How to compute the PDF of a sum of bernoulli and normal variables analytically? Can convolution be applied to get a closed form expression for $Z = X + N$ where $X$ is a Bernoulli random variable and $N$ is a zero mean normal random variable independent of $X$?
 A: Compute the CDF of $X+N$ using convolution, then differentiate the result.
The CDF of $X$ is 
$$F_X(x) = (1-p)\theta(x) + p\theta(x-1)$$
where $\theta$ is the Heaviside theta function (the indicator function of the nonnegative reals),
$$\theta(x) = 1\text{ if }x \ge 0,\ 0\text{ otherwise}.$$
By definition, the CDF of $X+N$ is
$$F_{X+N}(y) = \Pr(X+N \le y) = \Pr(X \le y-N) =\mathbb{E}(F_X(y-N)).$$
The last equality computes $F_X(y-N)$ for each possible $N=n$ and integrates over them all, weighting them by their probabilities $f_N(n)dn$.  It is a convolution, written as
$$\mathbb{E}(F_X(y-N)) = \int_\mathbb{R} F_X(y-n) f_N(n)dn = (F_X\star f_N)(y).$$
Using the expression of $F_X$ in terms of Heaviside functions, linearity of integration breaks this integral into two convolutions of multiples of $\theta$ against $f_N$.  But computing such convolutions is trivial, because for any distribution function $f$ with integral $F$,
$$(\theta \star f)(y) = \int_\mathbb{R} \theta(y-x)f(x)dx = \int_{-\infty}^y 1 f(x)dx + \int_{y}^\infty 0 f(x)dx = F(y).$$
It should be apparent that the CDF of $X+N$ is a linear combination of the CDF of $N$ and the CDF of $N-1$.  Thus differentiation of the CDF to obtain the PDF will obtain the same linear combination of the PDFs.  At this point you could simply write down the answer.
A: $X$ is Bernoulli distributed with probability $p$.  $N$ has mean zero and variance $\sigma^2$.  So, with probability $1-p$, $Z=X+N$ has mean zero and variance $\sigma^2$ and with probability $p$ it has unit mean and variance $\sigma^2$.  That looks like a mixture of Gaussians to me.
