Stock Loan Valuation Under Brownian-Motion Based and ...Norberg R, The Markov chain market, ASTIN...
Transcript of Stock Loan Valuation Under Brownian-Motion Based and ...Norberg R, The Markov chain market, ASTIN...
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Stock Loan Valuation Under Brownian-MotionBased and Markov Chain Stock Models
David Prager1
1Associate Professor of MathematicsAnderson University (SC)
Based on joint work with Professor Qing Zhang, University of Georgia
IMA Workshop: Financial and Economic ApplicationsJune 11, 2018
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1 What Is a Stock Loan?
2 History and Background
3 A Markov Chain Model
4 Specific Examples
5 Conclusion and Directions for Further Study
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What is a Stock Loan?
• Client (borrower) owns share of stock.• Use as collateral to obtain, for a fee, loan from bank
(lender).• Upon loan maturity (or before, for American maturity),
client may either:• Repay the loan (principal and interest).• Default (surrender the stock).
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Stock Loan Problem
• For a given stock, maturity, principal, and loan interest rate,what is the fair value of the fee charged by bank?
• Notations:• q = Loan Principal• γ = Loan Interest Rate• r = Risk-Free Rate• c = Bank Service Fee• Amount borrower gets = q − c
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Stock Loan Problem
• For a given stock, maturity, principal, and loan interest rate,what is the fair value of the fee charged by bank?
• Notations:• q = Loan Principal• γ = Loan Interest Rate• r = Risk-Free Rate• c = Bank Service Fee• Amount borrower gets = q − c
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Two Examples: Borrower’s Perspective
Stock ClosingPrice6/5/17
RepaymentAmount6/5/18*
Apple(AAPL)
$ 155.99 $172.40
SouthernCo.(SO)
$ 50.81 $ 56.15
*Assumes γ = 0.1 and q = share price.
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Two Examples: Borrower’s Perspective
Stock ClosingPrice6/5/17
RepaymentAmount6/5/18*
ClosingPrice6/5/18
Borrower’sDecision
Apple(AAPL)
$ 155.99 $ 172.40 $ 191.83 Repay
SouthernCo.(SO)
$ 50.81 $ 56.15 $ 43.93 Default
*Assumes γ = 0.1 and q = share price.
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Two Examples: Lender’s Perspective
Stock ClosingPrice6/5/17
CashPaidOut*
Borrower’sDecision
Lender’sNominalProfit
Apple(AAPL)
$ 155.99 $ 153.99 Repay 172.40-153.99=18.41
SouthernCo.(SO)
$ 50.81 $ 48.81 Default 43.93-48.81=(4.88)
*Assumes c = $2 and q = share price.
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Perpetual Stock Loans (Xia and Zhou, 2007)
• Stock obeys Geometric Brownian Motion:
St = x exp((
r − δ − (σ2/2))
t + σWt
)where δ is the dividend yield and x = S0.
• Bank collects dividends during loan period.• The loan is perpetual.
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Evaluating the Stock Loan: Preliminaries
• Let V (x) ≡
supτ∈T0
E[e−rτ
(x exp
((r − δ − σ2
2
)τ + σWτ
)− qeγτ
)+
].
• Assumption: V (x) = x − q + c > 0
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Evaluating the Stock Loan: Preliminaries
For a ∈ R+, let:• τa ≡ inf
[t ≥ 0 : e−γtSt = a
]• g(a) ≡ E
[e−rτa (Sτa − qeγτa)+
]= (a− q)E
[e(γ−r)τa Iτa<∞
].
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Solving for the Value Function: Case 1
Case 1: If δ = 0 and γ − r ≤ σ2
2, then
1 g(a) =(a− q)x
aand
2 V (x) = x .
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Solving for the Value Function: Case 2
Let a0 ≡
(q
[√(σ2 −
γ−r+δσ
)2+ 2δ + σ
2 + γ−r+δσ
])(√(
σ2 −
γ−r+δσ
)2+ 2δ − σ
2 + γ−r+δσ
) .
Case 2: If δ > 0, or δ = 0 and γ− r >σ2
2, and q < a0 ≤ x , then
1 g(a) attains its maximum at a = x and2 V (x) = x − q.
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Solving for the Value Function: Case 3
Let a0 ≡
(q
[√(σ2 −
γ−r+δσ
)2+ 2δ + σ
2 + γ−r+δσ
])(√(
σ2 −
γ−r+δσ
)2+ 2δ − σ
2 + γ−r+δσ
) .
Case 3: If δ > 0, or δ = 0 and γ − r >σ2
2, and a0 > x , then
1 g(a) attains its maximum on [q ∨ x ,∞) at a = a0 and2 V (x) = g(a0).
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Finite Maturity Stock Loans, Mean-Reverting Model
• Assume the stock loan matures at time T <∞ andmaturity is European.
• Assume the stock price obeys the mean-reverting model.
St = eXt
dXt = a(L− Xt )dt + σdWt
where a > 0 is the rate of reversion and L is the equilibriumlevel.
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Finite Maturity Stock Loans, Mean-Reverting Model
• Assume the stock loan matures at time T <∞ andmaturity is European.
• Assume the stock price obeys the mean-reverting model.
St = eXt
dXt = a(L− Xt )dt + σdWt
where a > 0 is the rate of reversion and L is the equilibriumlevel.
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Key Idea: Change of time
• Let φt ≡∫ t
0
(1/α
(φs, Γ
φs))
ds and α(t , ω) = α(t) ≡ σeat .
• Then the mean-reverting model can be written explicitly:
Xt = e−at (log x − L) + L + e−atW(
1φt
).
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Solving for the Value Function (P. and Zhang, 2010)
Let u ≡ T − s. Under the mean-reverting model with Europeanmaturity,
V (u, x) =e(γ−r)u+B2
4A +C√(φu+s)−1
1√A
[1− Φ
(√A(
P − B2A
))]
− qe(γ−r)u√(φu+s)−1
1√2A
[1− Φ
(P√
2A)]
whereC ≡ e−a(u+s)(log x − L) + LB ≡ e−a(u+s)
A ≡ 12(φu+s)−1
P ≡ ea(u+s)(log q − L) + L− log x ,
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Markov Chain Model for Perpetual Case
• αt denote a Markov chain with state space {1,2} and
generator Q =
(−λ1 λ1λ2 −λ2
).
• Stock obeysdSt
St= µ(αt )dt , S0 = x ≥ 0, t ≥ 0
• µ1 = µ(1) > 0 and µ2 = µ(2) < 0 are given return rates.
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Markov Chain Model for Perpetual Case
• αt denote a Markov chain with state space {1,2} and
generator Q =
(−λ1 λ1λ2 −λ2
).
• Stock obeysdSt
St= µ(αt )dt , S0 = x ≥ 0, t ≥ 0
• µ1 = µ(1) > 0 and µ2 = µ(2) < 0 are given return rates.
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Value Function
• Stopping Time: τ (perpetual case)• Payoff Function:
J(x , i , τ) ≡ E[e−rτ (Sτ − qeγτ )+ Iτ<∞|S0 = x , α0 = i
],
where x+ = max{0, x}.• Value Function: V (x , i) = supτ J(x , i , τ), where the sup is
taken over all stopping times τ .
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Sufficient Conditions for a Closed-Form Solution
• µ2 < r < γ < µ1.• r > ρ0 where ρ0 ≡
12
(µ1 − λ1 + µ2 − λ2 +
√((µ1 − λ1)− (µ2 − λ2))2 + 4λ1λ2
)is the larger root of the equationΦ(x) = (x + λ1 − µ1)(x + λ2 − µ2)− λ1λ2.
• λi > γ − r , for i = 1,2.
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Key Change of Variables
• Xt ≡ e−γtSt , so that
dXt = Xt [−γ + µ(αt )] dt .
• Letting ξ ≡ γ − r > 0, the value function becomes
V (x , i) = supτ
E[eξτ (Xτ − q)+ Iτ<∞|X0 = x , α0 = i
]
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HJB Equation and Variational Inequalities
With fi ≡ µi − γ, the generator for this value function is
Ah(x , i) = xfih′(x , i) + Qh(x , ·)(i),
where Qh(x , ·)(1) = λ1(h(x ,2)− h(x ,1)), andQh(x , ·)(2) = λ2(h(x ,1)− h(x ,2)).
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HJB Equation and Variational Inequalities
The associated variational inequalities are
max{ξh(x ,1) +Ah(x ,1), (x − q)+ − h(x ,1)} = 0,
max{ξh(x ,2) +Ah(x ,2), (x − q)+ − h(x ,2)} = 0.
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Solution via Smooth-Fit Substitution
• Start on the region (0, x∗) with free boundary x∗, i.e. thecase in which (ξ +A)h(x , i) = 0, i = 1,2.
• Solve the case in which (ξ +A)h(x ,1) = 0 for h(x ,2) andsubstitute into (ξ +A)h(x ,2) = 0.
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Solution via Smooth-Fit Substitution
• Start on the region (0, x∗) with free boundary x∗, i.e. thecase in which (ξ +A)h(x , i) = 0, i = 1,2.
• Solve the case in which (ξ +A)h(x ,1) = 0 for h(x ,2) andsubstitute into (ξ +A)h(x ,2) = 0.
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Characteristic Equation
Substitution gives a 2nd order ODE with characteristic equation
φ(β) = f1f2β2+[f1(ξ − λ2) + f2(ξ − λ1)]β+[(ξ − λ1)(ξ − λ2)− λ1λ2] .
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Characteristic Equation: Solutions (P. and Zhang,2014)
β1 =−D1 +
√D2
1 − 4f1f2D2
2f1f2,
β2 =−D1 −
√D2
1 − 4f1f2D2
2f1f2,
where
D1 = f1(ξ − λ2) + f2(ξ − λ1),
D2 = (ξ − λ1)(ξ − λ2)− λ1λ2.
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Free Boundary Solution
x∗ =
(ξ − λ1 + f1ξ − λ1
)(qβ2
β2 − 1
).
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Stopping Time Solution
h(x ,1) =
{A2xβ2 if 0 ≤ x ≤ x∗,A0x + B0 if x > x∗,
h(x ,2) =
{κ2A2xβ2 if 0 ≤ x ≤ x∗,x − q if x > x∗.
A0 = − λ1
ξ − λ1 + f1
B0 =λ1qξ − λ1
κ2 =1λ1
[−(ξ − λ1)− f1β2]
A2 =A0x∗ + B0
(x∗)β2
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Verification Theorem
h(x , i) = V (x , i), i = 1,2. Moreover, let
D = (0,∞)× {1} ∪ (0, x∗)× {2}
denote the continuation region. Then
τ∗ = inf{t ≥ 0; (Xt , αt ) 6∈ D}
is an optimal exercising time.
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Brownian Motion
Given ε > 0, take
µ1 = r − σ2
2+
σ√ε,
µ2 = r − σ2
2− σ√
ε,
λ1 = λ2 =1ε.
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Brownian Motion
As ε→ 0,• St = Sε
t converges weakly to
S0t = S0 exp
((r − σ2
2
)t + σWt
).
• x∗ = x ε,∗ → x0 ≡ β0q/(β0 − 1)
• V (x ,1) and V (x ,2) both converge to
V 0(x) =
{A0xβ0 if x < x0,
x − q if x ≥ x0,
where A0 ≡ (β0 − 1)β0−1q1−β0
(β0)β0.
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Numerical Examples: Default Parameters and InitialConditions
Parameter Valuer 0.05q 30S0 33γ 0.1λ1 135.25λ2 130.95µ1 4.89µ2 −5.13
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Numerical Examples: γ versus S0
0.05
0.1
0.15
0.2
0.25 2040
6080
100
0
20
40
60
80
Initial Stock Price
Loan Interest Rate
Sta
te 1
Loa
n V
alue
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Numerical Examples: γ versus S0
0.05
0.1
0.15
0.2
0.25 2040
6080
100
0
20
40
60
80
Initial Stock Price
Loan Interest Rate
Sta
te 2
Loa
n V
alue
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Numerical Examples: q versus λ2
110120
130140
1020
3040
50
0
5
10
15
20
25
Loan Principal
State 2 Switching Rate
Sta
te 1
Loa
n V
alue
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Numerical Examples: q versus λ2
110115
120125
130135
1020
3040
50
0
5
10
15
20
25
Loan Principal
State 2 Switching Rate
Sta
te 2
Loa
n V
alue
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Numerical Examples: γ versus µ2
0.050.1
0.150.2
0.25
−7
−6.5
−6
−5.5
−54
4.5
5
5.5
6
6.5
Loan Interest Rate
State 2 Return Rate
Sta
te 1
Loa
n V
alue
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Numerical Examples: γ versus µ2
0.050.1
0.150.2
0.25
−7−6.5
−6−5.5
−53
3.5
4
4.5
5
5.5
6
Loan Interest Rate
State 2 Return Rate
Sta
te 2
Loa
n V
alue
![Page 42: Stock Loan Valuation Under Brownian-Motion Based and ...Norberg R, The Markov chain market, ASTIN Bulletin, 2003, 33: 265–287. Prager D and Zhang Q, Stock loan valuation under a](https://reader033.fdocuments.in/reader033/viewer/2022041818/5e5c5eb6dc9b9d052e5658a0/html5/thumbnails/42.jpg)
Numerical Examples: λ1 versus µ2
130
140
150
160
−7−6.5
−6−5.5
−54
4.5
5
5.5
6
6.5
7
7.5
State 1 Switching Rate
State 2 Return Rate
Sta
te 1
Loa
n V
alue
![Page 43: Stock Loan Valuation Under Brownian-Motion Based and ...Norberg R, The Markov chain market, ASTIN Bulletin, 2003, 33: 265–287. Prager D and Zhang Q, Stock loan valuation under a](https://reader033.fdocuments.in/reader033/viewer/2022041818/5e5c5eb6dc9b9d052e5658a0/html5/thumbnails/43.jpg)
Numerical Examples: λ1 versus µ2
130140
150160
−7−6.5
−6−5.5
−53
3.5
4
4.5
5
5.5
6
6.5
7
State 1 Switching Rate
State 2 Return Rate
Sta
te 2
Loa
n V
alue
![Page 44: Stock Loan Valuation Under Brownian-Motion Based and ...Norberg R, The Markov chain market, ASTIN Bulletin, 2003, 33: 265–287. Prager D and Zhang Q, Stock loan valuation under a](https://reader033.fdocuments.in/reader033/viewer/2022041818/5e5c5eb6dc9b9d052e5658a0/html5/thumbnails/44.jpg)
Conclusions
• Combined continuous and discrete properties• Closed-form formulas for optimal stopping time and value
function• The stock loan valuation can be determined by the
corresponding exercise time, which is given in terms of asingle threshold level.
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Directions for Further Study
• Model calibration• Time variables
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References
• Xia, Jianming and Xun Yu Zhou, Stock Loans,Mathematical Finance, April 2007, 307-317.
• Norberg R, The Markov chain market, ASTIN Bulletin,2003, 33: 265–287.
• Prager D and Zhang Q, Stock loan valuation under aregime-switching model with mean-reverting and finitematurity, Journal of Systems Science and Complexity,2010: 572–583.
• Prager D and Zhang Q, Valuation of Stock Loans under aMarkov Chain Model, Journal of Systems Science andComplexity, 2014: 1–17.
• All stock market data is from Yahoo! Finance.• All figures were produced using MATLAB.