Using Itô’s Lemma to Find the Drift of an Exponential Brownian Process

Summary

The document derives the drift of a process defined as the exponential of a Wiener process. Applying Itô’s lemma to the exponential function produces both a stochastic term proportional to the process and a time-dependent correction from the second derivative. Since the squared Brownian increment contributes a time increment, the drift is one half of the process value, while the diffusion coefficient equals the process value.

The answer confirms the proposed derivation and gives a broader result for an exponential process with constant time growth and a scaled Brownian component. In that general form, the drift combines the direct time-growth parameter with half the squared diffusion scale, multiplied by the process itself. This is a concise mathematical explanation of Itô’s correction; it assumes the standard Wiener process and does not cover changes of measure, alternative stochastic dynamics, or applications to a particular trading model.

Key ideas

• Itô’s lemma adds a drift correction when a nonlinear function is applied to Brownian motion.
• For the exponential of a standard Wiener process, the drift is half the process value.
• The stochastic coefficient for that process equals its current value.
• An exponential process with constant growth and scaled Brownian motion has drift equal to growth plus half the squared scale, times the process.

Tags

# Calculation of a process's drift


# Calculation of a process's drift












> Let $X_t:=e^{W_t}$ where $W_t$ follows the Wiener process. Calculate the drift.

The answer is given as $X_t/2$. My attempt at a solution (which I'm afraid is poor from a mathematical standpoint):

I applied Ito's lemma as $$dX_t=\frac{\partial X_t}{\partial W_t}dW_t+\frac{1}{2}\frac{\partial^2 X_t}{\partial W_t^2}(dW_t)^2$$ and using the fact that $(dW_t)^2=dt$, we get: $$dX_t=\frac{e^{W_t}}{2}dt+e^{W_t}dW_t$$ Therefore the drift is indeed $X_t/2$.

Is my derivation correct? I would appreciate any input on that.

## Answer by Daneel Olivaw (score 3, accepted)

https://quant.stackexchange.com/a/53059

Your solution is correct. Generally speaking, for any $\alpha,\beta\in\mathbb{R}$, the drift $\mu_{X^{\alpha\beta}}$ of the process: $$X_t^{\alpha\beta}:=e^{\alpha t+\beta W_t}$$ will be equal to: $$\mu_{X^{\alpha\beta}}=\left(\alpha+\frac{\beta^2}{2}\right)X_t^{\alpha\beta}$$

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