Solving $dS^x(t)=S^x(t)(r(t)dt+\gamma(t,S^x(t))dW(t))$

Question

Consider the stochastic process $S^x(t)$ with $S^x(0)=x$ $$dS^x(t)=S^x(t)(r(t)dt+\gamma(t,S^x(t))dW(t))$$and define $D^x(t)=(\partial/\partial x)S^x(t)$. Then $$dD^x(t)=D^x(t)\Big[r(t)dt+\frac{\partial}{\partial s}\rho(t,S^x(t))dW(t)\Big]$$ ($W$ is a 1-d Brownian motion)and therefore, $$D^x(t)=exp\Big[\int_0^tr(u)du\Big]exp\Big[\int_0^t \frac{\partial}{\partial s}\rho(u,S^x(u))dW(u)-\frac{1}{2}\int_0^t\Big( \frac{\partial}{\partial s}\rho(u,S^x(u))\Big)^2du\Big]$$

where $\rho (t,s):=s\gamma(t,s)$. All the above I have been able to derive by myself. I need help proving the following identity. $$\frac{S^y(t)-S^x(t)}{M(t)}=y-x+\int_0^t \frac{1}{M(u)}[\rho(u,S^y(u))-\rho(u,S^x(u))]dW(u)$$

where $M(t)=exp[\int_0^t r(u)du]$

I was wondering if the best way is to find the integral of $D^x$ w.r.t $x$ which gives $$S^y(t)-S^x(t)=\int_x^yD(z)dz$$

However, this seems too complicated. Is there a better way or can this integral be evaluated? Thank you!

Answer 1

By Itô's formula, we have

$$f(t) X(t) - f(0) X(0) = \int_0^t f(s) \, dX(s) + \int_0^t X(s) f'(s) \, ds \tag{1}$$

for any Itô process $(X_t)_t$ and any function $f \in C_b^1(\mathbb{R})$. Using this identity for $$f(t) := \exp \left(- \int_0^t r(u) \, du \right) \quad \text{and} \quad X(t) := S^x(t)$$ we get

\begin{align*} \exp\left(-\int_0^t r(u) \, du\right) S^x(t) - x &= \int_0^t \exp\left(-\int_0^s r(u) \, du\right) \, dS^x(s) \\& \quad - \int_0^t r(s) \exp\left(-\int_0^s r(u) \, du\right) S^x(s) \, ds. \end{align*}

As $$dS^x(s) = S^x(s) r(s) \, ds + S^x(s) \gamma(s,S^x(s)) \, dW(s)$$

by the very definition of $(S^x(t))_{t \geq 0}$, this gives

\begin{align*} \exp\left(-\int_0^t r(u) \, du\right) S^x(t) - x &= \int_0^t \exp \left(- \int_0^s r(u) \, du \right) S^x(s) \gamma(s,S^x(s)) \, dW(s) \tag{2} \end{align*}

i.e.

$$\frac{1}{M_t} S^x(t)-x = \int_0^t \frac{1}{M(s)} \varrho(s,S^x(s)) \, dW(s)$$ where we have set $$M(t) = \exp \left( \int_0^t r(u) \, du \right) \quad \text{and} \quad \varrho(s,x) := s \gamma(s,x).$$

Since $(2)$ holds for any initial point $x \in \mathbb{R}$ we conclude that

$$\frac{1}{M_t} (S^y(t)-S^x(t)) = y-x + \int_0^t \frac{1}{M(s)} \left( \varrho(s,S^y(s))-\varrho(s,S^x(s) \right) \, dW(s).$$