Solving a Geometric Brownian Motion SDE with Deterministic Coefficients
Summary
The document derives the solution to an SDE in which an asset’s drift and volatility are deterministic functions of time. Applying Itô’s lemma to the logarithm of the asset value introduces a second derivative term. The quadratic variation of the diffusion contributes a negative half-variance adjustment to the drift, resolving the apparent extra factor involving the asset level.
Integrating the log process gives the asset value as its initial level multiplied by an exponential containing the integrated adjusted drift and a stochastic integral. This is the time-varying-coefficient form of the geometric Brownian motion solution. The derivation assumes deterministic coefficients and a positive initial value so the logarithm is defined; the note does not discuss broader existence conditions or extensions to state-dependent coefficients.
Key ideas
- Itô’s lemma applied to the logarithm contributes a second derivative correction.
- The diffusion’s quadratic variation cancels the apparent inverse-square asset term.
- The log drift equals the original drift minus half the instantaneous variance.
- Integrating the log process yields an exponential solution with a stochastic integral.
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# How to solve $dX_t = X_t(\sigma_t dW_t + \mu_t dt)$?
# How to solve $dX_t = X_t(\sigma_t dW_t + \mu_t dt)$?
> Solve the SDE $$dX_t = X_t(\sigma_t dW_t + \mu_t dt)$$ where $\sigma_t$,$\mu_t$ are deterministic.
Attempted solution
We have $$dX_t = X_t(\sigma_t dW_t + \mu_t dt)$$ Let $f(x) = \log X$, applying Ito formula we have
$$d \log(X_t) = \sigma_t dW_t X_t \frac{1}{X_t} + \mu_t dt X_t \frac{1}{X_t} + \frac{1}{2}\sigma^2 (-\frac{1}{X_t^2}$$
Cancelling terms we have $$d \log(X_t) = \sigma_t dW_t + \mu_t dt - \frac{1}{2}\sigma^2\frac{1}{X_t^2}$$
This is where I am lost I am pretty sure this is how we solve for this SDE but the $\frac{1}{X_t^2}$ term is throwing me off.
## Answer by user16651 (score 2, accepted)
https://quant.stackexchange.com/a/31147
let $$dY_t=\mu(t,Y_t)dt+\sigma(t,Y_t)dW_t$$ and $f\in \mathbb{C^2}$.By application of Ito's lemma, we have $$df(Y_t)=\frac{\partial f}{\partial y}dY_t+\frac{1}{2}\frac{\partial^2 f}{\partial y^2}d[Y_t,Y_t]$$ where $$\color{red}{d[Y_t,Y_t]=\sigma^2(t,Y_t)dt}$$ We have $$dX_t=\mu_t X_tdt+\sigma_t X_t dW_t$$ thus $$d \ln(X_t)=\frac{1}{X_t}(\mu_t X_tdt+\sigma_t X_t dW_t)+\frac{1}{2}\left(\frac{-1}{X_t^2}\right)(\sigma_t X_t)^2 dt$$ therefore $$d \ln(X_t)=\left(\mu_t-\frac{1}{2}\sigma_t^2 \right)dt+\sigma_t dW_t$$ by integration on $[0,t]$ we have $$\ln\left(\frac{X_t}{X_0}\right)=\int_{0}^{t}\left(\mu_s-\frac{1}{2}\sigma_s^2 \right)ds+\int_{0}^{t}\sigma_s dW_s$$ in other words $$X_t=x \exp\left(\int_{0}^{t}\left(\mu_s-\frac{1}{2}\sigma_s^2 \right)ds+\int_{0}^{t}\sigma_s dW_s\right)$$ where $X_0=x$
Useful link
- What is Ito's lemma used for in quantitative finance.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.