Solving Geometric Brownian Motion by Applying Itô’s Lemma to Log Price

For $X_0=x>0$ and constant coefficients, the solution is

X_t=x\exp\!\left((\mu-\tfrac12\sigma^2)t+\sigma W_t\right).

Apply Itô's lemma to

\log X_t

to obtain a constant drift and diffusion, integrate those terms, and exponentiate. Applying Itô's lemma back to the proposed exponential verifies that it solves the original SDE and stays positive.

Why the logarithm removes the level from the SDE

Write $dX_t=\mu X_t\,dt+\sigma X_t\,dW_t$ . For $f(x)=\log x$ , the first derivative is $1/x$ and the second is $-1/x^2$ . Apply the Itô formula where $X_t>0$ :

d\log X_t=\frac{1}{X_t}\,dX_t-\frac{1}{2X_t^2}\,d[X]_t =\left(\mu-\tfrac12\sigma^2\right)dt+\sigma\,dW_t,

because $d[X]_t=\sigma^2X_t^2\,dt$ . Integrating with $W_0=0$ gives the claimed expression.

To avoid relying on positivity in advance, verify the candidate directly. Let

L_t=\log x+(\mu-\sigma^2/2)t+\sigma W_t

and apply Itô's lemma to

e^{L_t}

. The exponential's second derivative adds

\sigma^2e^{L_t}/2

to the drift and cancels the log correction. The resulting equation is the original one. The drift and diffusion are globally Lipschitz, so the candidate is the unique solution. For

x=0

the solution is identically zero. For

x<0

the same formula holds by multiplying the positive solution started at

|x|

-1

The quadratic variation of

X

is the time integral of

\sigma^2X_t^2

. Multiplying by

f''(X_t)=-1/X_t^2

cancels the level again and leaves the negative constant drift correction. Once you integrate the log equation, exponentiating gives a positive process for positive initial value. You can verify it without assuming positivity in advance by applying Itô's formula directly to that candidate exponential. This recovers the original coefficients and the initial condition, which identifies the solution by uniqueness for the linear coefficients.

Evaluate the solution on one observed path

For an original exercise, take $x=120$ , $\mu=0.07$ , and $\sigma=0.30$ . Compute the value at $t=0.5$ on a path with $W_{0.5}=-0.40$ . The log change is

(0.07-0.045)(0.5)+0.30(-0.40)=-0.1075.

Thus

X_{0.5}=120e^{-0.1075}\approx107.77

. You need only the Brownian endpoint for this calculation because the coefficients are constant and the transformed SDE has additive noise. The full path would matter for a payoff depending on the running maximum or minimum. Using

\mu

in the exponential without the adjustment would give

120e^{-0.085}\approx110.22

on the same path and fail the verification step.

For another useful check, take two times $s<t$ and divide the exponential at $t$ by the exponential at $s$ . The ratio depends on $W_t-W_s$ and $t-s$ . This recovers the exact one-step transition and explains how to reuse the solution when simulating on a grid. In the degenerate case $\sigma=0$ , the ratio reduces to the usual deterministic compound-growth factor.

Solving Geometric Brownian Motion by Applying Itô’s Lemma to Log Price

Why the logarithm removes the level from the SDE

Evaluate the solution on one observed path

Change the drift and volatility of geometric Brownian motion

References and next steps