Conditional density of Sum of two independent and continuous random variables Let X and Y be independent and continuous random variables. How do you solve for the conditional density of X+Y given X
 A: The identities
$$
P[X+Y\in\mathrm dz\mid X=x]=P[Y\in\mathrm dz-x\mid X=x]=P[Y\in\mathrm dz-x]
$$
imply that the conditional density of $X+Y$ given $X$ is
$$
f_{X+Y\mid X=x}(z)=f_Y(z-x).
$$
Edit: A rigorous approach to find the conditional density $(g_x)_x$ of $X+Y$ conditionally on $X$ is to ask that, for every bounded function $u$,
$$
E[u(X+Y)\mid X]=v(X),\qquad v(x)=\int u(z)g_x(z)\mathrm dz.
$$
To do that, recall that, by definition, $E[u(X+Y)\mid X]=v(X)$ if and only if, for every bounded function $w$, $E[u(X+Y)w(X)]=E[v(X)w(X)]$, that is,
$$
\iint u(x+y)w(x)f_X(x)f_Y(y)\mathrm dx\mathrm dy=\int v(x)w(x)f_X(x)\mathrm dx.
$$
By the change of variable $(x,z)=(x,x+y)$, the LHS is also
$$
\iint u(z)w(x)f_X(x)f_Y(z-x)\mathrm dx\mathrm dz,
$$
hence, by identification,
$$
v(x)=\int u(z)f_Y(z-x)\mathrm dz,
$$
and, by identification again, everything works fine with
$$
g_x(z)=f_Y(z-x).
$$
Edit2: A rigorous approach to find the joint density $f$ of $(X+Y,X)$ is to ask that, for every bounded function $u$,
$$
E[u(X+Y,X)]=\iint u(z,x)f(z,x)\mathrm dz\mathrm dx.
$$
By definition and by the change of variable $(x,z)=(x,x+y)$, the LHS is also
$$
\iint u(x+y,x)f_X(x)f_Y(y)\mathrm dx\mathrm dy=\iint u(z,x)f_X(x)f_Y(z-x)\mathrm dx\mathrm dz,
$$
hence, by identification,
$$
f(z,x)=f_X(x)f_Y(z-x).
$$
A: Hint: $P(X+Y) = P(X+Y|X)P(X)$ and use the independence assumption to write $P(X+Y)$ as a convolution.
