I am reading Oksendal's book on SDEs. He uses Ito's formula to an Ito integral in proving the Ito representation theorem but the calculations confuse me.
Let $Y_t = \exp^{\int_0^t h(s) \,dBs – \frac{1}{2} \int_0^t h^2(s)\,ds}$.
The book says: by Ito's formula:
$dY_t = Y_t(h(t) dB_t – \frac{1}{2}h^2(t)dt) + \frac{1}{2}Y_t(h(t) dB_t)^2$. I am confused how to obtain this expression. Note that at this point, the book has only introduced the basic Ito formula and no other techniques yet.
If $g(t,x) = \exp^{\int_0^t h(s) dx – \frac{1}{2}\int_0^t h^2(s)\,ds}$ then $Y_t = g(t,B_t)$.
I therefore need to compute $\frac{\partial g}{\partial t}, \frac{\partial g}{\partial x}, \frac{\partial^2 g}{\partial x^2}$ to use the formula.
Now, I am stuck with computing the following derivatives: $\frac{\partial g}{\partial t} \int_0^t h(s)dx$ and $\frac{\partial g}{\partial x} \int_0^t h(s) dx$.
Mainly, my confusion is that I don't know how to differentiate with respect to a variable that also appears in the integrator of the integral.
For the first derivative, informally, I think I can write $\int_0^t h(s) \,dx = \int_0^t h(s) x'(s) \,ds$ so that I can apply the fundamental theorem of calculus (ie. the integrator is $ds$, not $dx$). Is this correct? For the second derivative, I don't know how to differentiate with respect to $x$.
Help appreciated.
Best Answer
Letting $X_t=\int_0^t h(s) \,dBs - \frac{1}{2} \int_0^t h^2(s)\,ds$ we have that this could be written as $dX_t= h(t)dB_t-1/2 h^2(t)dt$.
Hence letting $Y_t=\exp(X_t)$ you can apply Itô formula:
\begin{align} dY_t&= Y_t dX_t+\frac 1 2 d\langle X\rangle_t\\ &=Y_t[ h(t)dB_t- \frac 1 2 h^2(t)dt]+\frac 1 2 h^2(t)dt. \end{align}
Notice that $(dB_t)^2$ is an euristic way to denote the (differential of the...) quadratic variation which equals $dt$.