Validity of integral formula for $E[|X + Y|] - E[|X - Y|]$

Question

$\newcommand{\real}{\mathbb{R}}$ I read this identity from the monograph Probability Inequalities by Zhengyan Lin and Zhidong Bai. The author proposed to use this identity to prove the inequality $E[|X - Y|] \leq E[|X + Y|]$ when $X$ and $Y$ are i.i.d. I quote:

An alternative proof is to use the formula \begin{align} & E[|X + Y|] - E[|X - Y|] \\ =& \int_{-\infty}^\infty(1 - F(u) - F(-u))(1 - G(u) - G(-u))du, \tag{1} \end{align} where $F$ and $G$ are the distribution functions of $X$ and $Y$ respectively.

The author didn't give a proof to $(1)$ (assume $X$ and $Y$ are independent). In the original text, the lower limit of the integral is "$0$", which has been corrected to "$-\infty$" in $(1)$.

My question: does $(1)$ hold for any distribution functions $F$ and $G$ (such that $E[|X|] < \infty, E[|Y|] < \infty$)? The author seems to imply this is the case. As shown below, $(1)$ is indeed true if $F$ and $G$ are absolutely continuous with densities $f$ and $g$, but I have difficulty to generalize it to any distribution functions. Interestingly, $E[|X - Y|] \leq E[|X + Y|]$ holds for general i.i.d. $X$ and $Y$, as proved elegantly in this answer.

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My attempt: When $F$ and $G$ have densities $f$ and $g$ respectively, direct simplification shows \begin{align} & E[|X + Y|] - E[|X - Y|] \\ =& 2E[X(1 - G(X) - G(-X))] + 2E[Y(1 - F(Y) - F(-Y))] \\ =& 2\int_\real x(1 - G(x) - G(-x))f(x)dx + 2\int_\real x(1 - F(x) - F(-x))g(x)dx. \end{align} By the change of variable theorem: \begin{align} I = \int_\real x(1 - G(x) - G(-x))f(x)dx = \int_\real (-t)(1 - G(-t) - G(t))f(-t)dt. \end{align} Hence \begin{align} & 2I = \int_\real x(1 - G(x) - G(-x))(f(x) - f(-x))dx \\ =& -\int_\real x(1 - G(x) - G(-x))d(1 - F(-x) - F(x)). \tag{2} \end{align} Similarly, if denote $E[Y(1 - F(Y) - F(-Y)]$ by $J$, then \begin{align} 2J = -\int_\real x(1 - F(x) - G(-x))d(1 - G(-x) - G(x)). \tag{3} \end{align}

Integrating by parts, $-2I$ becomes \begin{align} & x(1 - G(x) - G(-x))(1 - F(x) - F(-x))|_{-\infty}^\infty \\ -& \int_\real (1 - F(x) - F(-x))(1 - G(x) - G(-x))dx \\ -& \int_\real x(1 - F(x) - F(-x))d(1 - G(x) - G(-x)) \\ =& -\int_\real (1 - F(x) - F(-x))(1 - G(x) - G(-x))dx + 2J, \end{align} hence \begin{align} & E[|X + Y|] - E[|X - Y|] \\ =& 2(I + J) = \int_\real (1 - F(x) - F(-x))(1 - G(x) - G(-x))dx. \end{align} That $\lim_{x \to \pm\infty}x(1 - F(x) - F(-x))(1 - G(x) - G(-x)) = 0$ in the above derivation can be derived by integrability of $X$ or $Y$. For example, for sufficiently positive large $x$: \begin{align} 0 \leq x(1 - F(x) - F(-x))(1 - G(x) - G(-x)) \leq x(1 - F(x)) \to 0 \end{align} as $x \to \infty$ ($x(1 - F(x)) \to 0$ as $x \to +\infty$ is well known if $E[|X|] < \infty$).

I think the difficulty of generalizing the above argument to general $F$ and $G$ lies in deriving $(2)$ and $(3)$, in particular when $F$ and $G$ have discontinuities. Therefore, I doubt if $(1)$ can still hold for the general case (perhaps some extra discrete terms need to be appended).

Answer 1

$\newcommand{\real}{\mathbb{R}}$ Without assuming the differentiability of $F$ and $G$, $(1)$ can be proved by using an integration representation of $|X - Y|$ and $|X + Y|$ as follows (inspired by this answer).

Note that for real numbers $x$ and $y$, we have \begin{align} & |x - y| = \int_\real \left[I_{[x, \infty)}(u)I_{(-\infty, y)}(u) + I_{[y, \infty)}(u)I_{(-\infty, x)}(u)\right]du, \\ & |x + y| = \int_\real \left[I_{(-y, \infty)}(u)I_{(-\infty, x)}(u) + I_{[x, \infty)}(u)I_{(-\infty, -y]}(u)\right]du, \end{align} where $I_A(x)$ stands for the indicator function: $I_A(x) = 1$ if $x \in A$ and $0$ otherwise. Therefore, \begin{align} & |X - Y| = \int_\real\left[I_{[X, \infty)}(u)I_{(-\infty, Y)}(u) + I_{[Y, \infty)}(u)I_{(-\infty, X)}(u)\right]du, \tag{1} \\ & |X + Y| = \int_\real\left[I_{(-Y, \infty)}(u)I_{(-\infty, X)}(u) + I_{[X, \infty)}(u)I_{(-\infty, -Y]}(u)\right]du. \tag{2} \end{align} It thus follows by the independence of $X$ and $Y$, the change of variable theorem, and the Tonelli's theorem that \begin{align} & E[|X - Y|] \\ =& \int_\Omega\left\{\int_\real \left[I_{[X, \infty)}(u)I_{(-\infty, Y)}(u) + I_{[Y, \infty)}(u)I_{(-\infty, X)}(u)\right]du\right\} dP \\ =& \int_{\real^2}\left\{\int_\real \left[I_{[x, \infty)}(u)I_{(-\infty, y)}(u) + I_{[y, \infty)}(u)I_{(-\infty, x)}(u)\right]du\right\} dF(x)dG(y) \\ =& \int_{\real}\left\{\int_{\real^2} \left[I_{(-\infty, u]}(x)I_{(u, \infty)}(y) + I_{(-\infty, u]}(y)I_{(u, \infty)}(x)\right]dF(x)dG(y)\right\} du \\ =& \int_{\real}\left\{\int_\real I_{(-\infty, u]}(x)dF(x)\int_\real I_{(u, \infty)}(y)dG(y) + \int_\real I_{(-\infty, u]}(y)dG(y)\int_\real I_{(u, \infty)}(x)dF(x)\right\} du \\ =& \int_{\real}\left\{F(u)(1 - G(u)) + G(u)(1 - F(u)\right\} du. \tag{3} \end{align}

By $(2)$, similar calculation yields \begin{align} & E[|X + Y|] \\ =& \int_\Omega\left\{\int_\real\left[I_{(-Y, \infty)}(u)I_{(-\infty, X)}(u) + I_{[X, \infty)}(u)I_{(-\infty, -Y]}(u)\right]du\right\} dP \\ =& \int_{\real^2}\left\{\int_\real \left[I_{(-y, \infty)}(u)I_{(-\infty, x)}(u) + I_{[x, \infty)}(u)I_{(-\infty, -y]}(u)\right]du\right\} dF(x)dG(y) \\ =& \int_{\real}\left\{\int_{\real^2}\left[I_{(-u, \infty)}(y)I_{(u, \infty)}(x) + I_{(-\infty, u]}(x)I_{(-\infty, -u]}(y)\right]dF(x)dG(y)\right\} du \\ =& \int_{\real}\left\{\int_\real I_{(-u, \infty)}(y)dG(y)\int_\real I_{(u, \infty)}(x)dF(x) + \int_\real I_{(-\infty, u]}(x)dF(x)\int_\real I_{(-\infty, -u]}(y)dG(y)\right\} du \\ =& \int_{\real}\left\{(1 - F(u))(1 - G(-u)) + F(u)G(-u)\right\}du. \tag{4} \end{align}

Subtracting $(3)$ from $(4)$ gives the desired formula: \begin{align} E[|X + Y|] - E[|X - Y|] = \int_{-\infty}^\infty(1 - F(u) - F(-u))(1 - G(u) - G(-u))du. \end{align}

Therefore, the referenced formula is indeed sufficiently general.

Answer 2

Here is a more concise derivation, a modification of this answer: First verify the identity $$ |x+y|-|x-y|=2[\min(x^+,y^+)+\min(x^-,y^-)-\min(x^+,y^-)-\min(x^-,y^+)].$$ Next, use this result to argue that for nonnegative and independent $U$ and $V$: $$E\min(U,V)=\int_0^\infty P(\min(U,V)>t)\,dt=\int_0^\infty P(U>t)P(V>t)\,dt. $$ Apply this last to find, when $X$ and $Y$ are independent, $$ \begin{aligned} E\min(X^+,Y^+)&=\int_0^\infty P(X^+>t)P(Y^+>t)\,dt=\int_0^\infty P(X>t)P(Y>t)\,dt\\ E\min(X^-,Y^-)&=\int_0^\infty P(X^->t)P(Y^->t)\,dt=\int_0^\infty P(-X>t)P(-Y>t)\,dt\\ E\min(X^+,Y^-)&=\int_0^\infty P(X^+>t)P(Y^->t)\,dt=\int_0^\infty P(X>t)P(-Y>t)\,dt\\ E\min(X^-,Y^+)&=\int_0^\infty P(X^->t)P(Y^+>t)\,dt=\int_0^\infty P(-X>t)P(Y>t)\,dt \end{aligned}$$ Put everything together: $$E|X+Y|-E|X-Y|=2\int_0^\infty[P(X>t)-P(-X>t)][P(Y>t)-P(-Y>t)]\,dt.$$ This is the same as the desired identity (1), since $P(X>t)=1-F(t)$ and $P(-X>t)=F(-t)$ for almost every $t$ (similarly for $Y$, $G$), and the integrand in (1) is an even function of $u$.