Why Geometric Brownian Motion Requires Itô's Lemma to Solve

Summary

The document explains why the stochastic differential equation for a geometric Brownian motion cannot be solved by treating its two sides like ordinary differentials and directly integrating the relative change. In the proposed direct approach, the unknown price process remains inside both the time integral and the stochastic integral, so the expression does not isolate the terminal price.

The answer instead applies a logarithmic transformation and Itô's lemma. This changes the dynamics of the log price into a drift term adjusted by half the variance, plus a Brownian motion term; those transformed dynamics can then be integrated. The explanation motivates the transformation by showing how it removes the price process from the integrands. It is a brief conceptual account rather than a full derivation: it does not spell out initial-value terms or the full resulting price distribution, and the displayed direct-integration expression should not be read as a valid solution.

Key ideas

• Direct integration leaves the unknown price process inside the integrals.
• A logarithmic transformation changes the equation into dynamics for the log price.
• Itô's lemma introduces a drift adjustment involving the volatility squared.
• The transformed equation can be integrated because its terms no longer contain the price process as an integrand.

Tags

# Integrating log-normal


# Integrating log-normal












The usual log normal model in differential form is:

$dS = \mu S dt + \sigma S dX$

where $dX$ is the stochastic part, so

$\frac{dS}{S} = \mu dt + \sigma dX$ (1)

and we normally solve this by subbing in $Y=\log(S)$. What's to stop us just integrating (1) to get

$S = \exp\left(\mu t + \sigma\mathcal{N}(0,1)\right)$ ?

Why do we have to through all the business of subbing in for $Y$ and using Ito's lemma?

## Answer by SRKX (score 3)

https://quant.stackexchange.com/a/4221

What you have to start with is:

$$dS_t=\mu S_t dt + \sigma S_t dW_t$$

where $W_t$ is a standard brownian motion (SBM).

You want to solve for $S_t$, so how would you proceed?

If you integrate both sides of the equation between 0 and $T$, you get:

$$S_T - S_0= \mu \int_0^T S_t dt + \sigma \int_0^T S_t dW_t$$

Okay and then what? The fact that you have $S_t$ in both integral is problematic.

The thing is, to solve for $S_t$, you in fact need to use a bit of trickery, and the substitution and the application of Ito's lemma allows you to get rid of the $S_t$ as you get:

$$dY = d(\ln S_t)=\left(\mu - \frac{\sigma^2}{2} \right) dt +\sigma dWt$$

The integration afterwards is straightforward.

So, you use the trick to get rid of the $S_t$ in the integrals an to be able to solve easily.

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