How to find the distribution of the following random variable? I am studying a state-space model which is described as follows:
\begin{align*}
\boldsymbol{l}_t = \boldsymbol{F_t}\boldsymbol{l}_{t-1}+\boldsymbol{g}_t\varepsilon_t, \quad \varepsilon_t\sim \mathcal{N}(0,1)\\
z_t = \boldsymbol{a}_t^T\boldsymbol{l}_{t}+b_t+\sigma_t\epsilon_t, \quad \epsilon\sim \mathcal{N}(0,1).
\end{align*}
We initially assume that we have $\boldsymbol{l}_0\sim \mathcal{N}(\boldsymbol{\mu}_0,\boldsymbol{\Sigma}_0)$ where $I$ is the identity matrix. My goal is to find the distribution of $$p(z_t|z_{1:t-1}) \sim \mathcal{N}(\boldsymbol{\mu}_t,\boldsymbol{\Sigma}_t).$$
We assume that we can obtain the following filtered Gaussian distributions using Kalman Filter
$$p(\boldsymbol{l}_{t-1}|z_{1:t})\sim \mathcal{N}(\boldsymbol{f}_t, \boldsymbol{S}_t).$$
For time $t=1$, it is easy, I have computed that,
$$\boldsymbol{\mu}_1 = \boldsymbol{a}_1^T\boldsymbol{\mu}_0, \quad \boldsymbol{\Sigma}_1 = \boldsymbol{a}_1^T\boldsymbol{\Sigma}_0\boldsymbol{a}_1 + \sigma_1^2.$$
I am not sure how to find the result for $t>1.$ Any hints will be much appreciated. This calculation is done in this paper (in the Supplemental). I have attached a relevant picture of the result below. (Ignore the superscript $(i)$).

 A: It is given that
\begin{align}
p(\boldsymbol{l}_{t-1}|z_{1:t}) \sim \mathcal{N}(\boldsymbol{f}_t, \boldsymbol{S}_t), \tag{1} \\
\boldsymbol{l}_t = \boldsymbol{F}_t\,\boldsymbol{l}_{t-1} + \boldsymbol{g}_t\,\varepsilon_t, \quad \varepsilon_t \sim \mathcal{N}(0,1), \tag{2} \\
z_t = \boldsymbol{a}_t^\top\,\boldsymbol{l}_{t} + b_t + \sigma_t\,\epsilon_t, \quad \epsilon_t\sim \mathcal{N}(0,1). \tag{3}
\end{align}
In order to find $p(z_t|z_{1:t-1}) \sim \mathcal{N}(\boldsymbol{\mu}_t,\boldsymbol{\Sigma}_t)$ one can to translate $(1)$ in time such that it uses the same information $z_{1:t-1}$, yielding $p(\boldsymbol{l}_{t-2}|z_{1:t-1})\sim \mathcal{N}(\boldsymbol{f}_{t-1}, \boldsymbol{S}_{t-1})$. However, from $(3)$ it can be seen that $\boldsymbol{l}_t$ is need to calculate $z_t$. In order to obtain $p(\boldsymbol{l}_{t}|z_{1:t-1})$ from $p(\boldsymbol{l}_{t-2}|z_{1:t-1})$ requires doing two prediction steps using $(2)$. 
For these prediction steps it is useful to know that the mean $\mu$ and variance $\Sigma$ of a distribution $p(x)$ are defined as $\mu = E[x]$ and $\Sigma = \text{Var}(x) = E[(x - \mu)\,(x - \mu)^\top]$ respectively, where $E[y]$ means the expected value of $y$. A related concept to the variance is the covariance, which is defined as $\text{Cov}(x,y) = E[(x - E[x])\,(y - E[y])^\top]$. Using this and the assumption that the covariance between $\varepsilon_t$ and all other distributions is zero then it can be shown that $p(\boldsymbol{l}_{t-1}|z_{1:t-1})$ has a mean $\mu$ and variance $\Sigma$ of
\begin{align}
\mu =& E[\boldsymbol{F}_{t-1}\,\boldsymbol{l}_{t-2} + \boldsymbol{g}_{t-1}\,\varepsilon_{t-1}|z_{1:t-1}] \\
=& \boldsymbol{F}_{t-1}\,E[\boldsymbol{l}_{t-2}|z_{1:t-1}] + \boldsymbol{g}_{t-1}\,E[\varepsilon_{t-1}|z_{1:t-1}] \\
=& \boldsymbol{F}_{t-1}\,\boldsymbol{f}_{t-1} \\
\Sigma =& E[(\boldsymbol{F}_{t-1}\,(\boldsymbol{l}_{t-2} - \boldsymbol{f}_{t-1}|z_{1:t-1}) + \boldsymbol{g}_{t-1}\,\varepsilon_{t-1})\,
(\boldsymbol{F}_{t-1}\,(\boldsymbol{l}_{t-2} - \boldsymbol{f}_{t-1}|z_{1:t-1}) + \boldsymbol{g}_{t-1}\,\varepsilon_{t-1})^\top] \\
=& \boldsymbol{F}_{t-1} \text{Var}(\boldsymbol{l}_{t-2}|z_{1:t-1}) \boldsymbol{F}_{t-1}^\top + 
\boldsymbol{F}_{t-1} \text{Cov}(\boldsymbol{l}_{t-2}|z_{1:t-1},\varepsilon_{t-1}) \boldsymbol{g}_{t-1}^\top + 
\boldsymbol{g}_{t-1} \text{Cov}(\varepsilon_{t-1},\boldsymbol{l}_{t-2}|z_{1:t-1}) \boldsymbol{F}_{t-1}^\top + 
\boldsymbol{g}_{t-1}\text{Var}(\varepsilon_{t-1})\boldsymbol{g}_{t-1}^\top \\
=& \boldsymbol{F}_{t-1}\,S_{t-1}\,\boldsymbol{F}_{t-1}^\top + 
\boldsymbol{g}_{t-1}\,\boldsymbol{g}_{t-1}^\top
\end{align}
Similarly, the next prediction step can be shown to give
$$
p(\boldsymbol{l}_{t}|z_{1:t-1}) \sim \mathcal{N}(\boldsymbol{F}_{t}\,\boldsymbol{F}_{t-1}\,\boldsymbol{f}_{t-1}, \boldsymbol{F}_{t}(\boldsymbol{F}_{t-1}\,\boldsymbol{S}_{t-1}\,\boldsymbol{F}_{t-1}^\top + \boldsymbol{g}_{t-1}\,\boldsymbol{g}_{t-1}^\top)\boldsymbol{F}_{t}^\top + \boldsymbol{g}_{t}\,\boldsymbol{g}_{t}^\top)
$$
To simplify the final expressions I define this last distribution as $p(\boldsymbol{l}_{t}|z_{1:t-1}) \sim \mathcal{N}(\hat{\boldsymbol{l}}_{t|1:t-1}, \Sigma_{t|1:t-1})$ with
$$
\left\{
\begin{array}{l}
\hat{\boldsymbol{l}}_{t|1:t-1} := \boldsymbol{F}_{t}\,\boldsymbol{F}_{t-1}\,\boldsymbol{f}_{t-1}, \\
\Sigma_{t|1:t-1} := \boldsymbol{F}_{t}(\boldsymbol{F}_{t-1}\,\boldsymbol{S}_{t-1}\,\boldsymbol{F}_{t-1}^\top + \boldsymbol{g}_{t-1}\,\boldsymbol{g}_{t-1}^\top)\boldsymbol{F}_{t}^\top + \boldsymbol{g}_{t}\,\boldsymbol{g}_{t}^\top.
\end{array}\right. \tag{4}
$$
The distribution that you are interested in can be obtained by substituting $p(\boldsymbol{l}_{t}|z_{1:t-1})$ using $(4)$ into $(3)$ yielding
$$
p(z_t|z_{1:t-1}) \sim \mathcal{N}(\boldsymbol{a}_t^\top\,\hat{\boldsymbol{l}}_{t|1:t-1} + b_t, \boldsymbol{a}_t^\top\,\Sigma_{t|1:t-1}\,\boldsymbol{a}_t + \sigma_t\,\sigma_t^\top). \tag{5}
$$
However, the expressions from the paper seems to do one less prediction step, so maybe $(1)$ should instead be $p(\boldsymbol{l}_{t}|z_{1:t}) \sim \mathcal{N}(\boldsymbol{f}_t, \boldsymbol{S}_t)$. Using this instead of $(1)$ requires only one prediction step, changing $(4)$ to
$$
\left\{
\begin{array}{l}
\hat{\boldsymbol{l}}_{t|1:t-1} := \boldsymbol{F}_{t}\,\boldsymbol{f}_{t-1}, \\
\Sigma_{t|1:t-1} := \boldsymbol{F}_{t}\,\boldsymbol{S}_{t-1}\,\boldsymbol{F}_{t}^\top + \boldsymbol{g}_{t}\,\boldsymbol{g}_{t}^\top.
\end{array}\right. \tag{6}
$$
Using $(6)$ instead of $(4)$ in $(5)$ almost completely agrees with the expressions in the paper. The only things that seems to be different from the equation in the paper are the time index in $S_{t-1}$ and the $b_t$ term, which should definitely be included unless it is somewhere assumed to be zero.
