Originally Posted by

**matheagle** First of all this is discrete.

You need to proceed in this manner, and drop the X,Y they are annoying.

$\displaystyle P(V+W=b|U+V=a)={P(V+W=b,U+V=a)\over P(U+V=a)}$

The denominator is easy, it's a Poisson with mean $\displaystyle \lambda_U+\lambda_V$

The numerator isn't that easy.

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OR maybe we just attack the expected value.

$\displaystyle E(V+W|U+V) =E(V|U+V)+E(W|U+V) =E(V|U+V)+E(W)$

Since W is independent of U,V, so the second one is easy, but the first is...

So derive the distribution of V given U+V. I think that's been done a lot....

$\displaystyle P(V=b|U+V=a)={P(V=b,U+V=a)\over P(U+V=a)}$

$\displaystyle ={P(V=b)P(U=a-b)\over P(U+V=a)}$