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Taylor
Taylor
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Normally, the state value at $t$ should depend on its value at $t-1$. That's why we must consider the state at $t$ as being $X_t := \begin{bmatrix} x_t \\ x_{t-1}\end{bmatrix}$ so that we obtain the measurement equation: $$y_t = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}X_t+v_t$$ and the state equation: $$X_t = \begin{bmatrix}0 & -0.9\\ 1 & 0\end{bmatrix}X_{t-1}+\begin{bmatrix}z_t \\ z_{t-1}\end{bmatrix}.$$$$X_t = \begin{bmatrix}0 & -0.9\\ 1 & 0\end{bmatrix}X_{t-1}+\begin{bmatrix}z_t \\ 0\end{bmatrix}.$$ Especially, $X_0\sim \mathcal N\left(\begin{bmatrix} 0 \\ 0\end{bmatrix},\begin{bmatrix} \sigma_0^2 & 0 \\ 0 & \sigma_1^2 \end{bmatrix}\right)$ since $x_0\sim \mathcal N(0,\sigma_0^2)$ and $x_{-1}\sim \mathcal N(0,\sigma_1^2)$ are independent.

Normally, the state value at $t$ should depend on its value at $t-1$. That's why we must consider the state at $t$ as being $X_t := \begin{bmatrix} x_t \\ x_{t-1}\end{bmatrix}$ so that we obtain the measurement equation: $$y_t = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}X_t+v_t$$ and the state equation: $$X_t = \begin{bmatrix}0 & -0.9\\ 1 & 0\end{bmatrix}X_{t-1}+\begin{bmatrix}z_t \\ z_{t-1}\end{bmatrix}.$$ Especially, $X_0\sim \mathcal N\left(\begin{bmatrix} 0 \\ 0\end{bmatrix},\begin{bmatrix} \sigma_0^2 & 0 \\ 0 & \sigma_1^2 \end{bmatrix}\right)$ since $x_0\sim \mathcal N(0,\sigma_0^2)$ and $x_{-1}\sim \mathcal N(0,\sigma_1^2)$ are independent.

Normally, the state value at $t$ should depend on its value at $t-1$. That's why we must consider the state at $t$ as being $X_t := \begin{bmatrix} x_t \\ x_{t-1}\end{bmatrix}$ so that we obtain the measurement equation: $$y_t = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}X_t+v_t$$ and the state equation: $$X_t = \begin{bmatrix}0 & -0.9\\ 1 & 0\end{bmatrix}X_{t-1}+\begin{bmatrix}z_t \\ 0\end{bmatrix}.$$ Especially, $X_0\sim \mathcal N\left(\begin{bmatrix} 0 \\ 0\end{bmatrix},\begin{bmatrix} \sigma_0^2 & 0 \\ 0 & \sigma_1^2 \end{bmatrix}\right)$ since $x_0\sim \mathcal N(0,\sigma_0^2)$ and $x_{-1}\sim \mathcal N(0,\sigma_1^2)$ are independent.

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paf
paf
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Normally, the state value at $t$ should depend on its value at $t-1$. That's why we must consider the state at $t$ as being $X_t := \begin{bmatrix} x_t \\ x_{t-1}\end{bmatrix}$ so that we obtain the measurement equation: $$y_t = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}X_t+v_t$$ and the state equation: $$X_t = \begin{bmatrix}0 & -0.9\\ 1 & 0\end{bmatrix}X_{t-1}+\begin{bmatrix}z_t \\ z_{t-1}\end{bmatrix}.$$ Especially, $X_0\sim \mathcal N\left(\begin{bmatrix} 0 \\ 0\end{bmatrix},\begin{bmatrix} \sigma_0^2 & 0 \\ 0 & \sigma_1^2 \end{bmatrix}\right)$ since $x_0\sim \mathcal N(0,\sigma_0^2)$ and $x_{-1}\sim \mathcal N(0,\sigma_1^2)$ are independent.