# How to differentiate a brownian motion?

By definition a wiener process cannot be differentiated.

But when we use Ito's lemma on $F = X^2$, where X is wiener process

we have total change in

$$dF = 2XdX + dt$$

How can we calculate $\frac{dF}{dX}$ when by definition it cannot be differentiated? Isin't this contradiction by definition?

• Have you looked at this Wikipedia page? – Bob Jansen Apr 26 '15 at 9:14
• yes...wikipedia doesn't answer my question...see my comment below. Even for writing the integral form, how do I get dF/dX. – Animesh Saxena Apr 27 '15 at 11:11
• I think you are mixing things here. The function $F: x \mapsto x^2$ is differentiable but not the Wiener process itself :) – byouness May 15 '18 at 15:51

## 3 Answers

In order to apply Ito's lemma, your function needs to be a twice-differentiable function. There is no issue with the non-differentiability of the Wiener process. $\frac{dF}{dX}$ involves differentiating F, not the Wiener process X.

Using a simple analogy: instantaneous velocity ($\frac{dD}{dt}$) is the derivative of position (D) over time; what is differentiated is not time, but distance. I believe this is where your confusion stems from.

We write the differential form of Ito formula for simplification. Actually, the differential form for Ito formula $$dF(W(t)) = 2W(t)dW(t) + dt$$ means the integral form for Ito formula, $$\int{dF} = \int{2W(t)dW(t)} + \int{dt}$$ which make sense in mathemaitcs.

• k but for writing that formua where do you get the first term 2W, you get it by doing dF/dX or 2X, so why dF/dX is possible here? – Animesh Saxena Apr 27 '15 at 11:10
• We get $2W(t)$ by using Ito formula. And I think ocstl answered your question. – logistic Apr 27 '15 at 13:29

You are right that a Wiener process can not be differenciated in the conventional way since the derivative in respect to time does not exist. For this reason Ito lemma should be used to integrate and differenciate Brownian or Wiener processes as these are considered ito processes.