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Using Itô’s Lemma to Evaluate a Brownian Stochastic Integral

Article Quant Q&A · Author: Sanjay
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Summary

The note shows how Itô’s formula can turn an integral involving a function of Brownian motion into a difference of endpoint values plus a time integral. Its example applies the formula to the cube of a shifted Brownian motion, producing a stochastic differential with both a Brownian increment term and a drift term. Integrating over time gives the stated identity and illustrates why the second derivative contribution cannot be omitted.

The discussion is a brief worked hint rather than a full derivation of the original integral or a general treatment of stochastic integration. The question’s initial setup appears inconsistent with the example’s integrand, so readers should check that the target expression matches the function to which Itô’s formula is applied. No empirical evidence or trading application is provided; the value is mathematical technique relevant to stochastic models.

Key ideas

  • Itô’s formula applied to a cubic function of Brownian motion generates both stochastic and time-integral terms.
  • Integrating the differential over an interval yields a relationship between endpoint values and two integrals.
  • The second derivative term in Itô’s formula contributes to the drift and cannot be ignored.
  • The example is a mathematical illustration, and its connection to the question’s stated integrand should be checked.

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# Brownian motion. Solve stoc. integral by using Ito's lemma


# Brownian motion. Solve stoc. integral by using Ito's lemma












I want to show that following statement is true by using Ito's lemma to solve stochastic integrals:

I define the functions in Ito's model: a()=0, b()= (2wt-2)^2. f(t)=Integrate[(2wt-2)^2]

Then df=(b^2/2)(d^2/dwt^2)+(bdf/dst). But it doesn’t add up. How do I show it by using Ito's lemma?

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/22157

Try Ito's formula for $(2W_t+1)^3$, and then integrate. More specifically, note that \begin{align*} d\left( (2W_t+1)^3 \right) &= 6(2W_t+1)^2 dW_t + 12 (2W_t+1) dt, \end{align*} then \begin{align*} (2W_T+1)^3 - 1 = \int_0^T 6(2W_t+1)^2 dW_t + 12 \int_0^T (2W_t+1) dt. \end{align*} The remaining is obvious.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.