Markov Switching Forecast. How can I derive this? [closed]

Consider the autoregressive model,

$\left[ \begin{array}{l}
y^{\ast}_t\\
x_t^{\ast}
\end{array} \right] = \left[ \begin{array}{l}
a_{11}\\
a_{21}
\end{array} \begin{array}{l}
a_{12}\\
a_{22}
\end{array} \right] \left[ \begin{array}{l}
y^{\ast}_{t – 1}\\
x^{\ast}_{t – 1}
\end{array} \right] + \text{} \left[ \begin{array}{l}
\varepsilon_t\\
\upsilon_t
\end{array} \right],$

where $\{ \varepsilon_t \}$ and $\{ v_t \}$ are white-noise processes with
zero mean, $y_t^{\ast}$ and $x_t^{\ast}$ are given by

$\begin{array}{lll}
y^{\ast}_t & = & y_t – \alpha_1 – \alpha_2 S_t,\\
x^{\ast}_t & = & x_t – \alpha_3 – \alpha_4 S_t,
\end{array}$

and $\{ S_t \}$ follows a two-state Markov process with transition
probabilities

$\begin{array}{lll}
p & = & P ( S_t = 1 | S_{t – 1} = 1),\\
q & = & P ( S_t = 0 | S_{t – 1} = 0 ) .
\end{array}$

Derive the expected value of $y_{t + n}$ conditional on information available
at time $t$ about the current and past values of $( y_t, x_t)$ and the current
value of $S_t$, i.e.,

$E ( y_{t + n} |_{} y_t, y_{t – 1}, \ldots .,
y_1, x_t, x_{t – 1}, \ldots ., x_1, S_t)$.

Answer

My attemp is the following:

From the system i derived

$\begin{array}{lll}
y^{\ast}_{t + n} & = & a_{12} \sum_{j = 0}^{\infty} a_{11}^j x_{t + n – j –
1}^{\ast} + \sum_{j = 0}^{\infty} a_{11}^j \varepsilon_{t + n – j}\\
& = & a_{11}^n y^{\ast}_t + a_{12} \sum_{j = 0}^{n – 1} a_{11}^j x_{t + n –
j – 1}^{\ast} + \sum_{j = 0}^{n – 1} a_{11}^j \varepsilon_{t + n – j}\\
y_{t + n} – \alpha_1 – \alpha_2 S_{t + n} & = & a_{11}^n ( y_t – \alpha_1 –
\alpha_2 S_t) + a_{12} \sum_{j = 0}^{n – 1} a_{11}^j ( x_{t + n – j – 1} –
\alpha_3 – \alpha_4 S_{t + n – j – 1}) + \sum_{j = 0}^{n – 1} a_{11}^j
\varepsilon_{t + n – j}
\end{array}$

Then,

$y_{t + n} = \alpha_1 ( 1 – a_{11}^n) + \alpha_2 ( S_{t + n} – a_{11}^n S_t) +
a_{11}^n y_t + a_{12} \sum_{j = 0}^{n – 1} a_{11}^j ( x_{t + n – j – 1} –
\alpha_3 – \alpha_4 S_{t + n – j – 1}) + \sum_{j = 0}^{n – 1} a_{11}^j
\varepsilon_{t + n – j}$

Taking expectations conditional on information at time t:

\begin{equation}
E ( y_{t + n} | I_t) = \alpha_1 ( 1 – a_{11}^n) + \alpha_2 ( E (
S_{t + n} | I_t) – a_{11}^n S_t) + a_{11}^n y_t + a_{12} \sum_{j
= 0}^{n – 1} a_{11}^j ( E ( x_{t + n – j – 1} | I_t) – \alpha_3 –
\alpha_4 E ( S_{t + n – j – 1} | I_t ))
\end{equation}

Attribution
Source : Link , Question Author : Julian Lopez Baasch , Answer Author : Julian Lopez Baasch

Leave a Comment