When does a random variable admit a density? We know the sum of two independent, absolutely continuous random variables has a density which is simply the convolution of the two densities. But what if one is discrete and the other is absolutely continuous? E.g. Say $U \thicksim U(0,1)$ and $X \thicksim\frac{1}{2}\delta_0+\frac{1}{2}\delta_1$. Will $U+X$ then admit a density (with respect to the Lebesgue measure)? If so, how does one compute it? Thanks in advance!
 A: I've come across this problem in data communication where discrete symbols are disturbed by additive continuous noise. What we did there was simply use convolution (after transforming the density of the discrete variable to a 'continuous' density using Dirac delta functions). So if the density of the discrete variable is $p_X=\frac{1}{2}\delta(x)+\frac{1}{2}\delta(x-1)$ then the density of the sum $Z=X+Y$ (where $Y$ is the continuous variable with density $p_Y(x)$) would be
$$p_Z(x)=\frac{1}{2}p_Y(x) + \frac{1}{2}p_Y(x-1)$$
I realize that there is no mathematical rigor in my answer but I just wanted to show that in engineering this problem is successfully treated in this way.
A: Yes, it will admit a density. First of all it is clear that the range of the random variable will be $(0,2)$. Let $Z= U+X$. Then by conditioning:
\begin{equation}
P(Z \in \mathrm{d}z) = P(Z \in \mathrm{d}z| X=1)P(X=1) + P(Z \in \mathrm{d}z| X=0)P(X=0). 
\end{equation}
Since $X$ takes values $1$ and $0$ with probability $1/2$ and using independence we have
\begin{equation}
P(U\in \mathrm{d}(z-1))\frac{1}{2} + P(U\in \mathrm{d}z)\frac{1}{2} = \frac{1}{2}f_U(z-1)+\frac{1}{2}f_U(z) = \frac{1}{2}1{\hskip -2.5 pt}\hbox{I}_{(0,2)}(z).
\end{equation}
Hence $Z$ is Uniform on $(0,2)$.
