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Let $X,Y$ be two i.i.d.

I am trying to bound the expectation of how afar from one another they can get?

That is, $E[|X-Y|]$. I know that:

$$ E[X-Y] = E[X]- E[Y] = 0$$

But what about $|X-Y|$?

One approach I had in mind is to consider a bernouli r.v. $Z $~$ Ber(p)$ where $p=P(X-Y>0)=P(X>Y)$. I assume as $X,Y$ are i.i.d. that $P(X>Y)=\frac{1}{2}$? But this is a symmetry-based argument, and I don't know how to exactly prove it.

Then:

$$ E[|X-Y|] = E[X-Y|Z=1]\cdot P(Z=1) - E[X-Y|Z=0]\cdot P(Z=0)$$

But I'm not sure if this is a good approach or how to compute $E[X-Y|Z=1]\cdot P(Z=1)$.

For example, we can assume $X,Y$~$U[a,b]$ or $X,Y$~$B(n,q)$, and may assume they are discrete, if it simplifies things.

Edit: Looking specifically for $X,Y$~$U[a,b]$ or $X,Y$~$B(n,q)$,

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    $\begingroup$ In general you can't say much more than whatever triangle inequality gives you $\endgroup$ Commented 21 hours ago
  • $\begingroup$ @BrunoAndrades And for specific distributions? $X,Y$~$U[a,b]$? Or $X,Y$~$B(n,q)$? $\endgroup$ Commented 21 hours ago
  • $\begingroup$ Probably the best thing in that case is to calculate explicitly the distribution of $X-Y$ and then calculate explicitly the expected value. It will be a not too ugly explicit expression $\endgroup$ Commented 20 hours ago

3 Answers 3

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The quantity $\mathbb{E}\left[ \vert X - Y \vert\right]$, where $X, Y \stackrel{\text{iid}}{\sim} F$, is called the Mean absolute difference (MAD).

Interestingly, if $X$ (and therefore $Y$) is $L_2$-integrable, i.e $\sigma^2 = \text{Var}(X) < \infty$, then according to this question, we have the bound $$ \mathbb{E} \vert X - Y \vert \le \dfrac{2}{\sqrt{3}}\sigma $$

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For a standard normal distribution (mean zero, standard deviation 1), the difference between two samples is also normal with mean zero, standard deviation $\sqrt2$. To get the distribution of the absolute value, you fold the negative half onto the positive half and then double it to renormalize.

You can then compute the expected value in the usual way, which I assume you know, and it leads to the result

$E(|x-y|)=\sqrt{4/\pi}.$

This agrees with the bound given by Thành Nguyen, as the underlying normal distribution used here has unit standard deviation.

Standard charts used in statistical process control, such as that given here, include the expected range between minimum and maximum of $n$ samples taken from a standard normal distribution; this is commonly labeled $d_2$ in SPC nomenclature. Then the 1.128 value quoted below would be $\sqrt{4/\pi}$ if it were represented exactly.

enter image description here

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We have $|z|=\frac{2}{\pi}\int_0^{\infty}(1-\cos tz)\frac{dt}{t^2}$ (derive in $z$ to check this). Denote by $\varphi_X$ the characteristic function of $X.$ We get $$ E(|X-Y|)=\frac{2}{\pi}\int_0^{\infty}(1-E(\Re (\varphi_X(t)\varphi_Y(-t)))\frac{dt}{t^2}\leq \infty.$$

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