworth Come From?
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
0 =hbφ (t) � 1bφ(t)
�i(n)t=0
=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
bφ(j) (0) � 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=1
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0
hcfW (t) � 1bφ(t)�i(n)
t=0=
n
∑j=3
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0
because W is mean 0 variance 1
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 31
/ 36
Where Does the Edgeworth Come From?
So fW (w) =∞
∑n=0
1n!
hcfW (t) � 1bφ(t)�i(n)
t=0i�nφ(n) (w) and use Leibniz�s
rulehcfW (t) � 1bφ(t)�i(n)
t=0=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
cfW(j) (0) � 1bφ(t)�(n�j)t=0
0 =hbφ (t) � 1bφ(t)
�i(n)t=0
=�
1bφ(t)�(n)t=0
+n
∑j=1
n!j !(n�j)!
bφ(j) (0) � 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=1
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=3
n!j !(n�j)!
�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0
because W is mean 0 variance 1
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 31
/ 36
But derivatives of Fourier transforms evaluated at 0 are just momentsso
hcfW (t) � 1bφ(t)�i(n)
t=0=
n
∑j=3
n!(n�j)?j !(n�j)! i
�j �E �Wj�� j?
�and
fW (w) = φ (w) +∞
∑n=0
1n!
n
∑j=3
n!(n�j)?j !(n�j)! i
�n�j �E �Wj�� j?
�φ(n) (w)
=
φ (w) + limN!∞
N
∑j=3
1j !
�E�Wj�� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n+j φ(2n+j) (w)
But
φ(2n+j) (w) =
24n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
2n+j�k w2n+j�2k
35 φ (w)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 32
/ 36
But derivatives of Fourier transforms evaluated at 0 are just momentssohcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=3
n!(n�j)?j !(n�j)! i
�j �E �Wj�� j?
�and
fW (w) = φ (w) +∞
∑n=0
1n!
n
∑j=3
n!(n�j)?j !(n�j)! i
�n�j �E �Wj�� j?
�φ(n) (w)
=
φ (w) + limN!∞
N
∑j=3
1j !
�E�Wj�� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n+j φ(2n+j) (w)
But
φ(2n+j) (w) =
24n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
2n+j�k w2n+j�2k
35 φ (w)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 32
/ 36
But derivatives of Fourier transforms evaluated at 0 are just momentssohcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=3
n!(n�j)?j !(n�j)! i
�j �E �Wj�� j?
�and
fW (w) = φ (w) +∞
∑n=0
1n!
n
∑j=3
n!(n�j)?j !(n�j)! i
�n�j �E �Wj�� j?
�φ(n) (w)
=
φ (w) + limN!∞
N
∑j=3
1j !
�E�Wj�� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n+j φ(2n+j) (w)
But
φ(2n+j) (w) =
24n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
2n+j�k w2n+j�2k
35 φ (w)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 32
/ 36
But derivatives of Fourier transforms evaluated at 0 are just momentssohcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=3
n!(n�j)?j !(n�j)! i
�j �E �Wj�� j?
�and
fW (w) = φ (w) +∞
∑n=0
1n!
n
∑j=3
n!(n�j)?j !(n�j)! i
�n�j �E �Wj�� j?
�φ(n) (w)
=
φ (w) + limN!∞
N
∑j=3
1j !
�E�Wj�� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n+j φ(2n+j) (w)
But
φ(2n+j) (w) =
24n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
2n+j�k w2n+j�2k
35 φ (w)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 32
/ 36
But derivatives of Fourier transforms evaluated at 0 are just momentssohcfW (t) � 1bφ(t)
�i(n)t=0
=n
∑j=3
n!(n�j)?j !(n�j)! i
�j �E �Wj�� j?
�and
fW (w) = φ (w) +∞
∑n=0
1n!
n
∑j=3
n!(n�j)?j !(n�j)! i
�n�j �E �Wj�� j?
�φ(n) (w)
=
φ (w) + limN!∞
N
∑j=3
1j !
�E�Wj�� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n+j φ(2n+j) (w)
But
φ(2n+j) (w) =
24n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
2n+j�k w2n+j�2k
35 φ (w)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 32
/ 36
Where Does the Edgeworth Come From?
Finally, if Y = σW+ µ a change of variables gives the Edgeworthexpansion
fY (y) = 1σ φ�y�µ
σ
�+ limN!∞
N
∑j=3
1j !
�µjσj� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n �
n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
k�y�µ
σ
�2n+j�2k1σ φ�y�µ
σ
�where µj is the j-th central moment of YFor Esscher (aka Saddlepoint) Expansion, Taylor expandhcfW (t) � 1bφ(t)
�iaround a di¤erent point than 0
For something even more �exible, use a di¤erent function than φ; trylogistic, gamma, inverse gamma or inverse logistic
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 33
/ 36
Where Does the Edgeworth Come From?
Finally, if Y = σW+ µ a change of variables gives the Edgeworthexpansion
fY (y) = 1σ φ�y�µ
σ
�+ limN!∞
N
∑j=3
1j !
�µjσj� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n �
n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
k�y�µ
σ
�2n+j�2k1σ φ�y�µ
σ
�
where µj is the j-th central moment of YFor Esscher (aka Saddlepoint) Expansion, Taylor expandhcfW (t) � 1bφ(t)
�iaround a di¤erent point than 0
For something even more �exible, use a di¤erent function than φ; trylogistic, gamma, inverse gamma or inverse logistic
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 33
/ 36
Where Does the Edgeworth Come From?
Finally, if Y = σW+ µ a change of variables gives the Edgeworthexpansion
fY (y) = 1σ φ�y�µ
σ
�+ limN!∞
N
∑j=3
1j !
�µjσj� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n �
n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
k�y�µ
σ
�2n+j�2k1σ φ�y�µ
σ
�where µj is the j-th central moment of Y
For Esscher (aka Saddlepoint) Expansion, Taylor expandhcfW (t) � 1bφ(t)�iaround a di¤erent point than 0
For something even more �exible, use a di¤erent function than φ; trylogistic, gamma, inverse gamma or inverse logistic
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 33
/ 36
Where Does the Edgeworth Come From?
Finally, if Y = σW+ µ a change of variables gives the Edgeworthexpansion
fY (y) = 1σ φ�y�µ
σ
�+ limN!∞
N
∑j=3
1j !
�µjσj� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n �
n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
k�y�µ
σ
�2n+j�2k1σ φ�y�µ
σ
�where µj is the j-th central moment of YFor Esscher (aka Saddlepoint) Expansion, Taylor expandhcfW (t) � 1bφ(t)
�iaround a di¤erent point than 0
For something even more �exible, use a di¤erent function than φ; trylogistic, gamma, inverse gamma or inverse logistic
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 33
/ 36
Where Does the Edgeworth Come From?
Finally, if Y = σW+ µ a change of variables gives the Edgeworthexpansion
fY (y) = 1σ φ�y�µ
σ
�+ limN!∞
N
∑j=3
1j !
�µjσj� j?
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n �
n+b j2 c∑k=0
(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)
k�y�µ
σ
�2n+j�2k1σ φ�y�µ
σ
�where µj is the j-th central moment of YFor Esscher (aka Saddlepoint) Expansion, Taylor expandhcfW (t) � 1bφ(t)
�iaround a di¤erent point than 0
For something even more �exible, use a di¤erent function than φ; trylogistic, gamma, inverse gamma or inverse logistic
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 33
/ 36
How Do You Get The Moments?
E�rlt�= E
hel ln(rt )
i= E
�elY�=
∞Z�∞
e ly fY (y) dy
Substitute the Edgeworth expression, complete the square to integratejust as if you were integrating for the lognormal, and expand thebinomials that occur when you change variables and you wind up with
E�rlt�=
e lµ+12 (lσ)
2
8<:1+ limN!∞
N
∑j=3
l jj !
�µj � j?σj
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n (lσ)2n �
�n+b j2 c
∑m=0
(2n+j)!(2n+j�2m)! (lσ)
�2mm
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k
9=;Remarkably,
m
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k = 0 when m > 0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 34
/ 36
How Do You Get The Moments?
E�rlt�= E
hel ln(rt )
i= E
�elY�=
∞Z�∞
e ly fY (y) dy
Substitute the Edgeworth expression, complete the square to integratejust as if you were integrating for the lognormal, and expand thebinomials that occur when you change variables and you wind up with
E�rlt�=
e lµ+12 (lσ)
2
8<:1+ limN!∞
N
∑j=3
l jj !
�µj � j?σj
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n (lσ)2n �
�n+b j2 c
∑m=0
(2n+j)!(2n+j�2m)! (lσ)
�2mm
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k
9=;Remarkably,
m
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k = 0 when m > 0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 34
/ 36
How Do You Get The Moments?
E�rlt�= E
hel ln(rt )
i= E
�elY�=
∞Z�∞
e ly fY (y) dy
Substitute the Edgeworth expression, complete the square to integratejust as if you were integrating for the lognormal, and expand thebinomials that occur when you change variables and you wind up with
E�rlt�=
e lµ+12 (lσ)
2
8<:1+ limN!∞
N
∑j=3
l jj !
�µj � j?σj
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n (lσ)2n �
�n+b j2 c
∑m=0
(2n+j)!(2n+j�2m)! (lσ)
�2mm
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k
9=;
Remarkably,m
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k = 0 when m > 0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 34
/ 36
How Do You Get The Moments?
E�rlt�= E
hel ln(rt )
i= E
�elY�=
∞Z�∞
e ly fY (y) dy
Substitute the Edgeworth expression, complete the square to integratejust as if you were integrating for the lognormal, and expand thebinomials that occur when you change variables and you wind up with
E�rlt�=
e lµ+12 (lσ)
2
8<:1+ limN!∞
N
∑j=3
l jj !
�µj � j?σj
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n (lσ)2n �
�n+b j2 c
∑m=0
(2n+j)!(2n+j�2m)! (lσ)
�2mm
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k
9=;Remarkably,
m
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k = 0 when m > 0
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 34
/ 36
How Do You Get The Moments?
Why? 0 =hbφ (t) � 1bφ(t)
�i(2m)t=0
=m
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k
Finally, E�rlt�=
e lµ+12 (lσ)
2
8<:1+ limN!∞
N
∑j=3
l jj !
�µj � j?σj
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n (lσ)2n
9=;
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 35
/ 36
How Do You Get The Moments?
Why? 0 =hbφ (t) � 1bφ(t)
�i(2m)t=0
=m
∑k=0
(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)
k
Finally, E�rlt�=
e lµ+12 (lσ)
2
8<:1+ limN!∞
N
∑j=3
l jj !
�µj � j?σj
� b N�j2 c∑n=0
(2n)?(2n)! (�1)
n (lσ)2n
9=;
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 35
/ 36
What About the Main Results?
Proofs were inspired by techniques in Decoupling: From Dependenceto Independence, by de la Peña and Giné
Essential lemmata are:
As joint distributions fJ,�dJg ' fJ,�d1g and�dJ '�d1 '�d^ tConditional on J = j 0 > 1 the following are each independentsets:fJ,�d1g,
��d1,d2, ...,dj 0�1, �d2, ...,dj 0�1,�dj 0 and fJ,�dJgThis is enough independence to get a geometric series inside the main
expectation E
"∞
∑j=1
ν (ej )
#and to pull apart the two sides of the
correlation expectation for ρa,b , leaving a common term involving�d1and�dJ which can be evaluated by writing�d1 = t� (�dJ + dJ�1 + ...+ d2)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 36
/ 36
What About the Main Results?
Proofs were inspired by techniques in Decoupling: From Dependenceto Independence, by de la Peña and Giné
Essential lemmata are:
As joint distributions fJ,�dJg ' fJ,�d1g and�dJ '�d1 '�d^ tConditional on J = j 0 > 1 the following are each independentsets:fJ,�d1g,
��d1,d2, ...,dj 0�1, �d2, ...,dj 0�1,�dj 0 and fJ,�dJgThis is enough independence to get a geometric series inside the main
expectation E
"∞
∑j=1
ν (ej )
#and to pull apart the two sides of the
correlation expectation for ρa,b , leaving a common term involving�d1and�dJ which can be evaluated by writing�d1 = t� (�dJ + dJ�1 + ...+ d2)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 36
/ 36
What About the Main Results?
Proofs were inspired by techniques in Decoupling: From Dependenceto Independence, by de la Peña and Giné
Essential lemmata are:
As joint distributions fJ,�dJg ' fJ,�d1g and�dJ '�d1 '�d^ t
Conditional on J = j 0 > 1 the following are each independentsets:fJ,�d1g,
��d1,d2, ...,dj 0�1, �d2, ...,dj 0�1,�dj 0 and fJ,�dJgThis is enough independence to get a geometric series inside the main
expectation E
"∞
∑j=1
ν (ej )
#and to pull apart the two sides of the
correlation expectation for ρa,b , leaving a common term involving�d1and�dJ which can be evaluated by writing�d1 = t� (�dJ + dJ�1 + ...+ d2)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 36
/ 36
What About the Main Results?
Proofs were inspired by techniques in Decoupling: From Dependenceto Independence, by de la Peña and Giné
Essential lemmata are:
As joint distributions fJ,�dJg ' fJ,�d1g and�dJ '�d1 '�d^ tConditional on J = j 0 > 1 the following are each independentsets:fJ,�d1g,
��d1,d2, ...,dj 0�1, �d2, ...,dj 0�1,�dj 0 and fJ,�dJg
This is enough independence to get a geometric series inside the main
expectation E
"∞
∑j=1
ν (ej )
#and to pull apart the two sides of the
correlation expectation for ρa,b , leaving a common term involving�d1and�dJ which can be evaluated by writing�d1 = t� (�dJ + dJ�1 + ...+ d2)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 36
/ 36
What About the Main Results?
Proofs were inspired by techniques in Decoupling: From Dependenceto Independence, by de la Peña and Giné
Essential lemmata are:
As joint distributions fJ,�dJg ' fJ,�d1g and�dJ '�d1 '�d^ tConditional on J = j 0 > 1 the following are each independentsets:fJ,�d1g,
��d1,d2, ...,dj 0�1, �d2, ...,dj 0�1,�dj 0 and fJ,�dJgThis is enough independence to get a geometric series inside the main
expectation E
"∞
∑j=1
ν (ej )
#and to pull apart the two sides of the
correlation expectation for ρa,b , leaving a common term involving�d1and�dJ which can be evaluated by writing�d1 = t� (�dJ + dJ�1 + ...+ d2)
Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 36
/ 36