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

$Z$ is normal with mean $\mu_z$ and standard deviation $\sigma^2_z$. Contitional on $Z = z$, $W$ is normal with mean $z$ and variance $\sigma^2_w$.

Does it follow from these hypotheses that W is a normal variable with mean $\mu_z$ and variance $\sigma^2_w$?

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 $\sigma^2_{w^{(2)}}$.

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 $\frac{A*e^{(w-c)^2}}{2b}$ where A,b,c are constants, implying that $W$ is a normal variable (whose mean and expectations were already derived).

• Your last paragraph makes sense but could be better/more clearly stated -- you're computing the joint density of (W,Z) (when taking the product of the conditional and the marginal for Z) and then from that the marginal for W (by integration) which seems to be just what is required here. – Glen_b Sep 7 '17 at 3:22
• I think you could also do this using a convolution, since $f(w) = f(w | z) f(z)$. – combo Sep 7 '17 at 7:12
• The title is a bit misleading here as the answer to the title without the extra assumptions in the question body would be "no" (if the conditional variance is not constant or the conditional mean is not an affine function of $z$, $W$ is not normal). – Juho Kokkala Sep 9 '17 at 6:41

Conditionally to $Z=z$, the random variable $W$ has the same distribution as $z + G$ where $G \sim \mathcal{N}(0, \sigma^2_w)$.
We would get the same claim if we had $W = Z + G$ where $G \sim \mathcal{N}(0, \sigma^2_w)$ is independent of $Z$.
Therefore $W$ has the same law as $Z+G$ where $G \sim \mathcal{N}(0, \sigma^2_w)$ is independent of $Z$, and this law is $\boxed{\mathcal{N}(\mu_z, \sigma^2_z+\sigma^2_w)}$.