In a LGSSM how do we know that the prediction distribution is Gaussian? I am trying to follow lecture notes regarding the Kalman Filter from a course taught at Stanford. The lecture notes can be found here. 
The linear Gaussian state space model (LGSSM) is introduced as 
\begin{align*}
z_0 &\sim N(0, \Sigma_0) \\
z_t &= A z_{t-1} + w_{t-1} &&\text{ for independent } &&w_{t-1} \sim N(0,Q) &\forall t \geq 1 \\
x_t &= C z_t + v_t &&\text{ for independent } &&v_t \sim N(0, R) &\forall t \geq 0 
\end{align*}
Here $z_i$ represents the hidden state at time $t=i$ and $x_j$ represents the data at time $t=j$. Also, $x_{0:k} = (x_0, x_1, \ldots, x_k)$.

In section 11.2.1 Time Update, the author stats that "last time, we leveraged the fact that we know $z_{t+1} | x_{0:t}$ will take a normal distribution..." 

I imagine this detail was discussed during the actual lecture, but I was not in attendance as it dates back to 2014. 
I tried to work it out for $t=1$. According to the text I should be able to show that $p(z_2 | x_0, x_1)$ is Gaussian (or at least proportional). 
\begin{align}
p(z_2 | x_0, x_1) &= \frac{p(z_2, x_0, x_1)}{p(x_0, x_1)} && (1) \\[1ex]
&= \frac{p(x_1|z_2, x_0) \cdot p(z_2 | x_0) \cdot p(x_0)}{p(x_1|x_0) \cdot p(x_0)} && (2) \\[1ex]
&= \frac{p(x_1|z_2, x_0) \cdot p(z_2 | x_0)}{p(x_1|x_0)} && (3) \\[1ex]
\end{align} 
I can't make anything useful out this work however. How do we know that the prediction distribution, $p(z_{t+1}| x_{0:t})$, is Gaussian?
 A: I believe these two facts are what you need:


*

*$p(z_{t+1} | z_t) = p(z_{t+1} | z_t, x_{0:t}) = \mathcal N \left( z_{t+1} \;|\; Az_t, Q \right)$: this is due to the Markovian nature of LGSSMs. In particular, the future is independent of the past given the present.

*If $p({\bf x}) = \mathcal N (\bf x \;|\; m,\; P)$ and $p({\bf y \;|\; x}) = \mathcal N (\bf y \;|\; Hx,\: R)$, then $p({\bf y}) = \mathcal N \left( \bf Hm,\; \bf HPH^{\top} + R \right)$.
Given that the filtering distribution $p(z_t | x_{0:t})$ and the state transition distribution $p(z_{t+1} | z_t)$ are both Gaussian, and in fact,
$$p(z_t | x_{0:t}) = \mathcal N \left( z_t \;|\; \hat z_{t|t} , P_{t|t} \right)$$
$$p(z_{t+1} | z_t) = \mathcal N \left( z_{t+1} \;|\; Az_t, Q \right)$$
we can make use of the facts above to see that the prediction distribution is Gaussian.
Let 


*

*$p(z_t | x_{0:t})$ act as $p({\bf x})$

*$p(z_{t+1} | z_t) = p(z_{t+1} | z_t, x_{0:t})$ act as $p({\bf y \;|\; x})$

*$p(z_{t+1} | x_{0:t})$ act as $p({\bf y})$

*$A$ act as $\bf H$ 

*$Q$ act as $\bf R$
Then we can see that $p(z_{t+1} | x_{0:t})$ is Gaussian with mean $\hat z_{t+1|t+1} = A \hat z_{t|t}$ and covariance $P_{t+1|t+1} = A P_{t|t} A^{\top} + Q$.
