Why is it true that if two random variables have the same distribution, they have the same expected value?


closed as off-topic by Simply Beautiful Art, mechanodroid, José Carlos Santos, Did, uniquesolution Sep 28 '17 at 16:28

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From a given distribution, one performs an integral to get the expected value. If you give the same distribution twice, you get the same result of the integral both times.

Let $f$ be the PDF of the random variable $X$ taking values in $\mathbb{R}$. (We can be slightly more general, but let's not.) Then $\mathbb{E}(X) = \int_{-\infty}^{\infty} \; x f(x) \, \mathrm{d}x$. Every time we evaluate this integral for the same $f$, we get the same expected value.

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    $\begingroup$ What's an example of dependent variables with the same distribution but different expectations? $\endgroup$ – uniquesolution Sep 28 '17 at 15:35
  • $\begingroup$ @uniquesolution : It's early here and I haven't sufficiently coffeed, so I guarded against conflating equality of the marginal PDFs with exchange symmetry of the joint PDF. But several sips later, that was overcautious. $\endgroup$ – Eric Towers Sep 28 '17 at 15:43

In general we have $EX = \int X dP$, and a little works shows that if we define the cumulative distribution function $F_X(\alpha) = P \{ \omega | X(\omega) \le \alpha \}$ then we can write $EX = \int t dF_X(t)$.

Suppose we have two random variables, say $X,Y$ and that $F_X(\alpha) = F_Y (\alpha)$ for all $\alpha$.

The last expression for expectation then shows that $EX = EY$.

Note that $X,Y$ need not be defined on the same probability space.

  • $\begingroup$ FYI, the formula $\int X dP_X$ is absurd. Use either $\int_\Omega X dP$ or, for some real-valued $X$, $\int_\mathbb Rx dP_X(x)$ $\endgroup$ – Did Sep 28 '17 at 16:10
  • $\begingroup$ @Did: There was no meaning to the $X$ subscript other than to distinguish it from the corresponding measure on $Y$. I am not sure I see the absurdity? $\endgroup$ – copper.hat Sep 28 '17 at 17:14
  • $\begingroup$ The notation $P_X$ is universally used to denote the measure image of $P$ by $X$, that is, the measure defined by $P_X(B)=P(X\in B)$, aka the distribution of $X$. Hence a measure on the image set instead of the source set. Hence the absurdity (now corrected, which is good). $\endgroup$ – Did Sep 28 '17 at 17:20

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