Using Itô’s Formula to Integrate the Square of Brownian Motion

Summary

The document derives an identity for the Itô integral of squared Brownian motion. Applying Itô’s formula to the cube of a Brownian motion gives a differential with both a stochastic term and a drift term. Integrating that differential from zero to time t and rearranging yields the stated expression for the integral.

It also shows a separate Itô calculation for the reciprocal of time plus Brownian motion, expressing its dynamics in terms of the process itself. The main derivation assumes standard Brownian motion starting at zero; with a different initial value, the integrated identity would include the corresponding initial cube. The reciprocal example is only formal where its denominator is nonzero, and the document does not discuss how to handle times when that condition fails.

Key ideas

• Applying Itô’s formula to the cube of Brownian motion produces a drift term as well as a stochastic term.
• Integrating the differential and rearranging gives the identity for the integral of squared Brownian motion.
• The derivation assumes Brownian motion starts at zero.
• The reciprocal-process calculation requires its denominator to remain nonzero.

Tags

# How to prove $\int_0^t W_s^2dWs = \frac{1}{3}W_s^3 - \int_0^t W_s ds$ using Ito's formula?


# How to prove $\int_0^t W_s^2dWs = \frac{1}{3}W_s^3 - \int_0^t W_s ds$ using Ito's formula?












Please help me with this problem.

## Answer by Gordon (score 6, accepted)

https://quant.stackexchange.com/a/22131

> As @SRKX suggested, full answers are provided below.

(a). Using Ito's lemma \begin{align*} d\left(W_t^3\right) &= 3W_t^2 dW_t +3W_t dt. \end{align*} Integrating on both sides, we obtain that \begin{align*} W_t^3 = 3\int_0^t W_s^2 dW_s +3\int_0^t W_s ds. \end{align*} Consequently, \begin{align*} \int_0^t W_s^2 dW_s = \frac{1}{3}W_t^3 -\int_0^t W_s ds. \end{align*}

(b). Note that \begin{align*} df(t, W_t) = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial W_t} dW_t + \frac{1}{2}\frac{\partial^2 f}{\partial W_t^2} dt. \end{align*} Then, \begin{align*} dX_t &=-\frac{1}{(t+W_t)^2}dt -\frac{1}{(t+W_t)^2}dW_t +\frac{1}{(t+W_t)^3}dt\\ &=(-X_t^2+X_t^3)dt -X_t^2 dW_t. \end{align*}

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