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Using Itô’s Lemma to Derive the SDE of a Squared Brownian Motion

Article Quant Q&A · Author: MikeHeimlich
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Summary

The document derives the stochastic differential equation for a process defined as a constant times the square of Brownian motion. Applying Itô’s lemma to the function of time and the Brownian state produces a first-order stochastic term and a drift term from the second derivative. The resulting drift is the constant coefficient multiplied by time’s differential, while the random term scales with the current Brownian value and its increment.

The key identity is that the square of a Brownian increment contributes a time increment in Itô calculus. This second-order term explains why the differential includes a nonzero drift even though Brownian motion itself has no drift. The example illustrates a standard stochastic-calculus rule; it does not discuss market data, option pricing applications, or empirical validation, and it assumes the process is driven by standard Brownian motion.

Key ideas

  • Itô’s lemma applies to a squared Brownian process by differentiating its defining function.
  • The second derivative creates a drift term through the quadratic variation of Brownian motion.
  • The stochastic part of the differential is proportional to the current Brownian value.
  • The derivation assumes standard Brownian motion and does not provide a trading application.

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StatisticsDerivatives pricing
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# Problem finding correct SDE for Stochastic Process


# Problem finding correct SDE for Stochastic Process












I am really struggling to come up with the correct SDE for the stochastic process:

$Y(t) = a[Z(t)]^2$

where $Z(t)$ is a Brownian Motion. According to my Prof, the SDE is:

$dY(t) = adt + 2aZ(t)dZt $

Can anyone explain how he got to that solution? Thanks in advance, any help appreciated

## Answer by Sanjay (score 3, accepted)

https://quant.stackexchange.com/a/45549

If you apply Ito to $Y_t=aZ_t^2$ it simple to arrive at $dY_t$:

$$ f(t,z):=az^2 \\ dY_t = f_t(t,Z_t)dt+f_z(t,Z_t)dZ_t+\frac{1}{2}f_{zz}(t,Z_t)*[dZ_t]^2 \\ dY_t = 0*dt+ 2aZ_t dZ_t+\frac{1}{2}2a[dZ_t]^2 \\ dY_t = 2aZ_t dZ_t+adt \\ $$ Which is your desired result. Recall that $dZ_t^2=dt$.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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