Question

For the given distribution:

P(X=x)=(2/3)(1/3)^x ;x=0,1,2,.....,find moment generating function, mean and variance of X.

Answer 1

Solution:

$M_X(t)=E[e^{tx}]={\Sigma}_0^{\infty} e^{tx}\frac23(\frac13)^x \\=\frac23{\Sigma}_0^{\infty} (\frac{e^{t}}3)^x$

$=\dfrac23[(\frac{e^{t}}3)^0+(\frac{e^{t}}3)^1+(\frac{e^{t}}3)^2+...] \\=\dfrac23[\dfrac{1}{1-\frac{e^{t}}3}] \\=\dfrac2{3-e^t}$

$M'_X(t)=(2[3-e^t]^{-1})'=2(-1)(3-e^t)^{-2}(-e^t)=\dfrac{2e^t}{(3-e^t)^2}$

$E(X)=M'_X(0)=\dfrac{2e^0}{(3-e^0)^2}=\dfrac12$

$M''_X(t)=[\dfrac{2e^t}{(3-e^t)^2}]'=\dfrac{(3-e^t)^2(2e^t)-2e^t(2(3-e^t)(-e^t))}{(3-e^t)^4}$

$E(X^2)=M''_X(0)=\dfrac{(3-e^0)^2(2e^0)-2e^0(2(3-e^0)(-e^0))}{(3-e^0)^4} \\=\dfrac{(2)^2(2)-2(2)(2)(-1)}{(2)^4} \\=\dfrac{8+8}{16}=1$

$V(X)=E(X^2)-[E(X)]^2=1-(\frac12)^2=\frac34$