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

261
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-2 08:25:22
$$
{\kappa _j} = E{\left( {\left| {{z_{t - j}} - {\eta _{2j}}} \right| - {\eta _{1j}}\left( {{z_{t - j}} - {\eta _{2j}}} \right)} \right)^\delta } = \int\limits_{ - \infty }^\infty  {{{\left( {\left| {z - {\eta _{2j}}} \right| - {\eta _{1j}}\left( {z - {\eta _{2j}}} \right)} \right)}^\delta }f\left( {z,0,1,...} \right)dz}
$$

262
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-2 08:25:50
$$
\begin{gathered}
  \sigma _t^2 = {q_t} + \sum\limits_{j = 1}^q {{\alpha _j}\left( {\varepsilon _{t - j}^2 - {q_{t - j}}} \right) + } \sum\limits_{j = 1}^p {{\beta _j}\left( {\sigma _{t - j}^2 - {q_{t - j}}} \right)}  \hfill \\
  {q_t} = \omega  + \rho {q_{t - 1}} + \phi \left( {\varepsilon _{t - 1}^2 - \sigma _{t - 1}^2} \right) \hfill \\
\end{gathered}
$$

263
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-2 08:26:19
$$
\begin{gathered}
  {E_{t - 1}}\left( {\sigma _{t + n}^2} \right) = {E_{t - 1}}\left( {{q_{t + n}}} \right) + \sum\limits_{j = 1}^q {{\alpha _j}\left( {\varepsilon _{t + n - j}^2 - {q_{t + n - j}}} \right) + } \sum\limits_{j = 1}^p {{\beta _j}\left( {\sigma _{t + n - j}^2 - {q_{t + n - j}}} \right)}  \hfill \\
  {E_{t - 1}}\left( {\sigma _{t + n}^2} \right) = {E_{t - 1}}\left( {{q_{t + n}}} \right) + \sum\limits_{j = 1}^q {{\alpha _j}{E_{t - 1}}\left[ {\varepsilon _{t + n - j}^2 - {q_{t + n - j}}} \right] + } \sum\limits_{j = 1}^p {{\beta _j}{E_{t - 1}}\left[ {\sigma _{t + n - j}^2 - {q_{t + n - j}}} \right]}  \hfill \\
\end{gathered}
$$

264
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-2 08:26:40
$$
\begin{gathered}
  {E_{t - 1}}\left( {\sigma _{t + n}^2} \right) = {E_{t - 1}}\left( {{q_{t + n}}} \right) + \sum\limits_{j = 1}^q {{\alpha _j}{E_{t - 1}}\left[ {\sigma _{t + n - j}^2 - {q_{t + n - j}}} \right] + } \sum\limits_{j = 1}^p {{\beta _j}{E_{t - 1}}\left[ {\sigma _{t + n - j}^2 - {q_{t + n - j}}} \right]}  \hfill \\
  {E_{t - 1}}\left( {\sigma _{t + n}^2} \right) = {E_{t - 1}}\left( {{q_{t + n}}} \right) + {\left[ {\sum\limits_{j = 1}^{\max (p,q)} {\left( {{\alpha _j} + {\beta _j}} \right)} } \right]^n}\left( {\sigma _t^2 - {q_t}} \right) \hfill \\
\end{gathered}
$$

265
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-2 08:28:06
$$
\begin{gathered}
  {Y_t} = \left[ {\begin{array}{*{20}{c}}
   {\log \sigma _t^2}  \\
   .  \\
   .  \\
   .  \\
   {\log \sigma _{t - p + 1}^2}  \\
   {\log {r_t}}  \\
   .  \\
   .  \\
   {\log {r_{t - q + 1}}}  \\

\end{array} } \right],A = \left( {\begin{array}{*{20}{c}}
   {\left( {{\beta _1},...,{\beta _p}} \right)} & {\left( {{\alpha _1},...,{\alpha _q}} \right)}  \\
   {\left( {{I_{p - 1 \times p - 1}},{0_{p - 1 \times 1}}} \right)} & {{0_{p - 1 \times q}}}  \\
   {\delta \left( {{\beta _1},...,{\beta _p}} \right)} & {\delta \left( {{\alpha _1},...,{\alpha _q}} \right)}  \\
   {{0_{q - 1 \times p}}} & {\left( {{I_{q - 1 \times q - 1}},{0_{q - 1 \times 1}}} \right)}  \\
\end{array} } \right),b = \left( {\begin{array}{*{20}{c}}
   \omega   \\
   {{0_{p - 1 \times 1}}}  \\
   {\xi  + \delta \omega }  \\
   {{0_{q - 1 \times 1}}}  \\
\end{array} } \right) \hfill \\
  {\varepsilon _t} = \left( {\begin{array}{*{20}{c}}
   {{0_{p \times 1}}}  \\
   {\tau \left( {{z_t}} \right) + {u_t}}  \\
   {{0_{q \times 1}}}  \\
\end{array} } \right) \hfill \\
\end{gathered}
$$

266
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-2 08:28:48
$$
\begin{array}{l}
\sigma _t^2 = \omega {\left[ {1 - \beta \left( L \right)} \right]^{ - 1}} + \left\{ {1 - [1 - \beta {{\left( L \right)}^{ - 1}}\phi \left( L \right){{\left( {1 - L} \right)}^d}} \right\}\varepsilon _t^2\\
= {\omega ^*} + \lambda \left( L \right)\varepsilon _t^2\\
= {\omega ^*} + \sum\limits_{j = 1}^\infty  {{\lambda _i}{L^i}\varepsilon _t^2}
\end{array}
$$

where ${\lambda _1} = {\phi _1} - {\beta _1} + d$ and ${\lambda _k} = {\beta _1}{\lambda _{k - 1}} + \left( {\frac{{k - 1 - d}}{k} - {\phi _1}} \right){\pi _{d,k - 1}}$. For the FIGARCH(1,d,1) model, sufficient conditions to ensure positivity of the conditional variance are $\omega  > 0$,${\beta _1} - d \le {\phi _1} \le \left( {\frac{{2 - d}}{2}} \right)$ , and $d\left( {{\phi _1} - \frac{{\left( {1 - d} \right)}}{2}} \right) \le {\beta _1}\left( {{\phi _1} - {\beta _1} + d} \right)$.

267
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-2 08:31:29
$$
\begin{array}{l}
\phi \left( L \right){\left( {1 - L} \right)^d}\varepsilon _t^2 = \omega  + \left( {1 - \beta \left( L \right)} \right)\left( {\varepsilon _t^2 - \sigma _t^2} \right)\\
\phi \left( L \right){\left( {1 - L} \right)^d}\varepsilon _t^2 = \omega  - \sigma _t^2 + \varepsilon _t^2 + \beta \left( L \right)\sigma _t^2 - \beta \left( L \right)\varepsilon _t^2\\
\sigma _t^2 = \omega  + \varepsilon _t^2 + \beta \left( L \right)\sigma _t^2 - \beta \left( L \right)\varepsilon _t^2 - \phi \left( L \right){\left( {1 - L} \right)^d}\varepsilon _t^2\\
\sigma _t^2 = \omega  + \left\{ {1 - \beta \left( L \right) - \phi \left( L \right){{\left( {1 - L} \right)}^d}} \right\}\varepsilon _t^2 + \beta \left( L \right)\sigma _t^2\\
\sigma _t^2 = \omega  + \left\{ {1 - \beta \left( L \right) - \left( {1 - \alpha \left( L \right)} \right){{\left( {1 - L} \right)}^d}} \right\}\varepsilon _t^2 + \beta \left( L \right)\sigma _t^2
\end{array}
$$

Truncating the expansion to 1000 lags and setting ${\left( {1 - L} \right)^d}\varepsilon _t^2 = \varepsilon _t^2 + \left( {\sum\limits_{k = 1}^{1000} {{\pi _k}{L^k}} } \right)\varepsilon _t^2 = \varepsilon _t^2 + \bar \varepsilon _t^2$, we can rewrite the equation as:

268
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-2 08:31:46
$$
\begin{array}{l}
\sigma _t^2 = \omega  + \left\{ {\varepsilon _t^2 - \beta \left( L \right)\varepsilon _t^2 - {{\left( {1 - L} \right)}^d}\varepsilon _t^2 + \alpha \left( L \right){{\left( {1 - L} \right)}^d}\varepsilon _t^2} \right\} + \beta \left( L \right)\sigma _t^2\\
\sigma _t^2 = \omega  + \left\{ {\varepsilon _t^2 - \beta \left( L \right)\varepsilon _t^2 - \left( {\varepsilon _t^2 + \bar \varepsilon _t^2} \right) + \alpha \left( L \right)\left( {\varepsilon _t^2 + \bar \varepsilon _t^2} \right)} \right\} + \beta \left( L \right)\sigma _t^2\\
\sigma _t^2 = \omega  + \varepsilon _t^2 - \beta \left( L \right)\varepsilon _t^2 - \left( {\varepsilon _t^2 + \bar \varepsilon _t^2} \right) + \alpha \left( L \right)\left( {\varepsilon _t^2 + \bar \varepsilon _t^2} \right) + \beta \left( L \right)\sigma _t^2\\
\sigma _t^2 = \omega  - \bar \varepsilon _t^2 - \beta \left( L \right)\varepsilon _t^2 + \alpha \left( L \right)\left( {\varepsilon _t^2 + \bar \varepsilon _t^2} \right) + \beta \left( L \right)\sigma _t^2\\
\sigma _t^2 = \left( {\omega  - \bar \varepsilon _t^2} \right) - \sum\limits_{j = 1}^p {{\beta _j}\varepsilon _{t - j}^2 + \sum\limits_{j = 1}^q {{\alpha _j}\varepsilon _{t - j}^2}  + } \sum\limits_{j = 1}^q {{\alpha _j}\bar \varepsilon _{t - j}^2}  + \sum\limits_{j = 1}^p {{\beta _j}\sigma _{t - j}^2} \\
\sigma _t^2 = \left( {\omega  - \bar \varepsilon _t^2} \right) + \sum\limits_{j = 1}^q {{\alpha _j}\left( {\varepsilon _{t - j}^2 + \bar \varepsilon _{t - j}^2} \right)}  + \sum\limits_{j = 1}^p {{\beta _j}\left( {\sigma _{t - j}^2 - \varepsilon _{t - j}^2} \right)}
\end{array}
$$

269
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-8 19:13:36
$$
\sigma
$$

270
tulipsliu(未真实交易用户) 在职认证  发表于 2021-1-17 15:42:14
$$
d_1 = \frac{1}{\sigma\sqrt{T}} \left[\log\left(\frac{S}{K}\right) +
\left(r+\frac{\sigma^2}{2}\right) T\right]
$$

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