Download Chapter 5. Continuous Probability Distributions

Chapter 5. Continuous Probability
Distributions
Sections 5.10: Moment Generating Function
Jiaping Wang
Department of Mathematical Science
04/08/2013, Monday
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Moment-generating Function
The moment generating function of a continuous random
variable X with a probability density function is given by
∞
M 𝑡 = 𝐸 𝑒𝑡𝑡 = ∫−∞ 𝑒𝑡𝑡𝑓 𝑥 𝑑𝑑
When the integral exists.
Properties:
1. If a random variable X has MGF MX(t), then Y=aX+b for
constants a and b has the MGF
𝑀𝑌 𝑡 = 𝑒𝑒𝑒𝑀𝑀 (𝑎𝑎)
2. MGFs are unique; that is, if two random variables have the
same MGFs, then they have the same probability distribution.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
MGF of Exponential Distribution
∞
∞
1
𝑥
1
1
𝑡𝑡
𝑀 𝑡 =� 𝑒
exp − 𝑑𝑑 = � exp[−𝑥
− 𝑡 ] 𝑑𝑑
𝜃
𝜃
𝜃 0
𝜃
0
1
1
𝜃
=
.
= Γ 1
1 − 𝜃𝜃
θ
1 − 𝜃𝜃
So we have
E(X)=M’(0)=[-(1-θt)-1(-θ)]t=0=θ
Similarly, we can find E(X2) and then V(X).
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
MGF of Gamma Distribution
We are given that the MGF of Gamma distribution
M(t)=(1-βt)-α
From this, we can find
E(X)=M’(0)=[-α(1-βt)-α-1(-β)]t=0=αβ.
E(X2)=M(2)(0)=[αβ(-α-1)(1-βt)-α-2(-β)]t=0=α(α+1)β2.
E(X3)=M(3)(0)=[α(α+1)β2(-α-2)(1-βt)-α-3(-β)]t=0=α(α+1)(α+2)β3.
And so on.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
MGF of Normal Distribution
𝑴𝒁 𝒕 = 𝑬 𝒆𝒕𝒕
∞
𝟏
∞
𝟐
𝒛
𝟏
= � 𝒆𝒕𝒕
𝒆𝒆𝒆 −
𝒅𝒅 = (
) � 𝒆𝒆𝒆 �𝒕𝒕
𝟐𝝅
𝟐
𝟐𝝅 −∞
−∞
∞
𝒛𝟐
𝟏
𝒛−𝒕 𝟐
− � 𝒅𝒅 = (
) � 𝒆𝒆𝒆 �−
𝟐
𝟐
𝟐𝝅 −∞
𝒕𝟐
𝒕𝟐 ∞
𝟏
𝒚𝟐
𝒕𝟐
+ � 𝒅𝒅 = 𝒆𝒆𝒆( ) �
𝒆𝒆𝒆 −
𝒅𝒅 = 𝒆𝒆𝒆( )
𝟐
𝟐 −∞ 𝟐𝝅
𝟐
𝟐
Where we set y=z-t.
So for 𝑿 = 𝝁 + 𝒁𝝈, 𝑴𝑴(𝒕) = exp(𝒕𝝁 + 𝒕𝟐𝝈𝟐/𝟐).
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Additional Example 1
A random variable X has a uniform distribution over the interval (a, b):
1
, 𝑎<𝑥<𝑏
𝑓 𝑥 = �𝑏−𝑎
.
0, 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
Find the moment generating.
𝑏 exp 𝑡𝑡
𝑏−𝑎
Answer: 𝑀(𝑡) = 𝐸[exp(𝑡𝑡)] = ∫𝑎
𝑑𝑑 =
exp 𝑡𝑡 −exp 𝑡𝑡
𝑡 𝑏−𝑎
.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Additional Example 2
Find the moment generating function of the exponential distribution with mean
θ. And then find the mean and variance by differentiating the mgf.
Answer:
𝑀 𝑡 = exp 𝑡𝑡 =
∞
1
𝑥
exp − 𝑑𝑑 =
∫0 exp 𝑡𝑡
𝜃
𝜃
1
𝜃
∞
∫0 exp −
1
𝜃
− 𝑡 𝑥 𝑑𝑑 =
1
.
1−𝜃𝜃
M’(t)=θ/(1-θt)M’(0)=E(X)=θ.
M(2)(t)= 2θ2/(1-θt)M(2)(0)=E(X2)=2θ2 V(X)=E(X2)-E2(X)=θ2.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Additional Example 3
A random variable has a continuous distribution
𝑥𝑥𝑥𝑥 −𝑥 , 𝑥 > 0
𝑓 𝑥 =�
0, 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
Find the moment generating function and its mean and variance by
differentiating the mgf.
Answer:
=
∞
∫0 exp
𝑡𝑡 𝑥𝑥𝑥𝑥 −𝑥 𝑑𝑑 =
𝑀 𝑡 = 𝐸 exp 𝑡𝑡
∞
− 1 − 𝑡 𝑥 𝑥 = ∫0
∞
∫0 𝑥𝑥𝑥𝑥
M’(t)=2/(1-t)3M’(0)=E(X)=2.
M(2)(t)=6/(1-t)4M(2)(0)=E(X2)=6  V(X)=E(X2)-E2(X)=2.
1
1−𝑡
2 𝑦𝑦𝑦𝑦 −𝑦 𝑑𝑑 =
1
(1−t)2
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL