Answer to Question #173482 in Statistics and Probability for ANJU JAYACHANDRAN

Question

9 b) For the given distribution: (5) 

 ; ,2,1,0 ,

3

1

3

2

( )  = K











P X = x = x

x

 find moment generating function, mean and 

variance of X.

Answer 1

The pmf of random variable X 

$p(x)=\dfrac{e^{−λ}λ^x}{x!} ,for x=0,1,2,….$

 

So moment generating function is given by-

$M_X(t)=E[e^{tX}]=∑_{x=0}^∞e^{tx}⋅\dfrac{e^{−λ}λ^x}{x!}=e^{−λ}∑_{x=0}^∞\dfrac{(e^{tλ})^x}{x!}=e^{−λ}(e^{e^{t}λ})=e^{λ(e^t−1)}$

.

 

Now we take the first and second derivatives of  $M_X(t) .$ Remember we are differentiating with respect to t :

$M'_X(t)=\dfrac{d}{dt}[e^{λ(e^t−1)}]=λe^te^{λ(e^t−1)}$

 

$M''X(t)=\dfrac{d}{dt}[λe^te^{λ(e^t−1)}]=λe^te^{λ(e^t−1)}+λ^2e^{2t}e^{λ(e^t−1)}$

Next we evaluate the derivatives at t=0 to find the first and second moments of X :

$E[X]=M'_X(0)$

$E[X^2]=M''_X(0)=λe^0e^{λ(e^0−1)}=λ=λe^0e^{λ(e^0−1)}+λ^2e^0e^{λ(e^0−1)}=λ+λ^2$

 

Finally, in order to find the variance, we use the alternate formula:

$Var(X)=E[X^2]−(E[X])^2=λ+λ^2−λ^2=λ.$