Prove that $\Bbb{E}(|X-Y|) \le \Bbb{E}(|X+Y|)$ for i.i.d $X$ and $Y$

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.

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

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

Therefore

\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: $X \stackrel{d}{=} -X$.

• 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? – Ramil Apr 25 '17 at 13:26
• Can this argument be extended to $E[|X+Y|^p] -E[|X-Y|^p]$ ? – Boby Apr 25 '17 at 13:31
• @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. – Sangchul Lee Apr 25 '17 at 13:39
• Great. Thank you. It is interesting that it holds for $p<1$. Since, $|a|^p$ for $p<1$ no longer induces a norm. – Boby Apr 25 '17 at 13:46
• @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. – 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

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.