On this page, we learn how to transform a partial differential equation into a stochastic differential equation and vice versa. Dynkin formula and Feynman-Kac formula are defined in this section, but the proofs of these formulae are omitted. Let's denote a random variable \(\mathrm{X}_t\) starting at \(x\) (i.e. \(\mathrm{X}_0=x\))) by \(\mathrm{X}_t^x\), and the expectation with respect to \(\mathrm{X}_t^x\) by \(\mathrm{E}^x\).
Definition.
Let \({\mathrm{X}_t}\) be a time homogeneous Ito process in \(\mathfrak{R}^n\) such that \((\mathrm{X}_{t+\delta}-\mathrm{X}_{t})\) depends on \(\delta\), but not on \(t\). The generator of \(\mathrm{X}_t\) is then defined by $$ \mathcal{A} f(x)=\lim_{t\rightarrow 0}\frac{E^x[f(\mathrm{X}_t)]-f(x)}{t}; \text{ } x \in \mathfrak{R}^n. $$ \(D_\mathcal{A}\) denotes the set of functions \(f:\mathfrak{R}^n \rightarrow \mathfrak{R}\) for which the limit exists \(\forall x \in \mathfrak{R}^n\).
Theorem. Dynkin formula
Let \(X_t\) be a time homogeneous Ito process defined as $$ \mathrm{d} \mathrm{X}_t = b(\mathrm{X}_t)\mathrm{d}t + v(\mathrm{X}_t)\mathrm{d}B_t, $$ and let \(\tau\) be a stopping time for the process, \(\mathrm{E}^x[\tau]<\infty\). If \(f\in C^2(\mathfrak{R}^n)\) is a bounded function, and \(b(\mathrm{X}_t)\) and \(v(\mathrm{X}_t)\) satisfy bounded conditions , then \(f\in D_\mathcal{A}\) and $$ \mathcal{A} f(x) =\sum_{i=1}^n b_i(x) \frac{\partial f}{\partial x_i}+\frac{1}{2} \sum_{i,j}(v v^{T})_{i,j}(x) \frac{\partial ^2f}{\partial x_i\partial x_j}, $$ where \(v^T\) is the transpose of \(v\).
Exercise
Given the following \(2-\)dimensional stochastic process $$ \mathrm{d} \mathrm{X}_t= \left[ \begin{array}{c} \mathrm{d} \mathrm{X}_t^{[1]}\\ \mathrm{d} \mathrm{X}_t^{[2]} \end{array} \right] =\left[ \begin{array}{c} b_1\\ b_2 \end{array} \right]\mathrm{d}t + \left[ \begin{array}{cc} v_1 &v_2\\ v_3 &v_4 \end{array} \right] \left[ \begin{array}{c}\mathrm{d}B_t^{[1]}\\\mathrm{d}B_t^{[2]} \end{array} \right], $$ where \(B_t^{[1]}\) and \(B_t^{[2]}\) are independent Brownian motions.
Find the generator \(\mathcal{A}\) of the above stochastic process.
Solution.
We have$$ b= \left[ \begin{array}{cc} b_1 \\ b_2 \end{array} \right] ~~~\text{and}~~~ v= \left[ \begin{array}{cc} v_1 & v_2 \\ v_3 & v_4 \end{array} \right] $$ where \(b_i\)'s and \(v_i\)'s are functions of \(x \in \mathfrak{R}\). This gives the following partial differential equation $$ \mathcal{A} f(x)=b_1 \frac{\partial f}{\partial x_1}+b_2 \frac{\partial f}{\partial x_2}+\frac{1}{2} ( v_1^2+v_2^2 ) \frac{\partial ^2f}{\partial x_1\partial x_1} + (v_1 v_3 +v_2 v_4) \frac{\partial ^2f}{\partial x_1\partial x_2} + \frac{1}{2} (v_3^2+v_4^2 )\frac{\partial ^2f}{\partial x_2\partial x_2}. \nonumber $$
Theorem. Feynman-Kac formula
Let \(f\in C^2(\mathfrak{R}^n)\) be a bounded function and let \(\tau\) be a stopping time for the process. Assume that \(q\in C(\mathfrak{R}^n)\) is a lower bounded function and that $$ y(\tau,x)=E^x \left[\mathrm{exp}\left(-\int_0^{\tau} q(\mathrm{X}_z)\mathrm{d}z\right) f(\mathrm{X}_{\tau}) \right].~~~(1) $$ Then $$ \begin{array}{c c} q~ y+ \frac{\partial y}{\partial \tau}= \mathcal{A} y: & \tau>0,x\in \mathfrak{R}^n;~~~(2)\\ y(0,x)=f(x)~:& x \in \mathfrak{R}^n.~~~(3) \end{array} $$ If \(y(\tau,x)\) is bounded on \(\mathfrak{R} \times \mathfrak{R}^n\) and \(y\) satisfies the conditions in (2) and (3), then the solution for \(y\) is given by the formula in (1).
Let's take an example when the function \(y(\tau,x)\) is defined over \(\tau \in[0,T]\). We transform the variable \(\tau\) by \(t=T-\tau\) and replace \(x\) by the security underlying price \(S\). Assume that q is a constant risk-free interest rate with \(q(S_t)=r\). If a bounded function \(y(t,S_t)\) is defined over \([0,T] \times [0,\infty)\), and satisfy $$ \begin{array}{c c} r~ y-\frac{\partial y}{\partial t}= \mathcal{A} y: & t\in[0,T],~S_t \geq 0; \\ y(T,S_T)=f(S_T)~:& S_T \geq 0, \end{array} $$ where \(y(T,S_T)\) is the termination value of an option at maturity time \(T\). The solution of \(y(t,S_t)\) is then
$$ y(t,S_t)=E^{S_t} \left[\mathrm{exp}\left(-\int_0^{T-t} r\mathrm{d}z\right) f(S_{T-t}) \right]=e^{- r (T-t)} E^{S_t} \left[ f(S_{T-t}) \right]. $$
References
- Oksendal B.,
Stochastic Differential Equations: An Introduction With Applications
, 5th ed. Springer, 2000.
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