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Let $D_1$ and $D_2$ be probability distributions on $\mathbb{R} \times \mathbb{R}$ with identical marginal distributions (i.e. the distribution of the first component of $D_1$ is the same as the distribution of the first component of $D_2$, and similarly for the second components).

Let $e_1 = \mathbb{E}_{ D_1} [x - y]$ and $e_2 = \mathbb{E}_{ D_2} [x -y]$. (Assume that these quantities are defined -- they could be finite or infinite.) Does $e_1$ necessarily equal $e_2$?

(Note that this does not immediately follow from linearity of expectation, because $\mathbb{E}[x - y]$ could be defined even if $\mathbb{E}[x]$ and $\mathbb{E}[y]$ are not.)

shalin
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Eric Neyman
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2 Answers2

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You are basically asking the following question. Take two real-valued random variables $X,Y$ defined on the same probability space, and two other ones $Z,W$ also defined on the same probability space (though not necessarily on the same one as $X,Y$). Suppose $X \stackrel{d}{=}Z$ and $Y\stackrel{d}{=} W$. Assuming that $E[X-Y]$ and $E[Z-W]$ both exist as extended real numbers, it necessarily true that they are equal?

I think the answer is yes (so decision theory works as it should).

We introduce the notation $u_+:=\max\{u,0\}$ and $u_- =\max\{-u,0\}$. Note that $u = u_+-u_-$ and $|u|=u_++u_-$. We also define $$f_n(x) = \begin{cases} x & |x|<n \\ n & x \ge n \\ -n & x \le -n. \end{cases}$$Note that $f_n$ is $1$-Lipchitz (i.e., $|f_n(x)-f_n(y)| \leq |x-y|$ for all $n,x,y$) and moreover $f_n(x) \to x$ pointwise as $n \to \infty$. First we have the following lemma.

Lemma. Suppose that $E[(X-Y)_-]<\infty$ and $E[(Z-W)_+]<\infty$. Then $E[(Z-W)_-]<\infty$ and $E[(X-Y)_+]<\infty$.

Proof. Since $f_n$ is non-decreasing and $1$-Lipchitz, we note that $$(f_n(z)-f_n(w))_+ \leq (z-w)_+,\;\;\;\;\;\;\;\;\forall z,w \in \Bbb R,$$$$(f_n(x)-f_n(y))_- \leq (x-y)_-,\;\;\;\;\;\;\;\;\forall x,y\in \Bbb R,$$ so by dominated convergence we have $\lim_n E[(f_n(Z)-f_n(W))_+] = E[(Z-W)_+]$ and $\lim_n E[(f_n(X)-f_n(Y))_-] = E[(X-Y)_-]$. Note that \begin{align}E[(f_n(Z)-f_n(W))_-] &=E[(f_n(Z)-f_n(W))_+] -E[f_n(Z)-f_n(W)]\\ &= E[(f_n(Z)-f_n(W))_+]-E[f_n(X)-f_n(Y)] \\&\le E[(f_n(Z)-f_n(W))_+]+E[(f_n(X)-f_n(Y))_-] ,\end{align} where we used the bound $a-b \leq a+b_-$ in the last inequality. We already know that the last expression stays bounded (in fact converges to $E[(Z-W)_+]+[(X-Y)_-]$) as $n \to \infty$. Thus by Fatou, we conclude that $$E[(Z-W)_-] = E[ \liminf_n (f_n(Z)-f_n(W))_-]\leq \liminf_nE[(f_n(Z)-f_n(W))_-] < +\infty,$$ proving the lemma (the argument for $E[(X-Y)_+]<\infty$ is essentially identical by symmetry). $\Box$

Now we prove the original claim. From now on, we can assume that $E[|X-Y|]$ and $E[|Z-W|]$ are both finite (this must be true by the lemma, unless both $E[X-Y]$ and $E[Z-W]$ are infinite and of the same sign). By the $1$-Lipchitz property stated above, we have that $|f_n(X)-f_n(Y)| \leq |X-Y|$ and similarly for $Z,W$. Then by dominated convergence and linearity of expectation, it holds that \begin{align}E[X-Y] &= \lim_n E[f_n(X)-f_n(Y)] = \lim_n E[f_n(X)]-E[f_n(Y)] \\ &= \lim_n E[f_n(Z)]-E[f_n(W)] = \lim_nE[f_n(Z) - f_n(W)] = E[Z-W], \end{align} proving the claim.

shalin
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Here is an improvement of @Shalop's wonderful argument.

Let $f_n(x)$ be the truncation as in @Shalop's answer. Then by the Monotone Convergence Theorem,

\begin{align*} \mathbb{E}_{D_i}[(X - Y)_+] &= \mathbb{E}_{D_i}\left[\mathbf{1}_{\{X > Y\}} \int_{Y}^{X} \mathrm{d}t \right] \\ &= \lim_{n\to\infty} \mathbb{E}_{D_i}\left[\mathbf{1}_{\{X > Y\}} \int_{Y}^{X} \mathbf{1}_{[-n,n]}(t) \, \mathrm{d}t \right] \\ &= \lim_{n\to\infty} \mathbb{E}_{D_i}[ (f_n(X) - f_n(Y))_{+} ] \end{align*}

and likewise for the negative parts. Since $e_i$ exists, either $\mathbb{E}_{D_i}[(X - Y)_+]$ or $\mathbb{E}_{D_i}[(X - Y)_-]$ is finite, and so, we get

\begin{align*} e_i &= \mathbb{E}_{D_i}[(X - Y)_+ - (X - Y)_-] \\ &= \lim_{n\to\infty} \mathbb{E}_{D_i}[ (f_n(X) - f_n(Y))_{+} - (f_n(X) - f_n(Y))_{-} ] \\ &= \lim_{n\to\infty} \mathbb{E}_{D_i}[ f_n(X) - f_n(Y) ]. \end{align*}

The last line depends only on the marginal distributions of $D_i$, and so, the desired conclusion follows.

If we do not assume the existence of $e_i$'s, we can come up with an easy counter-example.

Let $Z, W$ be i.i.d. $\operatorname{Cauchy}(0,1)$ variables, and let $X = Y = Z$. Then all of $X, Y, Z, W$ have the same distribution, but $\mathbb{E}[X - Y] = 0$ while $\mathbb{E}[Z - W]$ is not defined.

Sangchul Lee
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