The moment-generating function for the gamma random

Chapter 4, Problem 143E

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QUESTION:

The moment-generating function for the gamma random variable is derived in Example 4.13.

Differentiate this moment-generating function to find the mean and variance of the gamma distribution.

Find the moment-generating function for a gamma-distributed random variable.

               \(m(t)=E\left(e^{ty}\right)=\int_0^{\infty}e^{ty}\left[\frac{y^{\alpha-1}e^{-y/\beta}}{\beta^{\alpha}\Gamma(\alpha)}\right]\ dy\)

                \(=\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int_0^{\infty}y^{\alpha-1}\exp\left[-y\left(\frac{1}{\beta}-t\right)\right]\ dy\)

                \(=\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int_0^{\infty}y^{\alpha-1}\exp\left[\frac{-y}{\beta/(1-\beta t)}\right]\ dy\).

Equation Transcription:

Text Transcription:

m(t)=E(e^ty)=integral 0 to infinity e^ty[y^alpha-1 e^-y/beta over beta^alpha Gamma(alpha)]dy

=1 over beta^alpha Gamma(alpha) integral 0 to infinity y^alpha-1 exp[-y(1 over beta -t)]dy

=1 over beta^alpha Gamma(alpha) integral 0 to infinity y^alpha-1 exp[-y over /(1-beta t)]dy

Questions & Answers

QUESTION:

The moment-generating function for the gamma random variable is derived in Example 4.13.

Differentiate this moment-generating function to find the mean and variance of the gamma distribution.

Find the moment-generating function for a gamma-distributed random variable.

               \(m(t)=E\left(e^{ty}\right)=\int_0^{\infty}e^{ty}\left[\frac{y^{\alpha-1}e^{-y/\beta}}{\beta^{\alpha}\Gamma(\alpha)}\right]\ dy\)

                \(=\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int_0^{\infty}y^{\alpha-1}\exp\left[-y\left(\frac{1}{\beta}-t\right)\right]\ dy\)

                \(=\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int_0^{\infty}y^{\alpha-1}\exp\left[\frac{-y}{\beta/(1-\beta t)}\right]\ dy\).

Equation Transcription:

Text Transcription:

m(t)=E(e^ty)=integral 0 to infinity e^ty[y^alpha-1 e^-y/beta over beta^alpha Gamma(alpha)]dy

=1 over beta^alpha Gamma(alpha) integral 0 to infinity y^alpha-1 exp[-y(1 over beta -t)]dy

=1 over beta^alpha Gamma(alpha) integral 0 to infinity y^alpha-1 exp[-y over /(1-beta t)]dy

ANSWER:

Solution:

Step 1 of 2:

 From exercise 4.13. 

     We have the moment generating function for the gamma distribution is

                      

Here we have to differentiate this mgf and to find the mean and variance.

     That is,

                           (since

                                

                    If t = 0,

                                            =

                                            = E(Y)  

                   

                               

                      If t = 0,

                               

                                            =

                                           


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