楼主: tulipsliu
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[学科前沿] [QuantEcon]MATLAB混编FORTRAN语言 [推广有奖]

181
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:26:10
$$
\begin{align}
y_t &= \left[ 1    \phi \right] {x}_{t-1} +
\varepsilon_t\label{obseq}\\
{x}_t &= \left[\begin{array}{ll}
                        1 & \phi\\
                        0 & \phi
                   \end{array}\right]{x}_{t-1}
            + \left[\begin{array}{l}
                          \alpha\\
                          \beta
                    \end{array}\right]\varepsilon_t.\label{stateeq}
\end{align}
$$

182
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:27:49
The model is fully specified once we state the distribution of the
error term $\varepsilon_t$. Usually we assume that these are
independent and identically distributed, following a normal
distribution with mean 0 and variance $\sigma^2$, which we write as
$\varepsilon_t \sim\mbox{NID}(0, \sigma^2)$.

$$
\begin{align*}
y_t &= (\ell_{t-1} + \phi b_{t-1})(1 + \varepsilon_t)\\
\ell_t &= (\ell_{t-1} + \phi b_{t-1})(1 + \alpha \varepsilon_t)\\
b_t &= \phi b_{t-1} + \beta(\ell_{t-1}+\phi b_{t-1})\varepsilon_t,
\end{align*}$$
$$
\begin{align*}
y_t &= \left[ 1    \phi \right] \bm{x}_{t-1}(1 + \varepsilon_t)\\
{x}_t &= \left[\begin{array}{ll}
                        1 & \phi \\
                        0 & \phi
                   \end{array}\right]\bm{x}_{t-1} + \left[ 1
                   \phi
                   \right] {x}_{t-1}
             \left[\begin{array}{l}
                          \alpha\\
                          \beta
                    \end{array}\right]\varepsilon_t.
\end{align*}
$$

183
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:28:27
We constrain the initial states $\bm{x}_0$ so that the seasonal
indices add to zero for additive seasonality, and add to $m$ for
multiplicative seasonality. There have been several suggestions for
restricting the parameter space for  $\alpha$, $\beta$ and $\gamma$.
The traditional approach is to ensure that the various equations can
be interpreted as weighted averages, thus requiring $\alpha$,
$\beta^*=\beta/\alpha$, $\gamma^*=\gamma/(1-\alpha)$ and $\phi$ to
all lie within $(0,1)$. This suggests
$$0<\alpha<1,\qquad 0<\beta<\alpha,\qquad 0<\gamma <
1-\alpha,\qquad\mbox{and}\qquad 0<\phi<1.
$$

184
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:29:11
$$\mbox{AIC} = L^*(\hat{\theta},\hat{x}_0) + 2q,
$$

185
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:29:52
Choosing the model order using unit root tests and the AIC

A non-seasonal ARIMA($p,d,q$) process is given by
$$
\phi(B)(1-B^d)y_{t} = c + \theta(B)\varepsilon_t
$$

where $\{\varepsilon_t\}$ is a white noise process with mean zero
and variance $\sigma^2$, $B$ is the backshift operator, and
$\phi(z)$ and $\theta(z)$ are polynomials of order $p$ and $q$
respectively. To ensure causality and invertibility, it is assumed
that $\phi(z)$ and $\theta(z)$ have no roots for $|z|<1$

186
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:30:36
The seasonal ARIMA$(p,d,q)(P,D,Q)_m$ process is given by
$$
\Phi(B^m)\phi(B)(1-B^{m})^D(1-B)^dy_{t} = c + \Theta(B^m)\theta(B)\varepsilon_t
$$
where $\Phi(z)$ and $\Theta(z)$ are polynomials of orders $P$ and
$Q$ respectively, each containing no roots inside the unit circle.
If $c\ne0$, there is an implied polynomial of order $d+D$ in the
forecast function.

The main task in automatic ARIMA forecasting is selecting an
appropriate model order, that is the values $p$, $q$, $P$, $Q$, $D$,
$d$. If $d$ and $D$ are known, we can select the orders $p$, $q$,
$P$ and $Q$ via an information criterion such as the AIC:
$$\mbox{AIC} = -2\log(L) + 2(p+q+P+Q+k)$$
where $k=1$ if $c\ne0$ and 0 otherwise, and $L$ is the maximized
likelihood of the model fitted to the \emph{differenced} data
$(1-B^m)^D(1-B)^dy_t$.  The likelihood of the full model for $y_t$
is not actually defined and so the value of the AIC for different
levels of differencing are not comparable.

187
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:30:59
For seasonal data, we consider ARIMA$(p,d,q)(P,D,Q)_m$ models where
$m$ is the seasonal frequency and $D=0$ or $D=1$ depending on an
extended Canova-Hansen test [@CH95]. Canova and Hansen only
provide critical values for $2<m<13$. In our implementation of their
test, we allow any value of $m>1$. Let $C_m$ be the critical value
for seasonal period $m$. We plotted $C_m$ against $m$ for values of
$m$ up to 365 and noted that they fit the line $C_m = 0.269
m^{0.928}$ almost exactly. So for $m>12$, we use this simple
expression to obtain the critical value.

188
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:51:39
$$
\begin{align*}
\frac{{dy_1}}{{dt}} &=  - k_1 \cdot y_1 + k_2 \cdot y_2 \cdot y_3 \\
\frac{{dy_2}}{{dt}} &= k_1 \cdot y_1 - k_2 \cdot y_2 \cdot y_3 - k_3 \cdot y_2 \cdot y_2 \\
\frac{{dy_3}}{{dt}} &= k_3 \cdot y_2 \cdot y_2 \\
\end{align*}
$$

189
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 17:54:57
$$
\begin{align*}
    \frac{dX}{dt} &=  a \cdot X + Y \cdot Z \\
    \frac{dY}{dt} &=  b \cdot (Y - Z) \\
    \frac{dZ}{dt} &=  - X \cdot Y + c \cdot Y - Z
\end{align*}
$$

190
tulipsliu(未真实交易用户) 在职认证  发表于 2020-12-25 18:07:46
$$
D(Y | X) = D(Y | X_1, \dots, X_m) = D(Y | f( X_1, \dots,
X_m)),
$$

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