So I think I have the pieces, just having trouble putting the puzzle together.
$P(Y=2^{n})=P(Y=-2^{n}) = \frac{1}{2^{n+2}}$
\begin{align*} \sum_{n=0}^{\infty} p_y(y) &=\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}\\ &=\frac{1}{4} \sum_{n=0}^{\infty} \frac{1}{2^n}\\ &=1/2 \end{align*}
Is this on the right track? Is it $\sum_{n=0}^{\infty} p_y(y) =2*\sum_{n=0}^{\infty} \frac{1}{2^{n+2}} =\frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n}=1$ instead.
I also need to show that the expectation does not exist. So I have
So far I have that $E(Y) = \sum_i y_ip_y(y_i)$, thus \begin{align*} E(Y)= \sum_{n=0}^{\infty} 2^n \frac{1}{2^{n+2}} = \infty\\ E(Y)= \sum_{n=0}^{\infty} -2^n \frac{1}{2^{n+2}} = -\infty \end{align*} and thus the expectation does not exist
As always, thanks for any help.