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Applying Itô’s Formula to an Exponential of Brownian Motion Integrals

Article Quant Q&A · Author: Lior
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Summary

The document works through an Itô-calculus exercise involving two independent Brownian motions. It defines a process as their product minus one half of the time integral of the sum of their squared values, then differentiates that process using the product rule for stochastic differentials. Independence makes the cross-variation term between the Brownian motions vanish.

The solution computes the process’s quadratic variation from its two stochastic-integral components and applies Itô’s formula to its exponential. The drift term from the exponential’s second derivative cancels the finite-variation term in the original process, leaving a stochastic differential expressed in terms of the exponential and the two Brownian motions. This illustrates a useful technique: when a process contains an ordinary time integral, include that integral in the definition of an auxiliary process and differentiate it before applying Itô’s formula. The document gives an algebraic derivation but does not include the original exercise statement or discuss broader conditions for the stochastic integrals.

Key ideas

  • The solution combines the Brownian product and its compensating time integral into one process.
  • The product rule includes stochastic terms from each Brownian motion and a quadratic covariation term.
  • Independence of the Brownian motions makes their cross-variation vanish.
  • The process’s quadratic variation is the time integral of the sum of the squared Brownian values.
  • Applying Itô’s formula to the exponential cancels the drift terms and leaves a stochastic differential.

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# How to tackle this exercise about Ito's formula?


# How to tackle this exercise about Ito's formula?












In the following exercise, I can't get started on question 2) as I am not sure what to do when there is an integral inside:

Could you help me out?

## Answer by Gordon (score 6)

https://quant.stackexchange.com/a/18370

Let \begin{align*} X_t = W(t)W_*(t) - \frac{1}{2}\int_0^t\big(W_*(u)^2+ W(u)^2\big)du. \end{align*} Then, \begin{align*} dX_t &= W(t) dW_*(t) + W_*(t) dW(t) -\frac{1}{2}\left(W_*(t)^2+ W(t)^2\right)dt, \end{align*} as $W$ and $W_*$ are independent. Consequently, \begin{align*} X_t = \int_0^t \big[W(s) dW_*(s) + W_*(s) dW(s)\big] -\frac{1}{2}\int_0^t\left(W_*(s)^2+ W(s)^2\right)ds. \end{align*} Moreover, \begin{align*} \langle X, X\rangle_t = \int_0^t\left(W_*(s)^2+ W(s)^2\right)ds. \end{align*} That is, \begin{align*} d\langle X, X\rangle_t &= \left(W_*(t)^2+ W(t)^2\right)dt. \end{align*} Since $R_2(t) = e^{X_t}$, \begin{align*} dR_2(t) &= e^{X_t} dX_t + \frac{1}{2}e^{X_t}d\langle X, X\rangle_t\\ &= R_2(t)\Big( W(t) dW_*(t) + W_*(t) dW(t) \Big). \end{align*}

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