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Let's say we have two independent random variables, $x_1$ and $x_2$, both have a probability mass function $X$ defined as

$$X(n) = \begin{cases} 2^{-m} & \text{if $n=2^m$ for $1 \le m \in \mathbb Z$} \\ 0 & \text{otherwise} \end{cases}$$

So $X(2)=\frac 12$, $X(4) = \frac 14$, $X(8) = \frac 18$, and so on (this is the distribution used in the St. Petersburg paradox).

The expected value is undefined for both $x_1$ and $x_2$. My question is, does $E[x_1-x_2]$ exist? I know that if it exists, it must be $0$, by symmetry. I have a feeling it does equal $0$, but I'm not sure.

In general, if two random variables have the same probability distribution, when will the expected value of their difference be $0$? (I think there would be counter examples for this.)

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In probability theory, we use Lebesgue integration to define the expected value. In this sense your expected value is not defined, even as $\pm \infty$, because it is a $\infty - \infty$ form. That is, in Lebesgue integration you define $\int f = \int f^+ - \int f^-$, where $f^+$ is the positive part and $f^-$ is the negative part. Here both are $\infty$, so you get $\infty - \infty$ which we decline to define.

You could define integration through a notion like the Cauchy principal value; in this case the "symmetry argument" you mentioned would say that you do indeed get zero. This is often a bad idea because if you take the limit with broken symmetry (say summing from $-n$ to $2n$ and then sending $n \to \infty$) then you get a different limit.

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  • $\begingroup$ @PyRulez 1. It doesn't matter that it's discrete, Lebesgue integration uses measures, which can be discrete. 2. In Lebesgue integration the definition of $\int f d \mu$ is $\int f^+ d \mu - \int f^- d \mu$, where $f^+$ is the positive part and $f^-$ is the negative part. So add up the portion of the sum where $x_1-x_2$ is positive and the portion where it's negative, and see that they are both infinite. $\endgroup$ – Ian Jan 24 '16 at 17:06
  • $\begingroup$ @PyRulez In limits, $\infty - \infty$ is indeterminate. That is, if $x_n \to \infty$ and $y_n \to \infty$ then $\lim_{n \to \infty} (x_n-y_n)$ can exist. In Lebesgue integration, the two are separated by definition, so you get $\infty - \infty$ as a number, not as a form. We decline to define this quantity. By doing so, we get that the Lebesgue integral is always absolutely convergent (unlike the improper Riemann integral). $\endgroup$ – Ian Jan 24 '16 at 17:07
  • $\begingroup$ Why the downvote here? $\endgroup$ – copper.hat Jan 24 '16 at 17:09
  • $\begingroup$ @Ian do you have a proof or link? $\endgroup$ – PyRulez Jan 24 '16 at 17:09
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    $\begingroup$ The linearity of expectation relies on both expectations being finite. Surely you would agree that the expectation of $x_1-x_1$ is 0, for example. $\endgroup$ – Kevin O'Bryant Jan 24 '16 at 17:53

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