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Applying Itô’s Lemma to Solve the Geometric Brownian Motion SDE

Article Quant Q&A · Author: Alessandro
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Summary

The document explains why the time derivative of the function used to solve the geometric Brownian motion stochastic differential equation is zero. Taking the logarithm of the process gives a function of both time and the process, but this particular function has no explicit time dependence. It is still differentiable with respect to time, and that partial derivative is simply zero.

The answer clarifies that Itô’s lemma applies to functions of time and a stochastic process; a function may belong to this class even when its time derivative contributes nothing. The post states the familiar exponential solution to the GBM equation, but gives no derivation beyond this point about the lemma’s applicability. It is a concise conceptual clarification rather than a full walkthrough of the stochastic calculus or a discussion of the model’s assumptions and limits.

Key ideas

  • The logarithm of the process has no explicit dependence on time.
  • Its partial derivative with respect to time is zero because it is constant in that argument.
  • Itô’s lemma applies to time-and-process functions even when their time derivative vanishes.
  • The post clarifies one step in solving the GBM stochastic differential equation rather than presenting a complete derivation.

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# Solving the SDE for GBM


# Solving the SDE for GBM












Let's assume that we have the following stochastic differential equation:

$dX_t = \mu X_t dt + \sigma X_tdW_t$

and that we have to prove that this is its solution:

$X_t = X_0 \exp\left(\left(\mu -{\frac {\sigma ^{2}}{2}}\right)t+\sigma W_t\right)$

I can solve this thanks to the application of Ito's lemma on this deterministic regular function defined in two variables (time and process),

$f(t, X_t) = \ln(X_t)$

The point is that I can't understand why the partial derivative of that function with respect to t is equal to zero. Why do we use a function defined in t if it is not differentiable in t?

## Answer by Rylan (score 1)

https://quant.stackexchange.com/a/77089

$f$ as you've described it is differentiable in $t$ -- the derivative is just equal to zero.

The larger point is that Ito's lemma applies to a broader class of functions $f(t, X_t)$ -- the one we've chosen happens to have $\frac{df}{dt} = 0$ everywhere.

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.