Skip to content
← All questions

What is the closed-form value of the Itô integral ∫\int ₀ᵀ W(t)W(t)dW(t) for standard Brownian motion?

The closed-form value of the Itô integral of a standard Brownian motion with respect to itself is 12W(T)2−12T\frac{1}{2}W(T)^2 - \frac{1}{2}T. This result is derived by applying Itô's lemma to W2W^2, integrating the resulting differential identity over the interval [0,T][0, T], and using the initial condition W(0)=0W(0)=0.

Conditions

  • W(t)W(t) is a standard Brownian motion.
  • The integral is evaluated from 00 to TT.
  • The integral is interpreted in the Itô sense.

Reasoning, step by step

  1. Apply Itô's lemma to the square of Brownian motion to get the differential identity d(W2)=2W dW+dtd(W^2)=2W\,dW+dt.
  2. Integrate both sides of the identity over the interval from 00 to TT, yielding W(T)2−W(0)2=2∫0TW dW+TW(T)^2-W(0)^2=2\int_0^T W\,dW+T.
  3. Use the standard initial condition W(0)=0W(0)=0 to simplify the equation to W(T)2=2∫0TW dW+TW(T)^2=2\int_0^T W\,dW+T.
  4. Rearrange the equation and divide by 2 to isolate the stochastic integral, resulting in ∫0TW dW=12W(T)2−12T\int_0^T W\,dW=\frac{1}{2}W(T)^2-\frac{1}{2}T.

Example

The video explicitly derives this result, stating: 'Rearranging and dividing by 2 yields the main result ∫0TW dW=12W(T)2−12T\int_0^T W\,dW=\frac12 W(T)^2-\frac12 T.'

Common misconceptions

  • Applying classical calculus rules and expecting the result to be simply 12W(T)2\frac{1}{2}W(T)^2, ignoring the extra −12T-\frac{1}{2}T term caused by the non-zero quadratic variation of Brownian motion.
  • Assuming the integral evaluates to a deterministic constant rather than a random variable dependent on the terminal value W(T)W(T).

Watch the explanation

Connected concepts

Explore next

Related questions

Understand why

↗
Find a method

↗
Find a method

↗
Meet the concept

↗
Meet the concept

↗

Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.