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Chapter 5. Continuous Probability Distributions Sections 5.10: Moment Generating Function. Jiaping Wang Department of Mathematical Science 04/08/2013, Monday. Moment-generating Function.
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Chapter 5. Continuous Probability DistributionsSections 5.10: Moment Generating Function Jiaping Wang Department of Mathematical Science 04/08/2013, Monday
Moment-generating Function The moment generating function of a continuous random variable X with a probability density function is given by M When the integral exists. Properties: If a random variable X has MGF MX(t), then Y=aX+b for constants a and b has the MGF 2. MGFs are unique; that is, if two random variables have the same MGFs, then they have the same probability distribution.
MGF of Exponential Distribution So we have E(X)=M’(0)=[-(1-θt)-1(-θ)]t=0=θ Similarly, we can find E(X2) and then V(X).
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.
MGF of Normal Distribution Where we set y=z-t. So for
Additional Example 1 A random variable X has a uniform distribution over the interval (a, b): . Find the moment generating. Answer:
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)=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.
Additional Example 3 A random variable has a continuous distribution Find the moment generating function and its mean and variance by differentiating the mgf. Answer: M’(t)=2/(1-t)3M’(0)=E(X)=2. M(2)(t)=6/(1-t)4M(2)(0)=E(X2)=6 V(X)=E(X2)-E2(X)=2.