Math 341: Probability Eighteenth Lecture (11/12/09) · Summary for the Day Review Accumulation and...
Transcript of Math 341: Probability Eighteenth Lecture (11/12/09) · Summary for the Day Review Accumulation and...
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Math 341: ProbabilityEighteenth Lecture (11/12/09)
Steven J MillerWilliams College
[email protected]://www.williams.edu/go/math/sjmiller/
public html/341/
Bronfman Science CenterWilliams College, November 12, 2009
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Summary for the Day
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Summary for the day
Complex Analysis:⋄ Review definitions / statements.⋄ Accumulation point theorems.
Clicker question:⋄ Statement of the CLT.⋄ Clicker question on rate of convergence.
Central Limit Theorem:⋄ Poisson example.⋄ Proof with MGFs.⋄ Proof with Fourier analysis.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Review
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Accumulation points and functions
TheoremLet f be an analytic function on an open set U, withinfinitely many zeros z1, z2, z3, . . . . If limn→∞ zn ∈ U, then fis identically zero on U. In other words, if a function iszero along a sequence in U whose accumulation point isalso in U, then that function is identically zero in U.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Schwartz Space
Schwartz space S(ℝ): all infinitely differentiable functionsf such that, for any non-negative integers m and n,
supx∈ℝ
∣∣∣∣(1 + x2)m dnfdxn
∣∣∣∣ < ∞.
Inversion Theorem for Fourier Transform: Let f ∈ S(ℝ).Then
f (x) =
∫ ∞
−∞f (y)e2�ixydy .
f , g ∈ S(ℝ) with f = g then f (x) = g(x).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Accumulationand Moments
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Accumulation Points and Moments
TheoremAssume the MGFs MX (t) and MY (t) exist in aneighborhood of zero (i.e., there is some � such that bothfunctions exist for ∣t ∣ < �). If MX (t) = MY (t) in thisneighborhood, then FX (u) = FY (u) for all u. As thedensities are the derivatives of the cumulative distributionfunctions, we have f = g.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Accumulation Points and Moments
TheoremLet {Xi}i∈I be a sequence of random variables with MGFsMXi (t). Assume there is a � > 0 such that when ∣t ∣ < � wehave limi→∞ MXi (t) = MX (t) for some MGF MX (t), and allMGFs converge for ∣t ∣ < �. Then there exists a uniquecumulative distribution function F whose moments aredetermined from MX (t) and for all x where FX (x) iscontinuous, limn→∞ FXi (x) = FX (x).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Accumulation Points and Moments
Theorem: X and Y continuous random variables on[0,∞) with continuous densities f and g, all of whosemoments are finite and agree, and
1 ∃C > 0 st ∀c ≤ C, e(c+1)t f (et) and e(c+1)t g(et ) areSchwartz functions.
2 The (not necessarily integral) moments
�′rn(f ) =
∫ ∞
0x rn f (x)dx and �′
rn(g) =
∫ ∞
0x rng(x)dx
agree for some sequence of non-negative realnumbers {rn}∞n=0 which has a finite accumulation point(i.e., limn→∞ rn = r < ∞).
Then f = g (in other words, knowing all these momentsuniquely determines the probability density).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Application to equal integral moments
Return to the two densities causing trouble:
f1(x) =1√
2�x2e−(log2 x)/2
f2(x) = f1(x) [1 + sin(2� log x)] .
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Application to equal integral moments
Return to the two densities causing trouble:
f1(x) =1√
2�x2e−(log2 x)/2
f2(x) = f1(x) [1 + sin(2� log x)] .
Same integral moments: ek2/2.Have the correct decay.Using complex analysis (specifically, contourintegration), we can calculate the (a + ib)thmoments:
For f1 : e(a+ib)2/2
For f2 : e(a+ib)2/2 +i2
(e(a+i(b−2�))2/2 − e(a+i(b+2�))2/2
).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Application to equal integral moments
Return to the two densities causing trouble:
f1(x) =1√
2�x2e−(log2 x)/2
f2(x) = f1(x) [1 + sin(2� log x)] .
No sequence of real moments having anaccumulation point where they agree.athmoment of f2 is
ea2/2 + e(a−2i�)2/2(1 − e4ia�
),
and this is never zero unless a is a half-integer.Only way this can vanish is if 1 = e4ia�.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
ClickerQuestions
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Normalization of a random variable
Normalization (standardization) of a random variableLet X be a random variable with mean � and standarddeviation �, both of which are finite. The normalization, Y ,is defined by
Y :=X − E[X ]
StDev(X )=
X − �
�.
Note that
E[Y ] = 0 and StDev(Y ) = 1.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Statement of the Central Limit Theorem
Normal distributionA random variable X is normally distributed (or has thenormal distribution, or is a Gaussian random variable)with mean � and variance �2 if the density of X is
f (x) =1√
2��2exp
(−(x − �)2
2�2
).
We often write X ∼ N(�, �2) to denote this. If � = 0 and�2 = 1, we say X has the standard normal distribution.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Statement of the Central Limit Theorem
Central Limit TheoremLet X1, . . . ,XN be independent, identically distributedrandom variables whose moment generating functionsconverge for ∣t ∣ < � for some � > 0 (this implies all themoments exist and are finite). Denote the mean by � andthe variance by �2, let
X N =X1 + ⋅ ⋅ ⋅+ XN
N
and set
ZN =X N − �
�/√
N.
Then as N → ∞, the distribution of ZN converges to thestandard normal.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Statement of the Central Limit Theorem
Why are there only tables of values of standard normal?
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Statement of the Central Limit Theorem
Why are there only tables of values of standard normal?
Answer: normalization. Similar to log tables (only needone from change of base formula).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Alternative Statement of the Central Limit Theorem
Central Limit TheoremLet X1, . . . ,XN be independent, identically distributedrandom variables whose moment generating functionsconverge for ∣t ∣ < � for some � > 0 (this implies all themoments exist and are finite). Denote the mean by � andthe variance by �2, let
SN = X1 + ⋅ ⋅ ⋅+ XN
and set
ZN =SN − N�√
N�2.
Then as N → ∞, the distribution of ZN converges to thestandard normal.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Key Probabilities:
Key probabilities for Z ∼ N(0, 1) (i.e., Z has the standardnormal distribution).
Prob(∣Z ∣ ≤ 1) ≈ 68.2%.
Prob(∣Z ∣ ≤ 1.96) ≈ 95%.
Prob(∣Z ∣ ≤ 2.575) ≈ 99%.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
Question:Let X1,X2, . . . be iidrv with mean 0 and variance 1, and letZN = X N/(1/
√N). By the CLT ZN → N(0, 1); which
choice converges fastest? Slowest?
1 Uniform: X ∼ Unif(−√
3,√
3).2 Laplace: fX (x) = e−
√2∣x∣/
√2.
3 Normal: X ∼ N(0, 1).4 Millered Cauchy: fX (x) =
4a sin(�/8)�
11+(ax)8 ,
a =√√
2 − 1.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
Question:Let X1,X2, . . . be iidrv with mean 0 and variance 1, and letZN = X N/(1/
√N). By the CLT ZN → N(0, 1); which
choice converges fastest? Slowest?
1 Uniform: X ∼ Unif(−√
3,√
3). Kurtosis: 1.8.2 Laplace: fX (x) = e−
√2∣x∣/
√2. Kurtosis: 6.
3 Normal: X ∼ N(0, 1). Kurtosis: 3.4 Millered Cauchy: fX (x) =
4a sin(�/8)�
11+(ax)8 ,
a =√√
2 − 1. Kurtosis: 1 +√
2 ≈ 2.414.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
Question:Let X1,X2, . . . be iidrv with mean 0 and variance 1, and letZN = X N/(1/
√N). By the CLT ZN → N(0, 1); which
choice converges fastest? Slowest?
1 Uniform: X ∼ Unif(−√
3,√
3). Kurtosis: 1.8.2 Laplace: fX (x) = e−
√2∣x∣/
√2. Kurtosis: 6.
3 Normal: X ∼ N(0, 1). Kurtosis: 3.4 Millered Cauchy: fX (x) =
4a sin(�/8)�
11+(ax)8 ,
a =√√
2 − 1. Kurtosis: 1 +√
2 ≈ 2.414.
log MX (t) = t2
2 + (�4−3)t4
4! + O(t6).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
Question:Let X1,X2, . . . be iidrv with mean 0 and variance 1, and letZN = X N/(1/
√N). By the CLT ZN → N(0, 1); which
choice converges fastest? Slowest?
1 Uniform: X ∼ Unif(−√
3,√
3). Excess Kurtosis: -1.2.2 Laplace: fX (x) = e−
√2∣x∣/
√2. Excess Kurtosis: 3.
3 Normal: X ∼ N(0, 1). Excess Kurtosis: 0.4 Millered Cauchy: fX (x) =
4a sin(�/8)�
11+(ax)8 ,
a =√√
2 − 1. Excess Kurtosis: 1 +√
2 − 3 ≈ −.586.
log MX (t) = t2
2 + (�4−3)t4
4! + O(t6).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
-4 -2 2 4
0.05
0.10
0.15
Figure: Convolutions of 5 Uniforms.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
-4 -2 2 4
0.05
0.10
0.15
Figure: Convolutions of 5 Laplaces.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
-4 -2 2 4
0.1
0.2
0.3
0.4
Figure: Convolutions of 5 Normals.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
-4 -2 2 4
0.05
0.10
0.15
0.20
0.25
0.30
Figure: Convolutions of 1 Millered Cauchy.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convergence to the standard normal
-4 -2 2 4
0.05
0.10
0.15
0.20
0.25
Figure: Convolutions of 2 Millered Cauchy.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Central Limit Theorem
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
MGF and the CLT
Moment generating function of normal distributionsLet X be a normal random variable with mean � andvariance �2. Its moment generating function satisfies
MX (t) = e�t+�2 t2
2 .
In particular, if Z has the standard normal distribution, itsmoment generating function is
MZ (t) = et2/2.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
MGF and the CLT
Moment generating function of normal distributionsLet X be a normal random variable with mean � andvariance �2. Its moment generating function satisfies
MX (t) = e�t+�2 t2
2 .
In particular, if Z has the standard normal distribution, itsmoment generating function is
MZ (t) = et2/2.
Proof: Complete the square.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Poisson Example of the CLT
ExampleLet X ,X1, . . . ,XN be Poisson random variables withparameter �. Let
X N =X1 + ⋅ ⋅ ⋅+ XN
N, Y =
X − E[X ]
StDev(X ).
Then as N → ∞, Y converges to having the standardnormal distribution.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Poisson Example of the CLT
ExampleLet X ,X1, . . . ,XN be Poisson random variables withparameter �. Let
X N =X1 + ⋅ ⋅ ⋅+ XN
N, Y =
X − E[X ]
StDev(X ).
Then as N → ∞, Y converges to having the standardnormal distribution.
Moment generating function: MX (t) = exp(�(et − 1)).Independent formula: MX1+X2(t) = MX1(t)MX2(t).Shift formula: MaX+b(t) = ebtMX (at).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
General proof via Moment Generating Functions
Xi ’s iidrv,
ZN =X − �
�/√
N=
N∑
n=1
Xi − �
�√
N.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
General proof via Moment Generating Functions
Xi ’s iidrv,
ZN =X − �
�/√
N=
N∑
n=1
Xi − �
�√
N.
Moment Generating Function is:
MZN (t) =N∏
n=1
e−�t�
√N MX
(t
�√
N
)= e
−�t√
N� MX
(t
�√
N
)N
37
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
General proof via Moment Generating Functions
Xi ’s iidrv,
ZN =X − �
�/√
N=
N∑
n=1
Xi − �
�√
N.
Moment Generating Function is:
MZN (t) =N∏
n=1
e−�t�
√N MX
(t
�√
N
)= e
−�t√
N� MX
(t
�√
N
)N
Taking logarithms:
log MZN (t) = −�t√
N�
+ N log MX
(t
�√
N
).
38
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
General proof via Moment Generating Functions (cont)
Expansion of MGF:
MX (t) = 1 + �t +�′
2t2
2!+ ⋅ ⋅ ⋅ = 1 + t
(�+
�′2t2
+ ⋅ ⋅ ⋅).
39
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
General proof via Moment Generating Functions (cont)
Expansion of MGF:
MX (t) = 1 + �t +�′
2t2
2!+ ⋅ ⋅ ⋅ = 1 + t
(�+
�′2t2
+ ⋅ ⋅ ⋅).
Expansion for log(1 + u) is
log(1 + u) = u − u2
2+
u3
3!− ⋅ ⋅ ⋅ .
40
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
General proof via Moment Generating Functions (cont)
Expansion of MGF:
MX (t) = 1 + �t +�′
2t2
2!+ ⋅ ⋅ ⋅ = 1 + t
(�+
�′2t2
+ ⋅ ⋅ ⋅).
Expansion for log(1 + u) is
log(1 + u) = u − u2
2+
u3
3!− ⋅ ⋅ ⋅ .
Combining gives
log MX (t) = t(�+
�′2t2
+ ⋅ ⋅ ⋅)−
t2(�+
�′2t2 + ⋅ ⋅ ⋅
)2
2+ ⋅ ⋅ ⋅
= �t +�′
2 − �2
2t2 + terms in t3 or higher.
41
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
General proof via Moment Generating Functions (cont)
log MX
(t
�√
N
)
=�t
�√
N+
�2
2t2
�2N+ terms in t3/N3/2 or lower in N.
42
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
General proof via Moment Generating Functions (cont)
log MX
(t
�√
N
)
=�t
�√
N+
�2
2t2
�2N+ terms in t3/N3/2 or lower in N.
Denote lower order terms by O(N−3/2). Collecting gives
log MZN (t) = −�t√
N�
+ N(
�t
�√
N+
t2
2N+ O(N−3/2)
)
= −�t√
N�
+�t√
N�
+t2
2+ O(N−1/2)
=t2
2+ O(N−1/2).
43
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Central Limit Theoremand Fourier Analysis
44
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convolutions
Convolution of f and g:
h(y) = (f ∗ g)(y) =
∫
ℝ
f (x)g(y − x)dx =
∫
ℝ
f (x − y)g(x)dx .
45
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convolutions
Convolution of f and g:
h(y) = (f ∗ g)(y) =
∫
ℝ
f (x)g(y − x)dx =
∫
ℝ
f (x − y)g(x)dx .
X1 and X2 independent random variables with probability density p.
Prob(Xi ∈ [x , x +Δx ]) =
∫ x+Δx
xp(t)dt ≈ p(x)Δx .
Prob(X1 + X2) ∈ [x , x +Δx ] =
∫ ∞
x1=−∞
∫ x+Δx−x1
x2=x−x1
p(x1)p(x2)dx2dx1.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Convolutions
Convolution of f and g:
h(y) = (f ∗ g)(y) =
∫
ℝ
f (x)g(y − x)dx =
∫
ℝ
f (x − y)g(x)dx .
X1 and X2 independent random variables with probability density p.
Prob(Xi ∈ [x , x +Δx ]) =
∫ x+Δx
xp(t)dt ≈ p(x)Δx .
Prob(X1 + X2) ∈ [x , x +Δx ] =
∫ ∞
x1=−∞
∫ x+Δx−x1
x2=x−x1
p(x1)p(x2)dx2dx1.
As Δx → 0 we obtain the convolution of p with itself:
Prob(X1 + X2 ∈ [a, b]) =
∫ b
a(p ∗ p)(z)dz.
Exercise to show non-negative and integrates to 1.47
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Statement of Central Limit Theorem
WLOG p has mean zero, variance one, finite third moment anddecays rapidly so all convolution integrals converge: p infinitelydifferentiable function satisfying∫ ∞
−∞xp(x)dx = 0,
∫ ∞
−∞x2p(x)dx = 1,
∫ ∞
−∞∣x ∣3p(x)dx < ∞.
X1,X2, . . . are iidrv with density p.
Define SN =∑N
i=1 Xi .
Standard Gaussian (mean zero, variance one) is 1√2�
e−x2/2.
48
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Statement of Central Limit Theorem
WLOG p has mean zero, variance one, finite third moment anddecays rapidly so all convolution integrals converge: p infinitelydifferentiable function satisfying∫ ∞
−∞xp(x)dx = 0,
∫ ∞
−∞x2p(x)dx = 1,
∫ ∞
−∞∣x ∣3p(x)dx < ∞.
X1,X2, . . . are iidrv with density p.
Define SN =∑N
i=1 Xi .
Standard Gaussian (mean zero, variance one) is 1√2�
e−x2/2.
Central Limit Theorem Let Xi ,SN be as above and assume the thirdmoment of each Xi is finite. Then SN/
√N converges in probability to
the standard Gaussian:
limN→∞
Prob
(SN√
N∈ [a, b]
)=
1√2�
∫ b
ae−x2/2dx .
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem
The Fourier transform: p(y) =∫∞−∞ p(x)e−2�ixy dx .
50
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem
The Fourier transform: p(y) =∫∞−∞ p(x)e−2�ixy dx .
Derivative of g is the Fourier transform of −2�ixg(x);differentiation (hard) is converted to multiplication (easy).
g′(y) =
∫ ∞
−∞−2�ix ⋅ g(x)e−2�ixy dx ;
g prob. density, g′(0) = −2�iE[x ], g′′(0) = −4�2E[x2].
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem
The Fourier transform: p(y) =∫∞−∞ p(x)e−2�ixy dx .
Derivative of g is the Fourier transform of −2�ixg(x);differentiation (hard) is converted to multiplication (easy).
g′(y) =
∫ ∞
−∞−2�ix ⋅ g(x)e−2�ixy dx ;
g prob. density, g′(0) = −2�iE[x ], g′′(0) = −4�2E[x2].
Natural: mean and variance simple multiples of derivatives of pat zero: p′(0) = 0, p′′(0) = −4�2.
52
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem
The Fourier transform: p(y) =∫∞−∞ p(x)e−2�ixy dx .
Derivative of g is the Fourier transform of −2�ixg(x);differentiation (hard) is converted to multiplication (easy).
g′(y) =
∫ ∞
−∞−2�ix ⋅ g(x)e−2�ixy dx ;
g prob. density, g′(0) = −2�iE[x ], g′′(0) = −4�2E[x2].
Natural: mean and variance simple multiples of derivatives of pat zero: p′(0) = 0, p′′(0) = −4�2.
We Taylor expand p (need technical conditions on p):
p(y) = 1 +p′′(0)
2y2 + ⋅ ⋅ ⋅ = 1 − 2�2y2 + O(y3).
Near origin, p a concave down parabola.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Prob(X1 + ⋅ ⋅ ⋅+ XN ∈ [a, b]) =∫ b
a (p ∗ ⋅ ⋅ ⋅ ∗ p)(z)dz.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Prob(X1 + ⋅ ⋅ ⋅+ XN ∈ [a, b]) =∫ b
a (p ∗ ⋅ ⋅ ⋅ ∗ p)(z)dz.
The Fourier transform converts convolution to multiplication. IfFT[f ](y) denotes the Fourier transform of f evaluated at y :
FT[p ∗ ⋅ ⋅ ⋅ ∗ p](y) = p(y) ⋅ ⋅ ⋅ p(y).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Prob(X1 + ⋅ ⋅ ⋅+ XN ∈ [a, b]) =∫ b
a (p ∗ ⋅ ⋅ ⋅ ∗ p)(z)dz.
The Fourier transform converts convolution to multiplication. IfFT[f ](y) denotes the Fourier transform of f evaluated at y :
FT[p ∗ ⋅ ⋅ ⋅ ∗ p](y) = p(y) ⋅ ⋅ ⋅ p(y).
Do not want the distribution of X1 + ⋅ ⋅ ⋅+ XN = x , but ratherSN = X1+⋅⋅⋅+XN√
N= x .
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Prob(X1 + ⋅ ⋅ ⋅+ XN ∈ [a, b]) =∫ b
a (p ∗ ⋅ ⋅ ⋅ ∗ p)(z)dz.
The Fourier transform converts convolution to multiplication. IfFT[f ](y) denotes the Fourier transform of f evaluated at y :
FT[p ∗ ⋅ ⋅ ⋅ ∗ p](y) = p(y) ⋅ ⋅ ⋅ p(y).
Do not want the distribution of X1 + ⋅ ⋅ ⋅+ XN = x , but ratherSN = X1+⋅⋅⋅+XN√
N= x .
If B(x) = A(cx) for some fixed c ∕= 0, then B(y) = 1c A
( yc
).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Prob(X1 + ⋅ ⋅ ⋅+ XN ∈ [a, b]) =∫ b
a (p ∗ ⋅ ⋅ ⋅ ∗ p)(z)dz.
The Fourier transform converts convolution to multiplication. IfFT[f ](y) denotes the Fourier transform of f evaluated at y :
FT[p ∗ ⋅ ⋅ ⋅ ∗ p](y) = p(y) ⋅ ⋅ ⋅ p(y).
Do not want the distribution of X1 + ⋅ ⋅ ⋅+ XN = x , but ratherSN = X1+⋅⋅⋅+XN√
N= x .
If B(x) = A(cx) for some fixed c ∕= 0, then B(y) = 1c A
( yc
).
Prob(
X1+⋅⋅⋅+XN√N
= x)
= (√
Np ∗ ⋅ ⋅ ⋅ ∗√
Np)(x√
N).
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Prob(X1 + ⋅ ⋅ ⋅+ XN ∈ [a, b]) =∫ b
a (p ∗ ⋅ ⋅ ⋅ ∗ p)(z)dz.
The Fourier transform converts convolution to multiplication. IfFT[f ](y) denotes the Fourier transform of f evaluated at y :
FT[p ∗ ⋅ ⋅ ⋅ ∗ p](y) = p(y) ⋅ ⋅ ⋅ p(y).
Do not want the distribution of X1 + ⋅ ⋅ ⋅+ XN = x , but ratherSN = X1+⋅⋅⋅+XN√
N= x .
If B(x) = A(cx) for some fixed c ∕= 0, then B(y) = 1c A
( yc
).
Prob(
X1+⋅⋅⋅+XN√N
= x)
= (√
Np ∗ ⋅ ⋅ ⋅ ∗√
Np)(x√
N).
FT[(√
Np ∗ ⋅ ⋅ ⋅ ∗√
Np)(x√
N)](y) =
[p(
y√N
)]N.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Can find the Fourier transform of the distribution of SN :[p(
y√N
)]N
.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Can find the Fourier transform of the distribution of SN :[p(
y√N
)]N
.
Take the limit as N → ∞ for fixed y .
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Can find the Fourier transform of the distribution of SN :[p(
y√N
)]N
.
Take the limit as N → ∞ for fixed y .
Know p(y) = 1 − 2�2y2 + O(y3). Thus study[1 − 2�2y2
N+ O
(y3
N3/2
)]N
.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Can find the Fourier transform of the distribution of SN :[p(
y√N
)]N
.
Take the limit as N → ∞ for fixed y .
Know p(y) = 1 − 2�2y2 + O(y3). Thus study[1 − 2�2y2
N+ O
(y3
N3/2
)]N
.
For any fixed y ,
limN→∞
[1 − 2�2y2
N+ O
(y3
N3/2
)]N
= e−2�2y2
.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
Can find the Fourier transform of the distribution of SN :[p(
y√N
)]N
.
Take the limit as N → ∞ for fixed y .
Know p(y) = 1 − 2�2y2 + O(y3). Thus study[1 − 2�2y2
N+ O
(y3
N3/2
)]N
.
For any fixed y ,
limN→∞
[1 − 2�2y2
N+ O
(y3
N3/2
)]N
= e−2�2y2
.
Fourier transform of 1√2�
e−x2/2 at y is e−2�2y2.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
We have shown:
the Fourier transform of the distribution of SN converges toe−2�2y2
;
the Fourier transform of 1√2�
e−x2/2 at y is e−2�2y2.
Therefore the distribution of SN equalling x converges to 1√2�
e−x2/2.
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Summary for the Day Review Accumulation and Moments Clicker Questions CLT and MGF CLT and Fourier Analysis
Proof of the Central Limit Theorem (cont)
We have shown:
the Fourier transform of the distribution of SN converges toe−2�2y2
;
the Fourier transform of 1√2�
e−x2/2 at y is e−2�2y2.
Therefore the distribution of SN equalling x converges to 1√2�
e−x2/2.We need complex analysis to justify this inversion. Must be careful:Consider
g(x) =
{e−1/x2
if x ∕= 0
0 if x = 0.
All the Taylor coefficients about x = 0 are zero, but the function is notidentically zero in a neighborhood of x = 0.
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