Find conditional distribution of a sum of two random variables For random variables $X, Y, Z, W$ the following is true: $f_{Z,W|X = x, Y = y}(z,w) = \frac{1}{\Gamma(x)\Gamma(y)}z^{x-1}w^{y-1}e^{-z-w}$ for $x,y,z,w > 0$. We have to find $f_{Z + W|X+Y=u}(v)$.
I calculated $f_{Z + W|X=x,Y=y}(v)$ with the following formula: $f_{Z + W|X=x,Y=y}(v) = \int_{-\infty}^{\infty}f_{Z,W|X = x, Y = y}(s,v-s)\,ds$ and got $Z+W|X=x,Y=y \sim \Gamma(x + y, 1)$. Then I calculated $f_{Z + W|X+Y=u}(v)$ using the following formula: $f_{Z + W|X+Y=u}(v) = \int_{-\infty}^{\infty}f_{Z + W|X = s, Y = u-s}(v)\,ds$ and got $f_{Z + W|X+Y=u}(v) = \frac{1}{\Gamma(u)}uv^{u-1}e^{-v}$.
I would just like to know whether both formulas that I used are correct and if they are not correct, how should I approach this problem? Because the integral over the real line of my result isn't equal to $1$, I assume that at least one of the formulas is incorrect.
 A: You have found that conditionally on $X=x,Y=y$, $Z+W \sim Γ(x+y,1)$, so intuitively we'd expect $Z+W \sim Γ(u,1)$ conditionally on $X+Y = u$.
The equality $f_{Z + W|X+Y=u}(v) = \int_{-\infty}^{\infty}f_{Z + W|X = s, Y = u-s}(v)\,ds$ does not hold: the lhs ought to be a density, while the rhs is not integrable. The mistake is analogous to writing $ \mathbb P[A|B] = \sum_n \mathbb P[A|B_n] $, where the events $(B_n)$ form a partition of $B$. Here, the correct way would be $\mathbb P[A|B] = \sum_n \mathbb P[A|B_n]\mathbb P[B_n|B]  $. When working with densities, it is tricky to translate the idea from the 'correction' above, since ${(X,Y)}$, given $X+Y=u$, does not have a density at all (w.r.t. the Lebesgue-measure on $\mathbb R^2$, since all the probability is concentrated on the line $x+y=u$).
But even without a density, we can work with a regular conditional distribution  $\mathbb P_{(X,Y)|X+Y = u}$ of $(X,Y)$ given $X+Y = u$ to obtain
\begin{align}
\mathbb P[Z+W ∈ B | X+Y = u] &= ∫_{ℝ^2} \mathbb P[Z+W ∈ B | X = x, Y = y] \; d\mathbb P_{(X,Y)|X+Y = u}(x,y) \\
&= ∫_{ℝ^2} \mathbb P_{Γ(x+y,1)}(B) \; d\mathbb P_{(X,Y)|X+Y = u}(x,y) \\
&= \mathbb P_{Γ(u,1)}(B) ∫_{ℝ^2}  d\mathbb P_{(X,Y)|X+Y = u}(x,y) = \mathbb P_{Γ(u,1)}(B),
\end{align}
confirming the intuition.

Regular conditional distributions have a flavor of 'abstract nonsense'. If you are happy with being less formal, you could argue that $$f_{Z + W|X+Y=u}(v) = \int_{-\infty}^{\infty}f_{Z + W|X = s, Y = u-s}(v) \frac{f_{(X,Y)}(s,u-s)}{∫_ℝ f_{(X,Y)}(t,u-t) dt}\,ds,$$ $f_{(X,Y)}$ denoting the (assuming it exists) joint density of $(X,Y)$. Since $f_{Z + W|X = s, Y = u-s}$ is just the density of $Γ(u,1)$, it can be taken out of the integral, and this shows the claim.
