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Sagot :
Answer:
$\begin{align*}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$
Explanation:
\eqalign{
P(Y-X=m\mid Y\gt X)
&=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr
&=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr
&=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr
}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$
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