# Show that an SDE is Satisfied by a Given Process

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.

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 $$dN_t = g(t) dW_t$$.
• 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 4 '19 at 20:24
• 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 $\sum^{n-1}_{k=0}[g_k (W_{k+1} - W_k)]? – strwars Apr 4 '19 at 21:08 • That depends on whether you meant to have some partition implicit in what you wrote. To be completely clear, it should be of the form$\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 4 '19 at 21:56