Proof of formula in State-space linear models A linear, discrete-time, stationary, state-space model is
$$
\begin{cases}
x_{t+1} = F x_t + v_t \ &t=0,1,2\dots \\
y_t = H x_t + w_t \ &t=1,2,3,\dots
\end{cases}
$$
where:


*

*$F \in \mathbb{C}^{n_x \times n_x}$ is a stable matrix ($|\lambda_i(F)| < 1$  for all $i$);

*$x_t \in \mathbb{C}^{n_x \times 1}$ is the state vector;

*$y_t \in \mathbb{C}^{n_y \times 1}$ is the measurement vector;

*$v_t, w_t$ are random vectors (additive noise) which admit a PDF (Probability Density Function);


The noise terms are zero-mean and:
$$
\forall t_1 \neq t_2 : \ v_{t_1} \perp v_{t_2}, w_{t_1} \perp w_{t_2} \\
\forall t_1, t_2 : \ v_{t_1} \perp w_{t_2} \\
\forall t : \ E[v_t v_t^\dagger] = Q, E[w_t w_t^\dagger] = R
$$
where $X \perp Y$ means that $X$ and $Y$ are independent, $^\dagger$ means conjugate-transpose, and $Q,R$ are assumed to be positive definite hermitian matrices. The initial condition $x_0$ is an absolutely continuous random vector satisfying $E[x_0] = \mathbf{0}, x_0 \perp v_t, x_0 \perp w_t$ for any $t$. 
The following formula requires proof:
$$
(*) \quad f(x_t, y_t | X_{0:t-1}, Y_{1:t-1}) = f(x_t|x_{t-1}) f(y_t|x_t)
$$
where $f(\cdot)$ denotes the PDF, $X_{0:t-1} = (x_{t-1}, x_t, \dots, x_0)$, $Y_{1:t} = (y_t, y_{t-1},\dots,y_1)$. How to prove it?

It is clear that $x_t$ is a Markov process, so
$$
f(x_t | X_{0:t-1}) = f(x_t|x_{t-1})
$$
but how can we use this to prove $(*)$ ?
 A: $$
\begin{align}
f(x_t, y_t | X_{0:t-1}, Y_{1:t-1}) &= f(x_t | X_{0:t-1}, Y_{1:t-1}) f(y_t |x_t, X_{0:t-1}, Y_{1:t-1}) \\
&= f(x_t | x_{t-1}) f(y_t |x_t),
\end{align}
$$
where the first line follows from the definition of conditional probability, and the latter line follows from the fact that once you know $x_{t-1}$, $X_{0:t-2}, Y_{1:t-1}$ is uninformative about $x_t$ (by the independence assumption), and, likewise, once you know $x_t$, $X_{0:t-1}, Y_{1:t-1}$ is uninformative about $y_t$ (again by the independence assumption).
Note that in the proof, you're not directly using the fact that $x_t$ is Markov, instead, you're using something that is proved in a similar way that you would establish that $x_t$ is Markov. I.e. we need that the distribution of $x_t|x_{t-1}$ is independent of $x_{t-2},x_{t-3},\dots$ and $y_{t-1},y_{t-2},\dots$, which follows directly from the equations of the state space model and the independence of $v_t$, $w_t$ across time. The independence of the distribution of $y_t|x_t$ from $x_{t-1},x_{t-2},\dots$ and $y_{t-1},y_{t-2},\dots$ also follows directly from the equations of the state space model and the independence of $v_t$, $w_t$ across time.
