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Consider the following:

Give examples of two discrete random variables $X$ and $Y$, whose expected value is infinite (i.e. don't have expected value). Also:

1) $X$ only gets positive values.

2) $Y$ ~ $-Y$, so $Y$ and $-Y$ are similarly distributed.

I think that one way of finding a random variable with no expected value, is to find a variable with value $x$ and probability $p$, so that infinite sum their product is infinite.

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    $\begingroup$ The meaning of "$X$ has no expected value" is $E(|X|)=\infty$, not that certain values are impossible for $X$. So for case (1) you want a discrete random variable $X$ such that $\sum_{x=1}^\infty x P(X=x)=\infty$. $\endgroup$ – grand_chat Apr 25 '18 at 16:30
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    $\begingroup$ Have you learnt anything from your suspension? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 30 '18 at 12:49
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No, not quite; that variable $X$ is extremely easy to compute the expectation of: it is $0$. What you need to look for here is more technical: remember, expectations are just integrals (or might look like sums, depending on the associated measure). What you want is to create variables whose expectations are improper integrals (or infinite series) that don't converge.

To that end, here are a couple of hints:

(1) Consider a random variable taking values in $\{1,2,3,\ldots\}$. If $p_n=P(X=n)$, then you want $$ \sum_{n=1}^{\infty}p_n=1\qquad\text{and}\qquad\sum_{n=1}^{\infty}np_n=\infty. $$ What if $p_n=C\frac{1}{n^k}$ for some constants $C$ and $k$? How could they be chosen to make the above results hold?

(2) This is a similar thought, but we can cheat and use part (1). Remember: an integral on $(-\infty,\infty)$ (or a sum for $n=-\infty$ to $n=\infty$) can't converge if the piece to the left of $0$ tends to $-\infty$ and the piece to the right tends to $\infty$. Could you combine a couple independent copies of the variable from (1) to get a variable that is symmetric and satisfies this property?

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  • $\begingroup$ What if I choose the following: random variable takes values n = 2^n and the probability is p_n = 2^-n ? The infinite sum of their product is infinity. $\endgroup$ – HaydenSpetting Apr 25 '18 at 17:26
  • $\begingroup$ @HaydenSpetting Yes, value $2^n$ with probability $2^{-n}$ is a classic coin-tossing example. $\endgroup$ – r.e.s. Apr 25 '18 at 18:12
  • $\begingroup$ Thanks, I now got the variable X. I still can't figure out a suitable variable Y. It's hard to image how Y and -Y are distributed. $\endgroup$ – HaydenSpetting Apr 25 '18 at 19:07
  • $\begingroup$ @Nick_Peterson OK, so it just the same for case (2) ? Random variable with value 2^n and probability 2^-n. Infinite sum to negativity of their product is negative infinity. $\endgroup$ – HaydenSpetting Apr 26 '18 at 14:44

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