Answer to Question #271705 in Statistics and Probability for nyabe

Question

A probability density function of a discrete random variable x is given by

f(x)={(1/2)^x x=1,2,3,... 0 elsewhere}

use pgf to find a)E(x) b)V(x)

Answer 1

$$\displaystyle\sum_{n=1}^{\infin}q^n=\dfrac{q}{1-q}, |q|<1$$

Differentiate both sides of the equality with respect to $q$

$$\displaystyle\sum_{n=1}^{\infin}\dfrac{nq^n}{q}=\dfrac{1}{(1-q)^2}$$

$$\displaystyle\sum_{n=1}^{\infin}nq^n=\dfrac{q}{(1-q)^2}$$

Similarly

Differentiate both sides of the equality with respect to $q$

$$\displaystyle\sum_{n=1}^{\infin}\dfrac{n^2q^n}{q}=\dfrac{1}{(1-q)^2}+\dfrac{2q}{(1-q)^3}$$

$$\displaystyle\sum_{n=1}^{\infin}n^2q^n=\dfrac{q}{(1-q)^2}+\dfrac{2q^2}{(1-q)^3}$$

a)

$$E(X)=\displaystyle\sum_{x=1}^{\infin}x(\dfrac{1}{2})^x$$ $$q=1/2, n=x$$

$$E(X)=\displaystyle\sum_{x=1}^{\infin}x(\dfrac{1}{2})^x=\dfrac{1/2}{(1-1/2)^2}=2$$

b)

$$E(X^2)=\displaystyle\sum_{x=1}^{\infin}x^2(\dfrac{1}{2})^x$$ $$q=1/2, n=x$$

$$E(X^2)=\displaystyle\sum_{x=1}^{\infin}x^2(\dfrac{1}{2})^x=\dfrac{1/2}{(1-1/2)^2}+\dfrac{2(1/2)^2}{(1-1/2)^3}=6$$

$$Var(X)=E(X^2)-(E(X^2))^2=6-(2)^2=2$$