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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.

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Overall you are doing the problem correctly. The probability mass function has been given to you explicitly. Perhaps you were asked to show that it is indeed a pmf, though that is not clear from the question.

If that was asked for, you need to sum the probabilities ("weights") and show that the sum is $1$. The sum of the weights at numbers of the form $2^n$ is $\sum_{n=0}^\infty \frac{1}{2^{n+2}}$, which you showed is $\frac{1}{2}$. The sum of the weights at numbers of the form $-2^n$ is also $\sum_{n=0}^\infty \frac{1}{2^{n+2}}$, another $\frac{1}{2}$, for a total of $1$. So we were indeed given a pmf.

For the expectation, we need to find $\sum_{n=0}^\infty 2^n \cdot\frac{1}{2^{n+2}} + \sum_{n=0}^\infty (-2^n) \cdot\frac{1}{2^{n+2}}.$ Both of the above sums diverge, since $2^n\cdot\frac{1}{2^{n+2}}=\frac{1}{4}$. Thus the expectation does not exist.

This is an interesting example of a pmf which is symmetrical about $x=0$, but such that the mean is not $0$. In all symmetrical situations, if the mean exists, then it is at the centre of symmetry.

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    You are correct, it was to show that what was given was indeed a pmf. Either way, thank you for the help. I will edit the question.2012-09-23