-1
$\begingroup$

I want to calculate $\int ^t _0 W_tdW_t$

I know that the reasoning is the following:

Let $x(t)=W(t)$ with $a=0$ and $b=1$ in the definition of an Ito Process, and $f(t,x)=x^2$.

Then, applying Ito's formula:

$df=\frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial x}dW+\frac{1}{2}\frac{\partial ^2f}{\partial x^2}dt$

$dW^2=dt+2WdW$

$\int dW^2 = \int dt + 2\int WdW$

$\int WdW=\frac{1}{2}W^2-\frac{t}{2}$

My problem is that I don't understand why $df$ is equal to $dW^2$ since for me it would be $df=d(x^2)=d(W^2)$ and considering that I don't know over which variable I am deriving it, I don't know how the results above are explained.

It may be a really easy question but I am starting to study these things and I want to make sure that I understand the basics of it.

Thanks!

$\endgroup$
1
  • 2
    $\begingroup$ Yes, it is $d(W^2)$ $\endgroup$
    – nbbo2
    Commented Nov 17, 2016 at 20:47

1 Answer 1

6
$\begingroup$

You are "deriving" with respect to $t$ (the time index in your stochastic process).

$f(t,x) = x^2$ so $f(t,W_t) = W_t^2$. And Ito's lemma tells you

$W_b^2 - W_a^2 = \int_{t=a}^b d(W_t^2) = \int_{t=a}^b df(t,W_t) = \int_{t=a}^b 2W_t dW_t + \int_{t=a}^b dt$ for all $0 \le a <= b$.

PS: Actually you are not deriving. The differential notation is just a notation to avoid having to write integrals. Ito's lemma is an identity between stochastic integrals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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