Conditional densities and their sums The question I'm working on boils down to finding the distribution of $Z_2$ if you know that $Z_1 \sim N(\mu_0,1)$ and $Z_2|Z_1 \sim N(Z_1,1)$ (more generally, we want to determine ${\mathrm{E}}[w_1\cdot Z_1 + w_2\cdot Z_2]$ for any $w_1, w_2$). 
I think I'm getting lost somewhere in actual computation, and am hoping someone knows a quick trick that I'm not seeing regarding both distributions being normal.
I'm essentially looking at 
$$
f_{Z_2}(z_2) 
= \int_\mathbb{R} f_{Z_2 \mid Z_1}(z_2\mid z_1) f_{Z_1}(z_1) dz_1 \\
= \frac{1}{2\pi}\int_\mathbb{R} e^{-\frac{(z_2-z_1)^2}{2}}e^{-\frac{(z_1-\mu_0)^2}{2}} dz_1 = \ldots
$$
and after completing the square for $z_1$, I can't recognize the distribution...
Any ideas?
 A: Well if you just want $E(w_1 Z_1 + w_2 Z_2)$, you don't need to find the distribution of $Z_2$. Instead just use the tower property to obtain
\begin{align*}
E(w_1 Z_1 + w_2 Z_2) &= w_1 E Z_1 + w_2 EZ_2 \\
&= w_1 \mu_0 + w_2 E(E(Z_2 \mid Z_1)) \\
&= w_1 \mu_0 + w_2 E(Z_1) \\
&= w_1 \mu_0 + w_2 \mu_0 \\
&= (w_1+w_2)\mu_0.
\end{align*}
If you are still interested in $f_{Z_2}(z_2)$, you can complete the square like this,
\begin{align*}
f_{Z_2}(z_2) &\propto \int \exp\left\{ -\frac{1}{2}\left(z_2^2 -2z_2z_1 + z_1^2 + z_1^2 - 2z_1\mu_0 + \mu_0^2 \right) \right\} dz_1 \\
&\propto \exp\left\{-\frac{z_2^2}{2}\right\}\int \exp\left\{-\frac{1}{2} \left(2z_1^2 - 2z_1(z_2+\mu_0) \right) \right\}dz_1 \\
&= \exp\left\{-\frac{z_2^2}{2}\right\}\int \exp\left\{-\frac{1}{2 \cdot (1/2)} \left(z_1^2 - 2\cdot \frac{z_1(z_2+\mu_0)}{2} \right) \right\}dz_1 \\
&=  \exp\left\{-\frac{z_2^2}{2}\right\} \exp\left\{ \frac{(z_2+\mu_0)^2}{4} \right\}\int \exp\left\{-\frac{1}{2 \cdot (1/2)} \left(z_1 - \frac{z_2+\mu_0}{2}\right)^2 \right\}dz_1 \\
&\propto \exp\left\{-\frac{z_2^2}{2} + \frac{z_2^2}{4} + \frac{2z_2 \mu_0}{4}\right\} \\
&= \exp\left\{-\frac{1}{4}\left(z_2^2 - 2z_2 \mu_0\right)\right\} \\
&\propto \exp\left\{-\frac{1}{4} (z_2 - \mu_0)^2 \right\}.
\end{align*}
Recognizing this as the kernel of a $N(\mu_0,2)$ distribution, we have that $Z_2 \sim N(\mu_0,2)$. Notice that this agrees with our calculation above,
\begin{align*}
w_1 EZ_1 + w_2 EZ_2 = w_1 E(N(\mu_0,1)) + w_2 E(N(\mu_0,2)) = (w_1+w_2)\mu_0.
\end{align*}
