W is normal conditional on a normal variable Z. Does it follow that W is unconditionally normal?

Question

$Z$ is normal with mean $\mu_z$ and standard deviation $\mu_z$. Contitional on $Z=z, W$ is normal with mean $z$ and standard deviation $\sigma_w$.

Does it follow from these hypotheses that $W$ is a normal variable with mean $\mu_z$ and standard deviation $\sigma_w$?

My work: Using law of iterated expectations I can take $$E[W]=E[E[W|Z]]=E[Z]=\mu_z.$$ Similarly, I can compute $E[W(2)]$ with iterated expectations to get $\sigma w^2 + \mu_z^2$, implying that the variance of $W$ is just $\sigma_w^2$.

Now can I conclude that $W$ is a normal variable unconditionally? My reasoning is that if I take the product of the conditional density and multiply by the pdf of $Z$, and integrate out the variable $Z$, the resulting functional form of the pdf will be $A\exp{\left(\frac{(w−c)^2}b\right)}$ where A,b,c are constants, implying that $W$ is a normal variable (whose mean and expectations were already derived).

Answer 1

No. Let $p(z) = .5\delta(z-1)+.5\delta(z+1)$, where $\delta$ is the Dirac delta function. Then $$p(w)=\int\!\mathrm{d}z\,\mathcal{N}(w; z, \sigma_w^2) p(z)=.5\mathcal{N}(w; -1, \sigma_w^2)+.5\mathcal{N}(w; 1, \sigma_w^2),$$ which is not Gaussian (instead, it is a mixture of Gaussians).

The flaw in your reasoning is that when you integrate out $z$, the result will in general not be Gaussian; in fact, it will be Gaussian only if $p(z)$ was itself Gaussian.

Answer 2

[I presume that the standard deviation of $Z$ should be $\sigma_Z$.] You have the right idea -- iterated expectations. Use this method to the compute the characteristic function of $W$. Notice that you will end up with the variance of $W$ equal to $\sigma^2_Z+\sigma^2_W$ ("variance equals the mean of the conditional expectation plus the expectation of the conditional mean").