1

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.

strwars
  • 271

1 Answers1

1

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 $dN_t = g(t) dW_t$.

Rhys Steele
  • 19,671
  • 1
  • 19
  • 50
  • 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? – strwars Apr 04 '19 at 20:24
  • 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. – Rhys Steele Apr 04 '19 at 20:27
  • 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)]? – strwars Apr 04 '19 at 21:08
  • 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 $0$. – Rhys Steele Apr 04 '19 at 21:56