1
$\begingroup$

I have to show that $$ Y_t = e^{\theta N_t - \frac{1}{2}\theta^2 \int^t_0 g(s)^2 ds} $$ where $ N_t=\int^t_0g(s)dW_s $ satisfies $$ dX_t = \theta f(t) X_t dW_t $$ and that $ N_t \sim N(0, \int^t_0 g(s)^2 ds) $.

I have been trying to follow along with the process given here; however, I am uncertain as to how to go about without having the $dt$ term and the intergral within the $Y_t$ exponential makes me a bit confused.

$\endgroup$
1
$\begingroup$

Let $f(x,y) = e^{\theta x - \frac12 \theta^2 y}$ so that $Y_t = f(N_t, \int_0^t g(s)^2 ds)$. Let's also notice here that $\int_0^t g(s)^2 ds = \langle N \rangle_t$. By Ito's lemma, applied to $f$ we find \begin{align} dY_t =& \frac{\partial f}{\partial x}(N_t, \langle N \rangle_t) dN_t + \frac{\partial f}{\partial y}(N_t, \langle N \rangle_t) d \langle N \rangle_t + \frac12 \frac{\partial^2 f}{\partial x^2}(N_t, \langle N \rangle_t)d \langle N \rangle_t \\ =& \theta Y_t dN_t - \frac12 \theta^2 Y_t d\langle N \rangle_t + \frac12 \theta^2 Y_t d \langle N \rangle_t \\=& \theta Y_t dN_t \\=& \theta Y_t g(t) dW_t \end{align} where we don't see any other terms in the first line since $\langle N, \langle N \rangle \rangle = \langle \langle N \rangle, \langle N \rangle \rangle = 0$ and the last line follows by associativity of the integral since $dN_t = g(t) dW_t$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the reply @RhysSteele , the step-by-step explanation was very helpful. Does $ N_t \sim N(0, \int^t_0 g(s)^2ds $ follow from this, or is it a completely separate issue? $\endgroup$ – strwars Apr 4 '19 at 20:24
  • $\begingroup$ It's a separate issue. You can see it's gaussian by writing it as a limit in probability of Riemann type sums. Then computing the mean and variance is straightforward, remembering the Ito isometry. $\endgroup$ – Rhys Steele Apr 4 '19 at 20:27
  • $\begingroup$ Sorry to bother you again @RhysSteele , but what would the sum contain exactly? I figure it would be along the lines of the limit of n going to infinity of $\sum^{n-1}_{k=0}[g_k (W_{k+1} - W_k)]? $\endgroup$ – strwars Apr 4 '19 at 21:08
  • $\begingroup$ That depends on whether you meant to have some partition implicit in what you wrote. To be completely clear, it should be of the form $\sum_{k=0}^{n-1} g_{t_k} (W_{t_{k+1}} - W_{t_k})$ where the $t_k$ form a partition of $[0,t]$ and then you consider a sequence of such sums with the mesh size of the partition going to $0$. $\endgroup$ – Rhys Steele Apr 4 '19 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.