Solving a Geometric Brownian Motion with Time-Varying Volatility

Summary

The document shows how to solve a stochastic differential equation whose process has zero drift and a time-varying adapted volatility coefficient. It applies Itô's lemma to the logarithm of the process, turning the multiplicative equation into an additive one with a stochastic integral and a quadratic-variation correction.

Integrating the resulting equation and exponentiating gives the exponential solution, with the initial condition determining the multiplicative constant. The derivation is a basic building block for stochastic modeling and can help explain why exponential martingales include a variance adjustment. It assumes the stochastic integral is well defined and that the process remains suitable for taking its logarithm; no applications, empirical evidence, or broader modeling limitations are discussed.

Key ideas

• Taking the logarithm converts the multiplicative stochastic equation into an additive one.
• Itô's lemma contributes a negative half variance term to the log process.
• Integrating the log equation and exponentiating recovers the process solution.
• The volatility coefficient may vary over time and be adapted, subject to conditions for the stochastic integral.

Tags

# Solving Stochastic Differential Equation for Geometric Brownian Motion with time-dependent drift


# Solving Stochastic Differential Equation for Geometric Brownian Motion with time-dependent drift












Given the stochastic differential equation:

$$dZ_t = -Z_t \theta_t dB_t, \quad Z_0 = 1.$$

for an adapted process $\theta_t$ and Brownian motion $B_t$, how exactly do I apply Itô's Lemma to obtain:

$$ Z_t = \exp\left(- \int_{0}^{t}\theta_u \;dB_u - \frac{1}{2}\int_{0}^{t}\theta_u^2 \;du\right)? $$

## Answer by Cettt (score 5)

https://quant.stackexchange.com/a/51378

This is the SDE for a geometric Brownian motion with time dependent volatility $\theta_t$. It can be easily solved with the substitution $$ X_t = \log Z_t =: f(Z_t). $$

According to Ito's Lemma we have that \begin{align} dX_t &= d f(Z_t) = \frac{\partial f}{\partial z}(Z_t) \;d Z_t + \frac 12 \frac{\partial^2 f}{\partial z^2}(Z_t) \;\bigl(dZ_t \bigr)^2 = \\[2mm] &= -\frac{1}{Z_t} Z_t \theta_t \; dB_t - \frac{1}{2} \frac{1}{\bigl(Z_t\bigr)^2} \bigl(Z_t \theta_t\bigr)^2 \; dt = \\[2mm] &=-\theta_t \; dB_t - \frac 12 \theta_t^2 \; dt. \end{align}

Therefore, $$ X_t = X_0 - \int_0^t \theta_u \; dB_u - \frac 12 \int_0^t \theta_u^2 \; du, $$ and the result follows by simply applying $\exp$ to both sides.

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