Using Itô’s Lemma to Evaluate the Brownian Integral of W dW

Summary

The document clarifies how Itô’s lemma evaluates the stochastic integral of Brownian motion against itself. Taking the function of time and state as the square of the state, and evaluating it at Brownian motion, gives the process W squared. Itô’s lemma then expresses the change in that process as a stochastic term involving twice W dW plus a time increment. Integrating over an interval and rearranging gives the integral of W dW as one half the change in W squared minus one half the elapsed time.

The key conceptual point is that the differential is applied to the composite process f(t, W_t), with time as the process index; it is not an ordinary derivative with respect to an unspecified variable. The answer also cautions that differential notation is shorthand for an identity between stochastic integrals. This is an elementary Brownian-motion result under Itô calculus. The document gives no broader trading application, estimation method, or discussion of other stochastic processes.

Key ideas

• Apply Itô’s lemma to the function that squares the Brownian state.
• The change in squared Brownian motion includes both a stochastic integral and a time term.
• Rearranging the integrated identity gives the value of the integral of W dW.
• The differential refers to the evolving process indexed by time, not ordinary differentiation in an unspecified variable.
• Itô differential notation represents an identity between stochastic integrals.

Tags

# How is the Wiener integral $\int{WdW}$ calculated?


# How is the Wiener integral $\int{WdW}$ calculated?












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!

## Answer by AFK (score 6)

https://quant.stackexchange.com/a/31061

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.

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