Question

A discrete random variable, X has probability mass function, P(X = n)=(1/2)n . Let Y= 1, if x is even and Y=-1, if x is odd. Find the expected value of Y, (E(Y)), to 2 decimal places.

Answer 1

We will compute probabilities $P(Y=1)$ and $P(Y=-1)$ . Namely, using the formula for infinite geometric series (see https://en.wikipedia.org/wiki/Geometric_progression), we have:

$P(Y=1)=\sum_{k=1}^{+\infty}P(X=2k)=\sum_{k=1}^{+\infty}\frac{1}{2^{2k}}=\frac{1}{4}\frac{1}{1-\frac{1}{4}}=\frac{1}{3}$ ;

$P(Y=-1)=\sum_{k=1}^{+\infty}P(X=2k-1)=\sum_{k=1}^{+\infty}\frac{1}{2^{{2k-1}}}=\frac{1}{2}\frac{1}{1-\frac{1}{4}}=\frac{2}{3}$ .

Using the formula for the expected value in finite case (see https://en.wikipedia.org/wiki/Expected_value#Finite_case), we get:

$E[Y]=P(Y=1)-P(Y=-1)=-\frac13\approx-0.33$

Answer:-0.33