# Stochastic differential equation of a Brownian Motion

I have two questions about Ito's Lemma with respect to calculating SDEs. The examples are simple enough, but I haven't found an answer yet.

Take $W_t$ as a standard Brownian motion and $g(s)$ as some function of $s$. Assume that all regularities etc. are fulfilled and take $F$ as some function. I know that if $F = \int_0^tg(s)dW_s$, then the corresponding SDE is $dF = g(t)dW_t$. However, applying Ito's Lemma, I'm not sure how this SDE is derived. I am unsure about the next part:

• $dF = \underbrace{\frac{\partial F}{\partial t}}_{=g(t)dW_t}dt + \underbrace{\frac{\partial F}{\partial W_t}}_{=g(t)}dW_t + \underbrace{\frac{1}{2}\frac{\partial^2 F}{\partial W_t^2}}_{=0}dt = g(t)dW_tdt + g(t)dW_t = g(t)dW_t$

Question 1: is $\frac{\partial F}{\partial t} = g(t)dW_t$ correct? Or should this be zero?

Now take $F=\int_0^tW_s^2dW_s$. My approach would be:

• $dF = \underbrace{\frac{\partial F}{\partial t}}_{=W_t^2dW_t}dt + \underbrace{\frac{\partial F}{\partial W_t}}_{=W_t^2}dW_t + \underbrace{\frac{1}{2}\frac{\partial^2 F}{\partial W_t^2}}_{=W_t}dt = W_t^2dW_t+W_tdt$

Question 2: Are the partial derivatives in the above example correct?

In stochastic calculus, only stochastic integrals are defined. The differential form is just a notation. That is, $$dF=g(t)dW_t$$ is just another expression for the integral $$F=\int_0^t g(s) dW_s.$$ See, for example, in this book or this book, all Ito's lemmas are expressed in integral forms.
For your question, note that $F$ is not a function of $t$ and $W_t$, that is, it is not of the form $F(t, W_t)$. In fact, it depends on the whole path of $W_s$ from $0$ to $t$. Then Ito's lemma can not be applied to $F$. The application for both of your questions are incorrect.
• To make sure I get it: now take $F_t=\int_0^tW_t^2W_s^2dW_s$. My approach would be: $F_t = \underbrace{W_t^2}_{X}\underbrace{\int_0^tW_s^2dW_s}_{Y} = X_tY$; $dF = \frac{\partial F}{\partial X_t}dX_t + \frac{\partial F}{\partial Y_t}dY_t = Y(dt+2W_tdW_t) + X_t(W_t^2dW_t)$ $dF = \int_0^tW_s^2dW_s(dt+2W_tdW_t) + W_t^4dW_t$ – Kami Nov 22 '16 at 21:05
• @Kami; This should be another question, but since you asked, I won't be mind providing you some clue. Note that $F_t = X_t Y_t$, then$dF_t = Y_tdX_t+X_t dY_t + d[X, Y]_t = Y_t(dt+2W_tdW_t) + X_t W_t^2 dW_t + 2W_t^3 dt$. – Gordon Nov 22 '16 at 21:12