# 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?

I believe these two facts are what you need:

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

2. 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 $$P_{t+1|t+1} = A P_{t|t} A^{\top} + Q$$.