Let $X$ and $Y$ be two independent identically distributed random variables with finite expectation $\Bbb{E}(X) = \Bbb{E}(Y) < \infty$. Prove that

$$\Bbb{E}(|X-Y|) \le \Bbb{E}(|X+Y|)$$

I think that this inequality may follow somehow from Jensen's inequality, but I failed to use it here. Or maybe it is worth considering an expression $|x+y|-|x-y|$ and making use of some of its properties?

I am interested to see a proof of this fact or some favorable ideas that may help here. Any suggestions would be greatly appreciated.

  • $\begingroup$ Note. For $X$ and $Y$ with symmetric distributions, it results in equality. $\endgroup$ – Boby Apr 25 '17 at 13:05
  • $\begingroup$ Maybe the fact that the expected of $X-Y $ is zerocould be useful. $\endgroup$ – AnyAD Apr 25 '17 at 13:06
  • $\begingroup$ @Boby Yes, it does. Does this fact help to solve the problem somehow? $\endgroup$ – Ramil Apr 25 '17 at 13:07
  • $\begingroup$ @Ramil It only points to a fact that if you want to find meaningful examples, you have to look at asymmetric distributions. $\endgroup$ – Boby Apr 25 '17 at 13:09

Taking integration by parts to the Dirichlet integral, it is easy to check that

$$ \int_{-\infty}^{\infty} \frac{1-\cos(at)}{t^2} \, dt = \pi|a|. \tag{1}$$

Taking advantage of the fact that the integrand of $\text{(1)}$ is non-negative, by the Tonelli's theorem, for any real-valued random variable $Z$ we have

$$ \pi \Bbb{E}[|Z|] = \Bbb{E}\left[ \int_{-\infty}^{\infty} \frac{1-\cos(Zt)}{t^2} \, dt \right] = \int_{-\infty}^{\infty} \frac{1-\Bbb{E}[\cos(Zt)]}{t^2} \, dt. $$


\begin{align*} \pi \Bbb{E}[|X+Y| - |X-Y|] &= \int_{-\infty}^{\infty} \frac{\Bbb{E}[\cos((X-Y)t)-\cos((X+Y)t)]}{t^2} \, dt \\ &= \int_{-\infty}^{\infty} \frac{\Bbb{E}[2\sin(Xt)\sin(Yt)]}{t^2} \, dt \\ &= \int_{-\infty}^{\infty} \frac{2\Bbb{E}[\sin(Xt)]^2}{t^2} \, dt \\ &\geq 0. \end{align*}

Moreover, notice that the equality holds if and only if $\Bbb{E}[\sin(Xt)] = 0$ for all $t$. This means that the c.f. $\varphi_X(t) = \Bbb{E}[e^{itX}]$ is real-valued, which is equivalent to the symmetry condition: $X \stackrel{d}{=} -X$.

  • $\begingroup$ Wow! You did it again! Could you please reveal how on earth did you come up with such a solution? It doesn't seem to be quite natural. What is the intuition behind such solutions? $\endgroup$ – Ramil Apr 25 '17 at 13:26
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    $\begingroup$ Can this argument be extended to $E[|X+Y|^p] -E[|X-Y|^p] $ ? $\endgroup$ – Boby Apr 25 '17 at 13:31
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    $\begingroup$ @Boby, The argument readily extends to $p \in (0, 2)$ from the identity $$\int_{0}^{\infty} \frac{1-\cos(at)}{t^{1+p}} \, dt = \frac{\pi}{2 \Gamma(1+p)\sin(\pi p/2)} |a|^p. $$ The case $p = 0, 2$ are just algebra. For $p > 2$, I am not sure even the inequality remains true. $\endgroup$ – Sangchul Lee Apr 25 '17 at 13:39
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    $\begingroup$ Great. Thank you. It is interesting that it holds for $p<1$. Since, $|a|^p$ for $p<1$ no longer induces a norm. $\endgroup$ – Boby Apr 25 '17 at 13:46
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    $\begingroup$ @Boby It is not true in some cases. Say, for $p=4$ we have $(X+Y)^4 - (X-Y)^4 = 12X^3Y + 12Y^3X$, so $\Bbb{E}((X+Y)^4 - (X-Y)^4) = 24\Bbb{E}(X^3)\Bbb{E}(Y)$ and, for example, for $X, Y$ that are always negative the expectation will be negative. $\endgroup$ – Ramil Apr 25 '17 at 13:50

Here's another argument. It doesn't seem to generalize to $p$ norms, but perhaps it is instructive anyway.

For independent variables $X,Y$ with finite expectation we can write

\begin{align*} \mathbb E|X-Y| &=\int_{-\infty}^\infty \mathbb P[X\leq t< Y]+\mathbb P[Y\leq t<X] dt\\ &=\int_{-\infty}^\infty F_X(t)(1-F_Y(t))+F_Y(t)(1-F_X(t)) dt\tag{1} \end{align*}

where $F_Z$ denotes the cumulative distribution function $F_Z(t)=\mathbb P[Z\leq t]$ of a random variable $Z.$

In the case at hand, $X$ and $Y$ are i.i.d., so $F_{-Y}(t)=1-F_X(-t).$ Using $-Y$ instead of $Y$ in (1) gives

$$\mathbb E|X+Y|=\int_{-\infty}^\infty F_X(t)F_X(-t)+(1-F_X(-t))(1-F_X(t)) dt\tag{2}$$

We can get a comparable integrand for $\mathbb E|X-Y|$ by substituting $t$ for $-t$ in the final term in (1), and using $F_Y=F_X$: $$ \mathbb E|X-Y|=\int_{-\infty}^\infty F_X(t)(1-F_X(t))+F_X(-t)(1-F_X(-t)) dt\tag{3} $$

Writing $a=F_X(t)$ and $b=F_X(-t),$ clearly $((1-a)-b)((1-b)-a)=(1-a-b)^2\geq 0,$ so $a(1-a)+b(1-b)\leq ab+(1-a)(1-b).$ Integrating over $t$ and applying (3) and (2) gives $\mathbb E|X-Y|\leq \mathbb E|X+Y|.$ The equality case is when $F_X(-t)+F_X(t)=1$ a.e., which is when $X$ is symmetric about zero.


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