Is the independent sum of a continuous random variable and mixed random variable continuous? Suppose X is a continuous random variable (has a pdf on the reals), and Y is a mixed random variable (its CDF has atoms). X and Y are independent. Is the sum, Z=X+Y a continuous random variable (i.e., has a pdf on the reals)?
 A: There is a relatively elementary demonstration that the sum is continuous.
Let $X$ have a probability distribution function $F_X$ with density function $f_X$ and let the distribution function of $Y$ be $F_Y.$  We do not assume $Y$ has a density function.  I claim that $X+Y$ has a density function (implying it is absolutely continuous) and its density can be expressed as an expectation,
$$f_{X+Y}(z) = E[f_X(z-Y)] = \int_\mathbb{R} f_X(z-y) \mathrm{d}F_Y(y).$$
To prove this claim, it suffices to show that integrating $f_{X+Y}$ indeed gives the desired probability function for $X+Y.$  The integration is performed by invoking Fubini's Theorem to change the order of the integrals, then changing the variable of integration from $w-y$ to $x,$ and finally expressing a probability in terms of an indicator function $\mathcal{I}.$  The remaining equations are just definitions of distribution functions and expectations as integrals:
$$\eqalign{
\int_{-\infty}^z f_{X+Y}(w)\mathrm{d}w &= \int_{-\infty}^z \int_\mathbb{R} f_X(w-y) \mathrm{d}F_Y(y)\ \mathrm{d}w \\
&= \int_\mathbb{R} \int_{-\infty}^z f_X(w-y) \mathrm{d}w\ \mathrm{d}F_Y(y) \\
&= \int_\mathbb{R} \int_{-\infty}^{z-y} f_X(x) \mathrm{d}x\ \mathrm{d}F_Y(y) \\
&= \int_\mathbb{R} F_X(z-y) \mathrm{d}F_Y(y) \\
&= E[F_X(z-Y)] \\
&= E_Y[\Pr(X \le z-Y)] \\
&= E_Y[E_X[\mathcal{I}(X+Y\le z)]] \\
&= \Pr(X+Y\le z) \\
&= F_{X+Y}(z).
}$$
For some intuition, think of adding $X$ to $Y$ as "smearing" every possible value of $Y$ according to the distribution of $X$ or, equivalently, as using $Y$ to weight a mixture of shifted versions of $X.$  In either case it's clear the result will have no atoms because $X$ has no atoms and so (of course) none of its shifted versions have atoms, either, whence no mixture of them will have any atoms.

In this figure, the left panel depicts the density of $X.$  The next panel shows the mass of $Y$ -- this variable has no density.  Nevertheless, as shown in the third panel, adding $Y$ to $X$ produces as many continuous components of $X$ as there are spikes in $Y,$ each one scaled by the height of its spike.  The density of $X+Y$ is the accumulated height of all these components.  Because it is formed from density functions, it too is a density, showing that $X+Y$ follows a continuous distribution even though $Y$ does not.
