Sum of two independent random variables (density) I remember seeing the statement the sum of two independent random variables has a density if one of them has a density somewhere but forgot where it is. Is this statement true? If it is can someone provide me with a hint or a proof? Thanks!
 A: Let $X,Y$ be random variables, such that $X$ has density $f_X$ and let $\mu_Y$ be the distribution of $Y$. If $X,Y$ are independent, then $Z = X+Y$ has density given by:
$$f_Z(z) = (f_X * \mu_Y)(z) := \int_{\mathbb R} f_X(z-y)d\mu_Y(y)$$
Proof:
We need to check that for any $A \in \mathcal B(\mathbb R)$ we have $\mu_Z(A) = \int_{A}f_Z(z)dz$.
We get: $$ \mu_Z(A) = \mathbb P(X+Y \in A) = \mathbb E[1_A(X+Y)] = \mathbb E[\psi(X,Y)]$$ where $\psi(X,Y) = 1_{A}(X+Y)$.
So that $$ \mu_Z(A) = \int_{\mathbb R^2} 1_A(x+y) d(\mu_X \otimes \mu_Y)(x,y) = \int_{\mathbb R} \int_{\mathbb R} 1_A(x+y)d\mu_X(x)d\mu_Y(y)$$
Where use of Fubinii due to non-negativity and independence of $X,Y$ (which implies that the measure $\mu_{(X,Y)}$ is in the product form $\mu_X \otimes \mu_Y$). Now since $\mu_X$ has density $f_X$, we get :
$$ \mu_Z(A) = \int_{\mathbb R} \int_{\mathbb R} 1_A(x+y) f_X(x)dx d\mu_Y(y) = \int_{\mathbb R}\int_{\mathbb R}1_A(z)f_X(z-y)dzd\mu_Y(y) $$
Where at the last equality we substitution $z=x+y$ in the inner integral. Using Fubinii last time
$$ \mu_Z(A) = \int_A \int_{\mathbb R} f_X(z-y)d\mu_Y(y) dz = \int_A f_Z(z)dz$$
Since $A$ was arbitrary, by definition $f_Z$ is density of $Z$ (it is easy to see that it integrates to $1$ and is non-negative almost everywhere)
