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This question seeks an elementary proof for the following simple and intuitive fact:

Let $X$ and $Y$ be identically distributed, not necessarily independent, and $\mathbb E[X-Y]$ exist. Then, $$\mathbb E[X-Y]=0.$$

I think it should be examined somewhere in the literature.

The proof is straightforward when both $\mathbb E[X]$ and $\mathbb E[Y]$ are finite. I have a proof based on non-elementary results for the general case where $X$ and $Y$ may have undefined or infinite mean (e.g., $X=Y$ with the Cauchy distribution) as follows:

Proof: Rewrite $\mathbb E[X-Y]$ as $\mathbb E[\log e^X-\log e^ Y]$. Since both $U=e^ X$ and $V=e^ Y$ are positive, then using Lemma 2 given in this elegant answer by Sangchul Lee for $U \sim V$ we have: $$ \mathbb E[X-Y]=\mathbb{E}[\log(U/V)] = \\ \int_{0}^{\infty} \frac{\mathbb{E}[e^{-Vs}] - \mathbb{E}[e^{-Us}]}{s} \, \mathrm{d}s=0. $$

To prove Lemma 2 used above, advanced results such as the Frullani's integral and the Tonelli's theorem are used.

PS: It is worth noting that any random variable with zero mean can be represented as the difference of two identically distributed random variables [2].

Amir
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  • The question has been answered by two senior members. Robert nicely proved the statement by implicitly assuming that the expectation is finite, $\mathbb{E}[X - Y] \in (-\infty, \infty)$ because of using the dominated convergence theorem. Lee cleverly used Robert's idea and the so-called inverse Fatou lemma to prove the statement by only assuming that the expectation exists, $\mathbb{E}[X - Y] \in [-\infty, \infty].$ I just accepted the second one, but a shorter or more elementary proof may be given by other members as the statement seems very intuitive. – Amir Mar 11 '25 at 18:53

2 Answers2

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Consider the truncations $X_N = f_N(X)$ and $Y_N = f_N(Y)$, where $$ f_N(x) = \cases{N & if $x \ge N$\cr -N & if $x \le -N$\cr x & otherwise}$$ Then $X_N$ and $Y_N$ are also identically distributed, and $\mathbb E[X_N - Y_N] = \mathbb E[X_N] - \mathbb E[Y_N] = 0$.
Moreover, $|X_N - Y_N| \le |X - Y|$. By the Dominated Convergence Theorem, $\mathbb E[X - Y] = \lim_{N \to \infty} \mathbb E[X_N - Y_N] = 0$.

Robert Israel
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  • Could you elaborate on $|X_N - Y_N| \le |X - Y|$? – Amir Mar 09 '25 at 14:43
  • @Amir: Here it goes again (with minor edit): If $X<-N$ and $|Y|\leq N$, $$|f_N(X)-f_N(Y)|=|-N-Y|=N+Y<Y-X=|X-Y|$$

    If $X>N$ and $|Y|\leq N$, $$|f_N(X)-f_N(Y)|=N-Y<X-Y=|X-Y|$$

     – Mittens Mar 09 '25 at 16:10
  • @Mittens Thanks for the edit! – Amir Mar 09 '25 at 16:14
  • @Amir: you're welcome! Bob's answers tend to be sparse on details, however, his his answers are impeccable. I don'y understand why someone decided to downvote his answer... – Mittens Mar 09 '25 at 16:16
  • Thanks Robert for your answer! It seems difficult to come up with a simpler or shorter proof. – Amir Mar 09 '25 at 16:21
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    Great observation! I was able to modify your argument to prove the same conclusion in the case $\mathbb{E}[X - Y] \in [-\infty, +\infty]$. – Sangchul Lee Mar 09 '25 at 18:19
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Here is an adaptation of @Robert Israel's argument, proving that the same conclusion holds true when $\mathbb{E}[X - Y]$ is only assumed to exist in $[-\infty, \infty]$.

By reversing the sign if necessary, we may assume $\mathbb{E}[X - Y] > -\infty$, and in particular, $\mathbb{E}[(X - Y)_-] < \infty$. Now let

$$ f_N(x) = \operatorname{mid}(-N, x, N)$$

be the truncation function, and note that $\frac{f_N(x) - f_N(y)}{x - y} \in [0, 1]$ for any $x \neq y$. Using this, it is easy to check that the following inequality holds.

$$ -(X - Y)_- \leq f_N(X) - f_N(Y) \leq (X - Y)_+. $$

Since $f_N(X) - f_N(Y)$ is bounded below by an integrable RV, a variant of Fatou's lemma is applicable, yielding

$$ \liminf_{n\to\infty} \mathbb{E}[f_N(X) - f_N(Y)] \geq \mathbb{E}[\liminf_{n\to\infty} (f_N(X) - f_N(Y))]. $$

However, the left-hand side is $0$ because $X \sim Y$, and the right-hand side tends to $\mathbb{E}[X - Y]$. Therefore we get

$$ 0 \geq \mathbb{E}[X - Y]. $$

This shows that $X - Y$ is in fact integrable, and then by interchanging the role of $X$ and $Y$, we have $\mathbb{E}[X - Y] = 0$ as desired.

Sangchul Lee
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  • Thank you for extending Robert's idea for the general case, where the existence of $\mathbb E[X-Y]$ is considred instead of the Lebesgue integrability of $X-Y$, $\mathbb E\left |X-Y \right |< +\infty$. – Amir Mar 09 '25 at 19:27