1
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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) $ $dF = \int_0^tW_s^2dW_s(dt+2W_tdW_t) + W_t^4dW_t$ $\endgroup$ – Kami Nov 22 '16 at 21:05
  • $\begingroup$ @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$dF_t = Y_tdX_t+X_t dY_t + d[X, Y]_t = Y_t(dt+2W_tdW_t) + X_t W_t^2 dW_t + 2W_t^3 dt$. $\endgroup$ – Gordon Nov 22 '16 at 21:12

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.