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$X, Y$ are two independent random variables which are both Geometry($p$). We also define random variables $Z = |X-Y|, W = \min \{X,Y\}$. Calculate $E [W | Z = 1]$.

I tried to use $\min\{X,Y\}=(|X+Y|-|X-Y|)/2$. At first I thought $Z,W$ are independent, because $Z$ is the absolute value of the difference, so $Z$ doesn't tell me which one of $X$ or $Y$ is the minimum. Am I wrong?

the answer is (1-p)/(2-p)^2 and I don't understand how to get to this

Ella
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2 Answers2

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As @StubbornAtom mention in the comment, $\min(X,Y)$ and $X-Y$ are independent. Then $\min(X,Y)$ and $|X-Y|$ are independent as well. So $$ \mathbb E[\min(X,Y)\mid |X-Y|=1] = \mathbb E[\min(X,Y)]. $$ The distribution of $\min(X,Y)$ is geometric too: $\min(X,Y)\sim Geo(2p-p^2)=Geo(p(2-p))$.

Then $\mathbb E[\min(X,Y)]=\dfrac{1}{p(2-p)}$, if we have geometric distribution with pmf $\mathbb P(X=k)=p(1-p)^{k-1}$, $k=1,2,\ldots$.

If we have geometric distribution with pmf $\mathbb P(X=k)=p(1-p)^{k}$, $k=0,1,2,\ldots$, then $$\mathbb E[\min(X,Y)]=\dfrac{1}{p(2-p)}-1 = \frac{(1-p)^2}{p(2-p)}.$$

Note that both answers do not coincide with given answer.

NCh
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  • I did mention to OP that the minimum and absolute difference are independent, but then deleted the comment thinking it wasn't correct. Thanks for confirming. – StubbornAtom Jan 19 '20 at 17:04
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    Any borel functions of two independent r.v.'s are independent too. I did not seen your deleted comment and tried to guess its contents from the remaining one. In any case, the link in the answer leads precisely to your proof of independence. – NCh Jan 19 '20 at 17:10
  • Why min(x,y) and x-y are independent? If x-y is negative I can tell for sure that min(x,y)=x – Ella Jan 19 '20 at 17:20
  • @Ella Look at the link above. The independence is proved there. – NCh Jan 19 '20 at 17:32
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For $Z$, first note that $\{|X-Y|\geqslant 0\}$ has probability one, and $|X-Y|= 0$ if and only if $X=Y$. So first we compute \begin{align} \mathbb P(Z=0) &= \mathbb P(X=Y)\\ &= \sum_{n=1}^\infty \mathbb P(X=Y\mid X=n)\mathbb P(X=n)\\ &= \sum_{n=1}^\infty \mathbb P(Y=n)\mathbb P(X=n)\\ &= \sum_{n=1}^\infty \mathbb P(X=n)^2\\ &= \sum_{n=1}^\infty (1-p)^{2(n-1)}p^2\\ &= \frac p{2-p}. \end{align} For $n\geqslant 1$, $$\{|X-Y|=n\} = \{X-Y=n\}\cup\{X-Y=-n\}.$$ Hence \begin{align} \mathbb P(Z = n) &= \mathbb P(X-Y=n) + \mathbb P(X-Y=-n)\\ &= \sum_{k=1}^\infty\mathbb P(X-Y=n\mid Y=k)\mathbb P(Y=k) + \sum_{k=n+1}^\infty \mathbb P(X-Y=-n)\mathbb P(Y=k)\\ &= \sum_{k=1}^\infty \mathbb P(X=n+k)\mathbb P(Y=k) + \sum_{k=n+1}^\infty \mathbb P(X=k-n)\mathbb P(Y=k)\\ &= \sum_{k=1}^\infty (1-p)^{n+k-1}p(1-p)^{k-1}p + \sum_{k=n+1}^\infty \mathbb (1-p)^{k-n-1}p(1-p)^{k-1}p\\ &= \frac{p (1-p)^n}{2-p} + \frac{p (1-p)^n}{2-p}\\ &= \frac{2p (1-p)^n}{2-p}. \end{align} For $W$, note that $$ \{X\wedge Y=n\} = \{X=n,Y=n\}\cup \{X=n,Y>n\}\cup \{X>n,Y=n\}. $$ By symmetry, $\mathbb P(X=n,Y>n) = \mathbb P(X>n,Y=n)$. So we have \begin{align} \mathbb P(W=n) &= \mathbb P(X=n,Y=n) + 2\mathbb P(X=n,Y>n)\\ &= \mathbb P(X=n)^2 + 2\sum_{k=n+1}^\infty \mathbb P(X=n,Y=k)\\ &= (1-p)^{2(n-1)}p^2 + 2\sum_{k=n+1}^\infty (1-p)^{n-1}p(1-p)^{k-1}p\\ &= (1-p)^{2(n-1)}p^2 + 2p (1-p)^{2 n-1}\\ &= p(2-p) (1-p)^{2 (n-1)}. \end{align} That should help you in computing the desired conditional expectation.

Math1000
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