Conditional expectation of a geometric variable

Question

$X, Y$ 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

Answer 1

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.

Answer 2

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.