Conditional expectation of a geometric variable 
$X, Y$ 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 
 A: 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. 
A: 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.
