Deriving the American Option PDE with Stochastic Volatility and Rates

# Hedging portfolio and extraction PDE of SV model with stochastic interest rate


# Hedging portfolio and extraction PDE of SV model with stochastic interest rate












How can I extraction this 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*} for option price $P(S,\sigma ,r ,t)$ from stochastic system

\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*} such that $$ dW^{(i)}_tdW^{(j)}_t=\rho_{ij}dt $$ for american option pricing ?

## Answer by user16651 (score 2, accepted)

https://quant.stackexchange.com/a/18501

First we write dynamic of ${{x}_{t}}=\ln ({{S}_{t}})$ \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} Let \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} Such that $dB_i dB_j=0$ then \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} Now we define

$$\Sigma (x(t),t)=\left( \begin{matrix} {{\sigma }_{t}} & 0 & 0 \\ b({{\sigma }_{t}},t){{\rho }_{12}} & b({{\sigma }_{t}},t)\sqrt{1-\rho _{12}^{2}} & 0 \\ \beta ({{r}_{t}},t){{\rho }_{13}} & \beta ({{r}_{t}},t)\frac{{{\rho }_{23}}-{{\rho }_{12}}{{\rho }_{13}}}{\sqrt{1-\rho _{12}^{2}}} & \beta ({{r}_{t}},t)\sqrt{1-\rho _{13}^{2}-\frac{{{({{\rho }_{23}}-{{\rho }_{12}}{{\rho }_{13}})}^{2}}}{1-\rho _{12}^{2}}} \\ \end{matrix} \right)$$ $$\Xi (x(t),t)=\left( \begin{matrix} ({{r}_{t}}-\delta -\frac{1}{2}\sigma _{t}^{2}) \\ a({{\sigma }_{t}},t) \\ \alpha ({{r}_{t}},t) \\ \end{matrix} \right)$$ $$B(t)=\left( \begin{matrix} {{B}_{1}}(t) \\ {{B}_{2}}(t) \\ {{B}_{3}}(t) \\ \end{matrix} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,X(t)=\left( \begin{matrix} {{x}_{1}}(t) \\ {{x}_{2}}(t) \\ {{x}_{3}}(t) \\ \end{matrix} \right)=\left( \begin{matrix} x(t) \\ \sigma (t) \\ r(t) \\ \end{matrix} \right)$$ then $$dX(t)=\Xi (x(t),t)dt+\Sigma (x(t),t)dB(t)$$ For simplicity,as we let $${{l}_{1}}=\sqrt{1-\rho _{12}^{2}}\,\,\,\,\,\,\,,\,\,\,\,\,{{l}_{2}}=\frac{{{\rho }_{23}}-{{\rho }_{12}}{{\rho }_{13}}}{\sqrt{1-\rho _{12}^{2}}}\,\,\,\,\,,\,\,\,\,\,\,\,{{l}_{3}}=\sqrt{1-\rho _{13}^{2}-\frac{{{({{\rho }_{23}}-{{\rho }_{12}}{{\rho }_{13}})}^{2}}}{1-\rho _{12}^{2}}}$$ $$\Sigma (x(t),t)=\left( \begin{matrix} \sigma & 0 & 0 \\ b{{\rho }_{12}} & b{{l}_{1}} & 0 \\ \beta {{\rho }_{13}} & \beta {{l}_{2}} & \beta {{l}_{3}} \\ \end{matrix} \right)$$ Simply it follows that $$\Sigma {{\Sigma }^{\text{T}}}=\left( \begin{matrix} {{\sigma }^{2}} & \sigma b{{\rho }_{12}} & \sigma \beta {{\rho }_{13}} \\ \sigma b{{\rho }_{12}} & {{b}^{2}}\rho _{12}^{2}-{{b}^{2}}l_{1}^{2} & b\beta ({{\rho }_{12}}{{\rho }_{13}}+{{l}_{1}}{{l}_{2}}) \\ \sigma \beta {{\rho }_{13}} & b\beta ({{\rho }_{12}}{{\rho }_{13}}+{{l}_{1}}{{l}_{2}}) & {{\beta }^{2}}(\rho _{13}^{2}+l_{2}^{2}+l_{3}^{2}) \\ \end{matrix} \right)$$ As a result $$\Sigma {{\Sigma }^{\text{T}}}=\left( \begin{matrix} {{\sigma }^{2}} & \sigma b{{\rho }_{12}} & \sigma \beta {{\rho }_{13}} \\ \sigma b{{\rho }_{12}} & {{b}^{2}} & b\beta {{\rho }_{23}} \\ \sigma \beta {{\rho }_{13}} & b\beta {{\rho }_{23}} & {{\beta }^{2}} \\ \end{matrix} \right)$$ Kolmogorov backward operator $$(A\,\,P)(t,x(t))=\sum\limits_{i=1}^{3}{{{\Xi }_{i}}\frac{\partial P}{\partial {{x}_{i}}}(t,x(t))+\frac{1}{2}}\sum\limits_{i,j=1}^{3}{{{(\Sigma {{\Sigma }^{\text{T}}})}_{i,j}}\frac{{{\partial }^{2}}P}{\partial {{x}_{i}}\partial {{x}_{j}}}(t,x(t))}$$ According to Feynman–Kac Theorem $${{P}_{t}}+A\,(P)-rP=0$$ Change PDE for $S(t)={{e}^{x(t)}}$ then \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}

## Answer by Lost1 (score 0)

https://quant.stackexchange.com/a/18494

The PDE only holds in t he continuation region, in the excerise region, P is just the pay off of the function. Let $\tau$ be the first time you enter the stopping region, then by the martingale property of the option price up to the first stopping time $$\mathbb{E}P(S_{t\wedge\tau},\sigma_{t\wedge\tau},r_{t\wedge\tau})e^{-r{t\wedge\tau}}=P(S_0,\sigma_0,r_0)$$

You apply ito's formula to the left hand side to get

$$P(S_{t\wedge\tau},\sigma_{t\wedge\tau},r_{t\wedge\tau})e^{-r{t\wedge\tau}}=P(S_0,\sigma_0,r_0)+\text{something dt}+\text{a local martingale} $$

Now you substitute the last equation to the left hand side of the 1st equation, note the expectation of the local martingale is 0, you are left with

$$\mathbb{E}\int^{\tau\wedge t}_0 ... \text{d}t$$

Now divide this by $1/t$ and take the limit $t\rightarrow 0$ and justify the exchange of appropriate limits. Note that $\tau$ can be ignored because $P(\tau<t)\rightarrow 0$ as you start in the continuous region and the process is continuous.

Note that this is exactly the same thing you would get if you deal with a European option.