This is a proof-verification request; I’m also recording this proof for my own later reference. Any feedback is appreciated.

Claim: Let $X$ and $Y$ be independent, real-valued random variables on some probability space $(\Omega,\Sigma,\mathbb P)$. If $X-Y$ has a finite mean, then $X$ and $Y$ have finite means as well.

Proof: Let $\mu$ and $\nu$ denote the probability measures induced by $X$ and $Y$, respectively, on $(\mathbb R,\mathscr B_{\mathbb R})$, where $\mathscr B$ stands for the Borel $\sigma$-algebra. Since $X$ and $Y$ are independent, the joint distribution of $(X,Y)$ induced on $(\mathbb R^2,\mathscr B_{\mathbb R^2})$ is given by the product measure $\mu\otimes\nu$.

Then, one has that \begin{align*} \infty>&\,\mathbb E|X-Y|=\int_{\omega\in\Omega}|X(\omega)-Y(\omega)|\,\mathrm d\mathbb P(\omega)=\int_{(x,y)\in\mathbb R^2}|x-y|\,\mathrm d(\mu\otimes\nu)(x,y)\\ =&\,\int_{y\in\mathbb R}\int_{x\in\mathbb R}|x-y|\,\mathrm d\mu(x)\,\mathrm d\nu(y), \end{align*} where the last equality stems from Tonelli’s theorem. This yields that the function $$y\mapsto\int_{x\in\mathbb R}|x-y|\,\mathrm d\mu(x)$$ is finite-valued almost everywhere (measured with respect to $\nu$). In particular, there is some $y_0\in\mathbb R$ such that $$\int_{x\in\mathbb R}|x-y_0|\,\mathrm d\mu(x)<\infty,$$ which furthermore yields $$\mathbb E|X|=\int_{\omega\in\Omega}|X(\omega)|\,\mathrm d\mathbb P(\omega)=\int_{x\in\mathbb R}|x|\,\mathrm d\mu(x)\leq \int_{x\in\mathbb R}|x-y_0|\,\mathrm d\mu(x)+\underbrace{\int_{x\in\mathbb R}|y_0|\,\mathrm d\mu(x)}_{=|y_0|}<\infty.$$ The argument showing that $\mathbb E|Y|<\infty$ is completely analogous. $\blacksquare$

  • 3
    $\begingroup$ It is a very well written proof. $\endgroup$
    – Zhanxiong
    Commented May 19, 2016 at 4:59
  • $\begingroup$ Just for my education, can you explain what happens when $X,Y$ are from the Cauchy distribution? $X-Y$ has infinite mean as well? I understand you cannot write $$\mathbb{E}[X-Y] = \mathbb{E}[X] - \mathbb{E}[Y] = 0$$ since the very first step is undefined (so is the second one). But it would be interesting to understand what actually happens... $\endgroup$
    – gt6989b
    Commented May 19, 2016 at 5:03
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    $\begingroup$ @gt6989b For your example, $X - Y$ does not have finite mean. $\endgroup$
    – Zhanxiong
    Commented May 19, 2016 at 5:06
  • $\begingroup$ Are $\mu$ and $\nu$ the borel measures induced by the distributions of $X$ and $Y$ ? $\endgroup$
    – alpastor
    Commented May 19, 2016 at 5:18
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    $\begingroup$ @gt6989b Suppose that $X$ and $Y$ are independent and have Cauchy distributions. Then $-Y$ also has a Cauchy distribution (by symmetry) and is independent of $X$. Now, the sum of two independent Cauchy variables is also Cauchy, so that it follows that $X-Y=X+(-Y)$ is Cauchy and hence does not have a finite mean. $\endgroup$
    – triple_sec
    Commented May 19, 2016 at 5:53


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