@Behrouz Maleki has provided the PDE of two factor model in other post so could anyone please provide Ito lemma of this equation and how this PDE was derived from Vasicek model. as far as I know it is by constructing a portfolio. right?
2 Answers
Let $V(t, r_t, S_t)$ be the convertible bond price at time $t$, where \begin{align*} dS_t &= S_t(r_t dt + \sigma dW_t^1)\\ dr_t &=\kappa(\theta-r_t)dt+\Sigma dW_t^2, \end{align*} and where $\{W_t^1, \, t\ge 0\}$ and $\{W_t^2, \, t\ge 0\}$ are two standard Brownian motions with $d\langle W^1, W^2\rangle_t = \rho dt$. Then, \begin{align*} &\ dV(t, r_t, S_t) \\ =&\ \frac{\partial V}{\partial t}dt + \frac{\partial V}{\partial r}dr_t + \frac{\partial V}{\partial S}dS_t \\ &\quad + \frac{1}{2}\frac{\partial^2 V}{\partial r^2}d\langle r, r\rangle_t + \frac{1}{2}\frac{\partial^2 V}{\partial S^2}d\langle S, S\rangle_t +\frac{\partial^2 V}{\partial S \partial r}d\langle S, r\rangle_t\\ =&\ \left(\frac{\partial V}{\partial t} + \kappa(\theta-r_t) \frac{\partial V}{\partial r} + r_tS_t \frac{\partial V}{\partial S} + \frac{1}{2}\Sigma^2\frac{\partial^2 V}{\partial r^2} + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 V}{\partial S^2} + \rho\sigma\Sigma\frac{\partial^2 V}{\partial S \partial r}\right)dt \\ &\ +\Sigma \frac{\partial V}{\partial r}dW_t^2 + \sigma S_t \frac{\partial V}{\partial S}dW_t^1. \end{align*} We note that, under the risk-neutral measure, $\{e^{-\int_0^t r_s ds} V_t, \, t \ge 0\}$ is a martingale. By Ito's lemma, \begin{align*} d\left(e^{-\int_0^t r_s ds} V_t \right) &= -r_t e^{-\int_0^t r_s ds} V_t dt + e^{-\int_0^t r_s ds} dV_t\\ &= e^{-\int_0^t r_s ds}\bigg[\bigg(-r_t V_t + \frac{\partial V}{\partial t} + \kappa(\theta-r_t) \frac{\partial V}{\partial r} + r_tS_t \frac{\partial V}{\partial S} \\ &\qquad\qquad\qquad + \frac{1}{2}\Sigma^2\frac{\partial^2 V}{\partial r^2} + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 V}{\partial S^2} + \rho\sigma\Sigma\frac{\partial^2 V}{\partial S \partial r}\bigg)dt\\ &\quad + \Sigma \frac{\partial V}{\partial r}dW_t^2 + \sigma S_t \frac{\partial V}{\partial S}dW_t^1\bigg]. \end{align*} By the martingality property, \begin{align*} & -r_t V_t + \frac{\partial V}{\partial t} + \kappa(\theta-r_t) \frac{\partial V}{\partial r} + r_tS_t \frac{\partial V}{\partial S} \\ &\qquad + \frac{1}{2}\Sigma^2\frac{\partial^2 V}{\partial r^2} + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 V}{\partial S^2} + \rho\sigma\Sigma\frac{\partial^2 V}{\partial S \partial r}=0. \tag{1} \end{align*} This is the PDE satisfied by the instrument price. If Brownian motions $\{W_t^1, \, t\ge 0\}$ and $\{W_t^2, \, t\ge 0\}$ are independent, that is, $\rho=0$, then Equation $(1)$ becomes \begin{align*} -r_t V_t + \frac{\partial V}{\partial t} + \kappa(\theta-r_t) \frac{\partial V}{\partial r} + r_tS_t \frac{\partial V}{\partial S} + \frac{1}{2}\Sigma^2\frac{\partial^2 V}{\partial r^2} + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 V}{\partial S^2} =0. \tag{2} \end{align*}
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$\begingroup$ Many thanks Gordon .. I need reference please? $\endgroup$– Fad FCommented Nov 14, 2016 at 13:34
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$\begingroup$ I do not have a specific reference, but most of the mathematical finance book should be fine. Which part is not clear to you? $\endgroup$– GordonCommented Nov 14, 2016 at 13:39
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$\begingroup$ Thanks for your reply .. How did you get (1) in details? how did you drop dt and stochastic dw? $\endgroup$– Fad FCommented Nov 14, 2016 at 13:44
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$\begingroup$ A martingale has a zero drift, that is, the coefficient of then $dt$ term is zero, which leads to our PDE. $\endgroup$– GordonCommented Nov 14, 2016 at 13:46
Gordon's Answer is nice (+1). I want to add the other solution.
Remark 1
Let $X=(X_1,X_2,...,X_n)$ where the component $X_i$ has a stochastic differential of the form $$dX_i(t)=\mu_i(t)dt+\sum_{j=1}^{d}\sigma_{ij}(t)dW_j(t)$$ where $dW_k(t)dW_j(t)=\rho_{kj}dt$, for all $k,j\in\{1,2,...,d\}$. Let $f:\mathbb{R}^+\times\mathbb{R}^n\to\mathbb{R}\in\mathbb{C}^{1,2}.$ By application of Ito's lemma, we have $$df(t,X_1,...,X_n)=\frac{\partial f}{\partial t}dt+\sum_{i=1}^{n}\frac{\partial f}{\partial x_i}dX_i+\frac 12\sum_{i=1}^{n}\sum_{l=1}^{n}\frac{\partial ^2f}{\partial x_i\partial x_l}dX_idX_l\tag 1$$
Remark 2
Assume that the short rate $r_t$ follows the Ito process as described by the following stochastic differential equation $$dr_t=\mu(t,r_t)dt+\sigma(t,r_t)dW_t$$ and $P(t,T)$ denotes the zero-coupon bond price with maturity $T$. We can show $$\frac{\partial P}{\partial t}+\mu(t,r_t)\frac{\partial P}{\partial r}+\frac{1}{2}\sigma^2(t,r_t)\frac{\partial^2 P}{\partial r^2}-r_tP=0\tag 2$$
Remark 3
Let $$\qquad dS_t=rS_t+\sigma S_tdW_1(t)\\ \quad\qquad\quad dr_t=\kappa(\theta-r_t)dt+\Sigma dW_2(t)\\ dW_1(t)dW_2(t)=0\tag 3$$
Now we form a portfolio consisting of one option $V=V(t,S,r,T,K)$ (Short position), $\Delta_1$ units of the stock (long position) and $\Delta_2$ units of the $T-$zero-coupon bond price (long position). The portfolio has value $$\Pi=\Delta_1S_t+\Delta_2 P(t,T)-V(t,S,r,T,K)$$ therefore $$d\Pi=\Delta_1dS_t+\Delta_2dP(t,T)-dV(t,S,r,T,K)\tag 4$$ By application of Ito's lemma, we have $$dV_t=\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial s}dS+\frac{\partial V}{\partial r}dr+\frac 12\left(\sigma^2\frac{\partial^2 V}{\partial s^2}+\Sigma^2\frac{\partial^2 V}{\partial r^2}\right)dt\tag 5$$ and $$dP=\frac{\partial P}{\partial t}dt+\frac{\partial P}{\partial r}dr+\frac{1}{2}\Sigma^2\frac{\partial^2 P}{\partial r^2}dt\tag 6$$ $(4)\,,\,(5)\,,(6)$ and $(3)$ $$d\Pi =-\left( \frac{\partial V}{\partial t}+\frac{1}{2}{{\sigma }^{2}}{{S}^{2}}\frac{{{\partial }^{2}}V}{\partial {{S}^{2}}} \right)dt+\left( \Delta _1-\frac{\partial V}{\partial S} \right)dS\\+\left( \Delta _2\frac{\partial P}{\partial r}\,-\frac{\partial V}{\partial r} \right)dr+\Delta _2\,\left( \frac{\partial P}{\partial t}+\frac{1}{2}{{\Sigma }^{2}}\frac{{{\partial }^{2}}P}{\partial {{r}^{2}}} \right)dt\,$$ We then find the values of $\Delta_1$ and $\Delta_2$ that makes the portfolio riskless. Indeed, we set $$ \Delta_1=\frac{\partial V}{\partial S}\\ \Delta_2=\frac{\frac{\partial V}{\partial r}}{\frac{\partial P}{\partial r}}$$ thus $$d\Pi =-\left( \frac{\partial V}{\partial t}+\frac{1}{2}{{\sigma }^{2}}{{S}_{t}}^{2}\frac{{{\partial }^{2}}V}{\partial {{S}^{2}}}+\frac{1}{2}{{\Sigma }^{2}}\frac{{{\partial }^{2}}V}{\partial {{r}^{2}}} \right)dt+\Delta {{}_{2}}\,\left( \frac{\partial P}{\partial t}+\frac{1}{2}{{\Sigma }^{2}}\frac{{{\partial }^{2}}P}{\partial {{r}^{2}}} \right)dt\tag 7$$ $(2)$ and $(7)$ $$d\Pi =-\left( \frac{\partial V}{\partial t}+\frac{1}{2}{{\sigma }^{2}}{{S}^{2}}\frac{{{\partial }^{2}}V}{\partial {{S}^{2}}}+\frac{1}{2}{{\Sigma }^{2}}\frac{{{\partial }^{2}}V}{\partial {{r}^{2}}} \right)dt+\Delta {{}_{2}}\,\left( {{r}_{t}}\,P-\kappa \left( \theta -r \right)\,\frac{\partial P}{\partial r} \right)dt\tag 8$$ The condition that the portfolio earn the risk-free rate, $r$, implies that the change in portfolio value is $$d\Pi=r\Pi dt$$ in other words $$d\Pi =(\Delta _1r_t\,S_t+\Delta_2r_tP-r_tV)dt\tag 9$$ $(8)$ and $(9)$ $$\frac{\partial V}{\partial t}+{{r}_{t}}\,{{S}_{t}}\frac{\partial V}{\partial S}+\kappa \left( \theta -r \right)\frac{\partial V}{\partial t}+\frac{1}{2}{{\sigma }^{2}}{{S}^{2}}\frac{{{\partial }^{2}}V}{\partial {{S}^{2}}}+\frac{1}{2}{{\Sigma }^{2}}\frac{{{\partial }^{2}}V}{\partial {{r}^{2}}}-{{r}_{t}}V=0$$
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$\begingroup$ You are really incredible guys .. Many thanks $\endgroup$– Fad FCommented Nov 14, 2016 at 13:33
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$\begingroup$ can I have a reference for that please? $\endgroup$– Fad FCommented Nov 14, 2016 at 13:33
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$\begingroup$ using this PDE, can we get the exact or closed solution of convertible bond with Vasicek? $\endgroup$– Fad FCommented Nov 14, 2016 at 13:54
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$\begingroup$ We sell one option and buy $\Delta _1$ unit Stock and $\Delta_2$ unit Zero-coupon bond $\endgroup$– user16651Commented Nov 14, 2016 at 14:56