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Let $X$ and $Y$ be two zero-mean independent and identically distributed random variables. Is there a bound for the following ratio,

$$\frac{\mathbb{E}[|X+Y|]}{\mathbb{E}[|X|+|Y|]}=\frac{\mathbb{E}[|X+Y|]}{2\mathbb{E}[|X|]} ,$$

where $\mathbb{E}$ and $|.|$ are the expectation and absolute value operations, respectively?

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    $\begingroup$ I can give a universal upper bound! $\endgroup$
    – Anthony Quas
    Commented Jul 26, 2017 at 15:43
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    $\begingroup$ Thank you. Obviously, $1$ is an upper bound. However, I mean a bound based on higher order statistics of $X$ and $Y$. $\endgroup$
    – Math_Y
    Commented Jul 26, 2017 at 15:47
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    $\begingroup$ In view of the upvoted answer and comment, I wonder if th OP was asking for an upper bound, better than $1$, depending on some moments of $X$. $\endgroup$
    – Jean Duchon
    Commented Jul 27, 2017 at 9:31

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Yes, there is a bound for this ratio -- it is always between $\frac{1}{2}$ and $1$. Upper bound is obvious since $|X+Y| \le |X| + |Y|$ so let us prove the lower bound.

Let $X$ be positive with probabiliy $p$ and negative with probability $q = 1 - p$. Let $A$ be conditional expected value of $X$ when $X > 0$ and $B$ be conditional expected value of $X$ when $X < 0$. Since X is zero-mean $pA + qB = 0$ and $E[|X|] = pA - qB = 2pA$. Let $C = \frac{1}{2}E[|X|] = pA$. Even looking only at events where $XY \ge 0$ we see that $E[|X + Y|] \ge 2p^2A - 2q^2B = 2C(p + q) = 2C$ and so $\frac{E[|X + Y|]}{E[|X| + |Y|]} \ge \frac{1}{2}$.

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    $\begingroup$ Also, this bound is tight. Suppose $X$ and $Y$ are $-1, +1$ with probability $1/2$. Then $|X+Y| = 2$ with probability $1/2$ and $0$ otherwise, while $|X| = |Y| = 1$. $\endgroup$
    – David Harris
    Commented Jul 26, 2017 at 16:02
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This is Jensen's inequality. Applied to the distribution of Y one gets $E|X+Y| \ge |X + E(Y)|$

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