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Let $X_i$ and $Y_i$, $i:1,\ldots,n$, be continuous i.i.d. random variables, uniformly distributed over $(0,1)$. Say we sample from these $RV$s and retain only values complying with $|X_i-Y_i|>\delta$, where $\delta$ is some given small positive constant.

I would like to prove that, EDIT: $$P\left(\frac{1}{n} \sum _{i=1}^n (Y_i-X_i)(1-2X_i)>\epsilon\right)\to1$$ when $n\to\infty$.

Where $\epsilon$ is a small positive constant which depends on $\delta$, but not on $n$.

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the pairs $(X_i,Y_i)$ are still i.i.d. so the limit is given by the SLLN, and you only need calculate the expected value given etc. –  mike Jul 11 '12 at 22:19
    
Could you elaborate...? And I also need the maximum value of epsilon, given delta. –  Omri Jul 12 '12 at 4:57
    
OK, I get it now, Thanks. –  Omri Jul 12 '12 at 20:20
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3 Answers

up vote 2 down vote accepted

It is most probable that for every $n$, $$ \mathrm P\left(\sum_{i=1}^n(X_i-Y_i)(1-2X_i)\gt0\right)\lt\frac12, $$ and, for every $\delta$, $$ \mathrm P\left(\sum_{i=1}^n(X_i-Y_i)(1-2X_i)\gt0\ \Bigg\vert\ \forall i,|X_i-Y_i|\gt\delta\right)\lt\frac12, $$ hence no $\epsilon\gt0$ will do.


Regarding the revised version, let $Z_i=(Y_i-X_i)(1-2X_i)$. Then $(Z_i)_i$ is i.i.d. with mean $\mathrm E(Z_i)=\frac16$ hence, for every $\epsilon\lt\frac16$, the (weak) law of large numbers shows that $$ \mathrm P\left(\frac1n\sum_{i=1}^nZ_i\gt\epsilon\right)\to1. $$ Likewise, let $(X,Y,Z)$ distributed like $(X_1,Y_1,Z_1)$, $A_\delta=[|X-Y|\gt\delta]$, $U^\delta$ any random variable distributed like $Z$ conditional on $A_\delta$, and $u_\delta=\mathrm E(U^\delta)=\mathrm E(Z\mid A_\delta)$. Then, for every $\epsilon\lt u_\delta$, the (weak) law of large numbers applied to an i.i.d. sequence $(U^\delta_i)_i$ distributed like $U^\delta$ shows that $$ \mathrm P\left(\frac1n\sum_{i=1}^nZ_i\gt\epsilon\ \Bigg\vert\ \forall i,|X_i-Y_i|\gt\delta\right)=\mathrm P\left(\frac1n\sum_{i=1}^nU^\delta_i\gt\epsilon\right)\to1. $$ To complete the proof, it remains to estimate $u_\delta$. First note that, by invariance of $A_\delta$ with respect to the symmetry $(X,Y)\to(Y,X)$, $\mathrm E(Y-X\,;\,A_\delta)=0$. Hence, $u_\delta=2\mathrm E((X-Y)X\mid A_\delta)$. Since $\mathrm E((X-Y)X\mid A_\delta)=\mathrm E((Y-X)Y\mid A_\delta)$, $u_\delta=\mathrm E((X-Y)^2\mid A_\delta)$. This proves the claim that $u_\delta\gt0$.

One can go further since the density of $|X-Y|$ is $2(1-x)\,[0\leqslant x\leqslant 1]$ and, consequently, the density of $|X-Y|$ conditionally on $A_\delta$ is $f_\delta(x)=2(1-\delta)^{-2}(1-x)\,[\delta\leqslant x\leqslant 1]$.

Thus, $u_\delta=\int\limits_\delta^1x^2f_\delta(x)\,\mathrm dx=\frac16(1+2\delta+3\delta^2)$. In particular, any $\epsilon\lt\frac16$ goes, for every $0\leqslant\delta\lt1$, and $\epsilon=\frac16$ goes, for every $0\lt\delta\lt1$.

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Thanks, but please see my answer below... –  Omri Jul 14 '12 at 21:50
    
You are welcome. –  Did Jul 14 '12 at 22:21
    
Thanks much (again), @did. One (last) comment: If we remove the condition $|X_i - Yi| > δ$ altogether (which of course is not the same as having $δ=0$), does my version of the answer (without the $I_A$ part) also suffice? And then we can simply take $\epsilon$ to be any value < $2Var(X_i)$? –  Omri Jul 15 '12 at 8:33
    
Removing the condition $|X_i-Y_i|\gt\delta$ is the same thing as having $\delta=0$. // Yes, one can choose any value $\epsilon\lt E((Y-X)(1-2X))=2E(X^2)-E(X)=\frac16$. –  Did Jul 15 '12 at 9:20
    
Of course, I meant $2(E(X^2)-E(X)^2)=2Var(X)$ –  Omri Jul 15 '12 at 11:58
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Edit: here is a new version, with lighter computations and, hopefully, no error. My thanks to Did for pointing out some problems in my former proof. Of course, my answer is now essentially the same as Did...

Let $(X,Y)$ be a couple of random variables uniformly distributed in $[0,1]^2$. Let $\delta \in [0,1)$. The event $\{|X-Y|>\delta\}$ has probability:

$$\int_{[0,1]^2} 1_{|x-y|> \delta} \ dx \ dy = 2 \int_{[0,1]} \int_{[0,1]} 1_{x > \delta + y} dx \ dy = (1-\delta)^2.$$

Let $(\tilde{X}, \tilde{Y})$ be the random variables $(X,Y)$ conditioned by the event $\{|X-Y|>\delta\}$. It has the following density with respect to the Lebesgue measure on $[0,1]^2$:

$$\frac{1_{x > y + \delta} + 1_{x < y - \delta}}{(1-\delta)^2}.$$

Hence:

$$\mathbb{E} ((\tilde{X}-\tilde{Y})(1-2\tilde{X})) = \frac{1}{(1-\delta)^2} \int_{[0,1]^2} (x-y)(1-2x) (1_{x > y + \delta} + 1_{x < y - \delta}) \ dx \ dy$$

$$\cdots = \frac{1}{(1-\delta)^2} \int_\delta^1 \int_0^{x-\delta} (x-y)(1-2x) \ dy \ dx + \frac{1}{(1-\delta)^2} \int_0^{1-\delta} \int_{x+\delta}^1 (x-y)(1-2x) \ dy \ dx $$

The change of variables $(u,v) = (1-x,1-y)$ shows that these two integrals are the same, so we only need to compute one of them. Thanks to Wolfram,

$$\int_\delta^1 \int_0^{x-\delta} (x-y)(1-2x) \ dy \ dx = -\frac{1}{2} \int_\delta^1 (2x-1) (x^2-\delta^2) \ dx = -\frac{(1-\delta)^2 (1+2\delta+3\delta^2)}{12}.$$

Thus:

$$\mathbb{E} ((\tilde{X}-\tilde{Y})(1-2\tilde{X})) = -\frac{1+2\delta+3\delta^2}{6}.$$

This formula gives the good limits when $\delta$ goes to $0$ or $1$. It is also always negative (even for $\delta = 0$). By the law of large numbers, you should expect the sum to be negative for large $n$ almost surely.

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Thanks, but please see my answer below... –  Omri Jul 14 '12 at 21:50
    
Hi, I found the problem - it was a typo, see the edited original question, $(Y_i-X_i)$ instead of $(X_i-Y_i)$... in this case do we just remove the minus sign from the fraction in the RHS of your answer's last line? –  Omri Jul 15 '12 at 6:48
    
@D.Thomine: Unless I am mistaken, the limit of $E_\delta((X-Y)(1-2X))$ when $\delta\to0$ should be $-\frac16$ but your formula yields $-\frac13$. (Also, since the result might be analytic with respect to $\delta$, I mention that when $\delta\to1$, the limit should be $-1$ and that the limit of your formula is $-\frac32$. Finally, I do not quite understand why the denominators $2-\delta$ survive until the end of your computations.) –  Did Jul 15 '12 at 7:32
    
@did: This computation is too complicated, it's not surprising that there are errors. I'm re-writing it in a simpler way. –  D. Thomine Jul 15 '12 at 11:36
    
No worry. $ $ $ $ –  Did Jul 15 '12 at 11:38
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[EDIT: the problem with this answer is that $I_A$ and $(Y_i-X_i)(1-2X_i)$ are dependent.]

Let $Z_i$ denote the summand multiplied by the indicator $I_A$ for $|X_i - Yi| > δ$, $Z_i=(Y_i-X_i)(1-2X_i)I_A$. We need to prove that $P\left(\bar{Z_n}>\epsilon\right)\to1$. Since the $Z_i$ are $i.i.d$ and bounded, from the weak law of large numbers $\bar{Z_n}$ converges in probability to $E(Z_i)$. It is therefore sufficient to show that $E(Z_i) > 0$. We proceed to evaluate $E(Z_i)$ analytically, $$E(Z_i)=E(I_A)\left(E(Y_i)-2E(Y_iX_i)-E(X_i)+2E(X_i^2)\right)=2E(I_A)\left(E(X_i^2)-E(X_iY_i)\right)=2E(I_A)Var(X_i)>0$$ since $0<E(I_A)<1$ by definition, $X_i$ and $Y_i$ are independent and with same means, and $Var(X_i)>0$.

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Since $A$ depends on $(X_i,Y_i)$, the factorization of $E(I_A)$ in the first displayed equality used to compute $E(Z_i)$ is wrong. In fact, $E(Z_i)=-2E((X_i-Y_i)X_iI_A)=-E((X_i-Y_i)^2I_A)\lt0$, where the second identity stems from a symmetry argument. –  Did Jul 14 '12 at 22:21
    
OK, I suspected $I_A$ was dependent... However, why do you have a minus sign in the RHS of $E(Z_i)=-2E((X_i-Y_i)X_iI_A)$? –  Omri Jul 14 '12 at 23:14
    
I think I will let you suspect that as well. –  Did Jul 14 '12 at 23:25
    
OK, I found the error - see my comment to D. Thomine. –  Omri Jul 15 '12 at 6:50
    
@did, could you just elaborate on your "symmetry argument" in your first comment here? –  Omri Jul 15 '12 at 6:54
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