Question

A discrete random variable has probability mass function, P(X = n) = (1/2)^n. Y=1 when x is even and Y= -1 when x is odd. find the expected value of Y

Answer 1

At first, we will find the probability of $Y$ to become $1$ or $-1$ . Namely, we have:

$P(Y=1)=P(\{X =2k|k\in {\mathbb{N}}\})=\sum_{k=1}^{+\infty}P(X=2k)=\sum_{k=1}^{+\infty}\frac{1}{2^{2k}}=\sum_{k=1}^{+\infty}\frac{1}{4^{k}}.$

Using the formula for an infinite geometric series, we have:

$P(Y=1)=\sum_{k=1}^{+\infty}\frac{1}{4^{k}}=\frac14\frac{1}{1-\frac14}=\frac13$ .

In the same way we obtain:

$P(Y=-1)=P(\{X =2k-1|k\in {\mathbb{N}}\})=\sum_{k=1}^{+\infty}P(X=2k-1)=\newline\sum_{k=1}^{+\infty}\frac{1}{2^{2k-1}}=2\sum_{k=1}^{+\infty}\frac{1}{4^{k}}=\frac23.$

We remind that the expected value for discrete random variable Y is(see e.g., https://en.wikipedia.org/wiki/Expected_value):

$E[Y]=P(Y=1)*1+P(Y=-1)*(-1)=\frac13-\frac23=-\frac13$

Answer:$-\frac13$