Q: how to think of/compute the marginal distribution of the sum of 2 random variables? I am tasked with finding $P(X\mid X+Y)$. I am having difficulty understanding the process of finding the marginal distribution for the denominator. If it were a simple $P(X\mid Y)$, I would know how to proceed, namely by finding $f_Y(y)$ whatever it’s distribution is. Is it the case that since we are asking about $P(X\mid X+Y)$, can we assume that the marginal distribution for the denominator is still in terms of $Y$ since $X$ is in the numerator? 
 A: As noted in the comments above, it is assumed that we have separate and independent probability density functions (PDFs) $f_X(x)$ and $f_Y(y)$ on $X$ and $Y$, respectively. I'm going to post a more general recipe for solving problems of this type that can be easily tailored to this specific case.
Suppose we have a pair of random variables (RVs) $X$ and $Y$ for which the joint PDF is $f_{XY}(x,y)$. Suppose also that some function $z = g(X,Y)$ of these RVs is given, and we wish to find the marginal distribution $f_X(x|z)$ of $X$ given the fixed value $z = g(X,Y)$ of this quantity. By fixing $g(X,Y)$, we are effectively reducing the range of $X$ and $Y$ from two dimensions down to a one dimensional manifold defined by $z = g(X,Y)$. The marginal PDF we seek will be proportional to the values of the joint PDF $f_{XY}(x,y)$ along this manifold. We imagine solving $z = g(x,y)$ for $y$ to yield $y(x, z)$. Then we have:
$$
f_X(x|z) \;=\; C(z)\, f_{XY}(x, y(x, z))\, ,
$$
where $C(z)$ is a normalization constant. This describes the variation of $f_X(x|z)$ with $x$, and all that is left is to find $C(z)$ by ensuring normalization in the variable $x$:
$$
1 \;=\; \int dx\, f_X(x|z) \;=\; C(z) \int dx\, f_{XY}(x, y(x, z))
\quad\longrightarrow\quad C(z) = \frac{1}{\int dx\, f_{XY}(x, y(x, z))}\, .
$$
Thus we are left with
$$
f_X(x|z) \;=\; \frac{f_{XY}(x, y(x, z))}{\int dx\, f_{XY}(x, y(x, z))}\, .
$$
In this particular case we have $z = g(x,y) = x+y$, so that $y(x,z) = z - x$, as well as $f_{XY}(x,y) = f_X(x)\, f_Y(y)$, so this reduces to:
$$
f_X(x|z) \;=\; \frac{f_X(x)\, f_Y(z-x)}{\int dx\, f_X(x)\, f_Y(z-x)}
$$
