I am new to Ito-calculus, so please forgive me if the question is stupid.

Let $W(t)$ be a Brownian-Motion and $f(W(t))=W(t)^2$. If I want to calculate the differential $\mathrm{d}f(W(t))$, Ito's lemma yields: \begin{align*} \mathrm{d}f(W(t))&=\frac{\mathrm{d} f(W(t))}{\mathrm{d} W(t)} \mathrm{d}W(t)+\frac{1}{2}\frac{\mathrm{d}^2 f(W(t))}{\mathrm{d} W(t)^2} \mathrm{d}W(t)^2 \\ &=2W(t)\mathrm{d}W(t)+\frac{1}{2}\cdot 2 \cdot \underbrace{\mathrm{d}W(t)^2}_{=\mathrm{d}t} \\ &=2W(t)\mathrm{d}W(t)+\mathrm{d}t \end{align*} Using the integral notation instead of the differential notation I get: \begin{align} W(t)^2-W(0)^2&=2\int_0^tW(u)\mathrm{d}W(u)+\int_0^t\mathrm{d}t \\ W(t)^2&=2\int_0^tW(u)\mathrm{d}W(u)+t \end{align} I assume that my calculations are correct up to this point. However, I wonder how we deal with functions that depend on $\mathrm{d}W(t)$. For instance, $$ W(t)\mathrm{d}W(t) $$ From my understanding, Ito's lemma only holds for functions involving $W(t)$ and not $\mathrm{d}W(t)$. From the calculations above I conclude that: $$ W(t)\mathrm{d}W(t)=\frac{1}{2}W(t)^2-\frac{t}{2} $$ But how do we to calculate $W(t)\mathrm{d}W(t)$ directly ? Did I misunderstood something or is there another way on how to approach this type of functions ?

Thanks in advance.


Ito's lemma is for twice differentiable functions of the form $f(t,W(t))$. You speak of $W(t)dW(t)$ - this is informal notation and doesn't have a mathematical meaning. Although once you put the integral sign, it becomes mathematically precise. So there's nothing known as $W(t)dW(t)$, but $I(t)=∫_0^{t}W(u)dW(u)$ is well defined via the definition of an Ito integral.

The more fundamental way to calculate this integral is to break up the time axis into slices and sum up the pieces, :

$W(t_j)[W(t_{j+1})-W(t_j)]$ across all $j$, and taking the limit as number of slices become infinite. This is a standard proof you can look up.

  • $\begingroup$ Ok got it. I thought that $W(t)dW(t)$ is a short notation for $\int_0^tW(u)dW(u)$, isn't it? $\endgroup$
    – Lars
    Jul 22 at 17:43
  • 2
    $\begingroup$ No, that is not right. $\endgroup$
    – Arshdeep
    Jul 22 at 17:44
  • 1
    $\begingroup$ The statement that $\int_0^t W_s dW_s = \frac{1}{2} W_t^2-\frac{1}{2}t$ is correct (see (7) (8) {9} on page 4 here math.nyu.edu/~goodman/teaching/DerivSec10/notes/week6.pdf But you left out the integral sign on the left side in your equation. $\endgroup$
    – noob2
    Jul 22 at 23:06
  • 1
    $\begingroup$ Q1: Is the notation correct? - Yes Q2. - Do I have to write it as $∫_0^{t}dW(u)2=∫_0^{t}W(u)dW(u)+∫_0^{t}du$ ? - Also yes, this is ultimately the formal form of the short hand notation (although you can work with the differential form as if it were mathematically precise). This doesn't mean your previous assertion (first comment) is true. $\endgroup$
    – Arshdeep
    Jul 23 at 16:07
  • 1
    $\begingroup$ Yes, I think you got the signs mixed up but more or less the idea is right. Assuming your integral form is true, the correct differential form is: $W(t)dW(t)=1/2(dW(t)+dt)$ $\endgroup$
    – Arshdeep
    Jul 24 at 8:22

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.