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 $X$?

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#### Best Answer

**Compute the CDF of $X+N$ using convolution, then differentiate the result.**

The CDF of $X$ is

$$F_X(x) = (1-p)theta(x) + ptheta(x-1)$$

where $theta$ is the Heaviside theta function (the indicator function of the nonnegative reals),

$$theta(x) = 1text{ if }x ge 0, 0text{ otherwise}.$$

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

$$mathbb{E}(F_X(y-N)) = int_mathbb{R} F_X(y-n) f_N(n)dn = (F_Xstar f_N)(y).$$

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.

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