隨機利率SV模型的套期保值及PDE提取
如何提取此 PDE
$$ \begin{align*} 0 =& P_t+P_SS(r-\delta)+P_\sigma a(\sigma)+P_r\alpha (r,t) \ +& \frac{1}{2}P_{SS}S^2\sigma ^2 + \frac{1}{2}P_{\sigma \sigma}b^2(\sigma)+\frac{1}{2}P_{rr}\beta^2(r) \ +& P_{S\sigma}\sigma Sb(\sigma)\rho {12}+P{Sr}\sigma S\beta(\sigma)\rho {13}+P{\sigma r}\beta(\sigma)b(\sigma)\rho _{23}-rP \end{align*} $$ 對於期權價格 $ P(S,\sigma ,r ,t) $ 從隨機系統 $$ \begin{align*} dS_t &= (r_t-\delta)S_tdt+\sigma _tS_tdW_t^{(1)} \ d\sigma _t &=a(\sigma _t)dt+b(\sigma _t)dW^{(2)}_t\ dr_t &= \alpha(r_t,t)dt+\beta (r_t)dW_t^{(3)} \end{align*} $$ 這樣 $$ dW^{(i)}_tdW^{(j)}t=\rho{ij}dt $$ 美式期權定價？
首先我們寫動態的 $ {{x}{t}}=\ln ({{S}{t}}) $
$$ \begin{align} & d{{x}{t}}=({{r}{t}}-\delta -\frac{1}{2}\sigma {t}^{2})t+{{\sigma }{t}}d{{W}{1}}(t) \ & d{{\sigma }{t}}=a({{\sigma }{t}},t)dt+b({{\sigma }{t}},t)d{{W}{2}}(t) \ & d{{r}{t}}=\alpha ({{r}{t}},t)dt+\beta ({{r}{t}},t)d{{W}{3}}(t) \ \end{align} $$ 讓 $$ \begin{align} & {{W}{1}}={{B}{1}} \ & {{W}{2}}={{\rho }{12}}{{B}{1}}+\sqrt{1-\rho {12}^{2}}{{B}{2}} \ & {{W}{3}}={{\rho }{13}}{{B}{1}}+\frac{{{\rho }{23}}-{{\rho }{12}}{{\rho }{13}}}{\sqrt{1-\rho {12}^{2}}}{{B}{2}}+\sqrt{1-\rho {13}^{2}-\frac{{{({{\rho }{23}}-{{\rho }{12}}{{\rho }{13}})}^{2}}}{1-\rho {12}^{2}}}{{B}{3}} \ \end{align} $$ 這樣 $ dB_i dB_j=0 $ 然後 $$ \begin{align} & d{{x}{t}}=({{r}{t}}-\delta -\frac{1}{2}\sigma {t}^{2})dt+{{\sigma }{t}}d{{B}{1}}(t) \ & d{{\sigma }{t}}=a({{\sigma }{t}},t)dt+b({{\sigma }{t}},t)({{\rho }{12}}d{{B}{1}}(t)+\sqrt{1-\rho {12}^{2}}d{{B}{2}}(t)) \ & d{{r}{t}}=\alpha ({{r}{t}},t)dt+\beta ({{r}{t}},t)\left( {{\rho }{13}}d{{B}{1}}(t)+\frac{{{\rho }{23}}-{{\rho }{12}}{{\rho }{13}}}{\sqrt{1-\rho {12}^{2}}}d{{B}{2}}(t)+\sqrt{1-\rho {13}^{2}-\frac{{{({{\rho }{23}}-{{\rho }{12}}{{\rho }{13}})}^{2}}}{1-\rho {12}^{2}}}d{{B}{3}}(t) \right)(t) \ \end{align} $$ 現在我們定義 $$ \Sigma (x(t),t)=\left( \begin{matrix} {{\sigma }{t}} & 0 & 0 \ b({{\sigma }{t}},t){{\rho }{12}} & b({{\sigma }{t}},t)\sqrt{1-\rho {12}^{2}} & 0 \ \beta ({{r}{t}},t){{\rho }{13}} & \beta ({{r}{t}},t)\frac{{{\rho }{23}}-{{\rho }{12}}{{\rho }{13}}}{\sqrt{1-\rho {12}^{2}}} & \beta ({{r}{t}},t)\sqrt{1-\rho {13}^{2}-\frac{{{({{\rho }{23}}-{{\rho }{12}}{{\rho }{13}})}^{2}}}{1-\rho {12}^{2}}} \ \end{matrix} \right) $$ $$ \Xi (x(t),t)=\left( \begin{matrix} ({{r}{t}}-\delta -\frac{1}{2}\sigma {t}^{2}) \ a({{\sigma }{t}},t) \ \alpha ({{r}{t}},t) \ \end{matrix} \right) $$ $$ B(t)=\left( \begin{matrix} {{B}{1}}(t) \ {{B}{2}}(t) \ {{B}{3}}(t) \ \end{matrix} \right),,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,X(t)=\left( \begin{matrix} {{x}{1}}(t) \ {{x}{2}}(t) \ {{x}{3}}(t) \ \end{matrix} \right)=\left( \begin{matrix} x(t) \ \sigma (t) \ r(t) \ \end{matrix} \right) $$ 然後 $$ dX(t)=\Xi (x(t),t)dt+\Sigma (x(t),t)dB(t) $$ 為簡單起見，我們讓 $$ {{l}{1}}=\sqrt{1-\rho {12}^{2}},,,,,,,,,,,,,{{l}{2}}=\frac{{{\rho }{23}}-{{\rho }{12}}{{\rho }{13}}}{\sqrt{1-\rho {12}^{2}}},,,,,,,,,,,,,{{l}{3}}=\sqrt{1-\rho {13}^{2}-\frac{{{({{\rho }{23}}-{{\rho }{12}}{{\rho }{13}})}^{2}}}{1-\rho {12}^{2}}} $$ $$ \Sigma (x(t),t)=\left( \begin{matrix} \sigma & 0 & 0 \ b{{\rho }{12}} & b{{l}{1}} & 0 \ \beta {{\rho }{13}} & \beta {{l}{2}} & \beta {{l}{3}} \ \end{matrix} \right) $$ 簡單地說 $$ \Sigma {{\Sigma }^{\text{T}}}=\left( \begin{matrix} {{\sigma }^{2}} & \sigma b{{\rho }{12}} & \sigma \beta {{\rho }{13}} \ \sigma b{{\rho }{12}} & {{b}^{2}}\rho {12}^{2}-{{b}^{2}}l{1}^{2} & b\beta ({{\rho }{12}}{{\rho }{13}}+{{l}{1}}{{l}{2}}) \ \sigma \beta {{\rho }{13}} & b\beta ({{\rho }{12}}{{\rho }{13}}+{{l}{1}}{{l}{2}}) & {{\beta }^{2}}(\rho {13}^{2}+l{2}^{2}+l_{3}^{2}) \ \end{matrix} \right) $$ 因此 $$ \Sigma {{\Sigma }^{\text{T}}}=\left( \begin{matrix} {{\sigma }^{2}} & \sigma b{{\rho }{12}} & \sigma \beta {{\rho }{13}} \ \sigma b{{\rho }{12}} & {{b}^{2}} & b\beta {{\rho }{23}} \ \sigma \beta {{\rho }{13}} & b\beta {{\rho }{23}} & {{\beta }^{2}} \ \end{matrix} \right) $$ Kolmogorov 後向運算元 $$ (A,,P)(t,x(t))=\sum\limits_{i=1}^{3}{{{\Xi }{i}}\frac{\partial P}{\partial {{x}{i}}}(t,x(t))+\frac{1}{2}}\sum\limits_{i,j=1}^{3}{{{(\Sigma {{\Sigma }^{\text{T}}})}{i,j}}\frac{{{\partial }^{2}}P}{\partial {{x}{i}}\partial {{x}{j}}}(t,x(t))} $$ 根據費曼-卡克定理 $$ {{P}{t}}+A,(P)-rP=0 $$ 更改 PDE 為 $ S(t)={{e}^{x(t)}} $ 然後 $$ \begin{align} & 0={{P}{t}}+({{r}{t}}-\delta )S{{P}{S}}+a({{\sigma }{t}},t){{P}{\sigma }}+\alpha ({{r}{t}},t){{P}{r}} \ & ,,,,,,,+\frac{1}{2}{{\sigma }^{2}}{{S}^{2}}{{P}{SS}}+\frac{1}{2}{{b}^{2}}{{P}{\sigma \sigma }}+\frac{1}{2}{{\beta }^{2}}{{P}{rr}} \ & ,,,,,,,+\sigma b{{\rho }{12}}S{{P}{S\sigma }}+\sigma \beta {{\rho }{13}}S{{P}{Sr}}+b\beta {{\rho }{23}}{{P}{\sigma r}}-rP \ \end{align} $$