Why is the following variable a mixed random variable?

Question

Let X∼N(0,1) and W∼Bernoulli(1/2) be independent random variables. Define the random variable Y as a function of X and W:

Find the PDF of Y and X+Y.

I don't understand why Z turned out to be a mixed RV. If it can take on a value of 0 with a probability of 0.5, can't we assume 0 itself is ∼N(0,0) instead of a step function?

Answer 1

You can certainly think of the degenerate random variable at $0$ as having distribution $N(0, 0)$, but it does not change the fact that $\mathsf{Z}$ is a mixed random variable. What makes $\mathsf{Z}$ a mixed random variable, is the fact that its distribution is one of two different distributions, depending on $\mathsf{W}$. Even if $(\mathsf{Z}\mid\mathsf{W}=1)\sim N(\mu, \sigma^2)$ for some $\sigma>0$ (and assuming $(\mu, \sigma)\neq(0, 2)$), $\mathsf{Z}$ would still be a mixed random variable, even though its distribution given $\mathsf{W}$ is normal either way.