Answer to Question #202810 in Statistics and Probability for Raj Kumar

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

$$M_X(t)=E[e^{tX}]=\displaystyle\sum_{x=0}^{\infin}e^{tx}(\dfrac{2}{3})(\dfrac{1}{3})^x=\dfrac{2}{3}\displaystyle\sum_{x=0}^{\infin}(\dfrac{e^t}{3})^x$$

$$=\dfrac{2}{3}\displaystyle\sum_{x=0}^{\infin}(\dfrac{e^t}{3})^x=\dfrac{2}{3}(\dfrac{1}{1-\dfrac{e^t}{3}})=\dfrac{2}{3-e^t},\ \dfrac{e^t}{3}<1$$

$$M_X(t)=\dfrac{2}{3-e^t},\ t<\ln3$$

$$\dfrac{dM}{dt}=\dfrac{2e^t}{(3-e^t)^2}$$

$$\dfrac{dM}{dt}|_{t=0}=\dfrac{2(1)}{(3-1)^2}=\dfrac{1}{2}$$

$$mean=E[X]=\dfrac{1}{2}$$

$$\dfrac{d^2M}{dt^2}=\dfrac{2e^t(3-e^t)^2+4e^{2t}(3-e^t)}{(3-e^t)^4}$$

$$=\dfrac{2e^t(3+e^t)}{(3-e^t)^3}$$

$$\dfrac{d^2M}{dt^2}|_{t=0}=\dfrac{2(1)(3+1)}{(3-1)^3}=1$$

$$Var(X)=1-(\dfrac{1}{2})^2=\dfrac{3}{4}$$