Time-Varying Volatility in Geometric Brownian Motion

Summary

The document explains how to find the distribution of a geometric Brownian motion whose volatility changes over time, using volatility proportional to time. Since the asset value is the exponential of a Gaussian process, its logarithm is normally distributed and the asset value is lognormally distributed. The log distribution’s mean is the drift term less half the accumulated variance; its variance is the integral of squared volatility over time.

Substituting the specified volatility function into those integrals gives the parameters needed to calculate probabilities, such as the chance that the value at a future time is below a threshold. The answer describes converting that probability into a standard normal probability. It does not carry out the integrals or provide a numerical probability, and it leaves the drift parameter unspecified. The result also relies on the stated Gaussian process model and volatility specification.

Key ideas

• The exponential of a Gaussian process has a lognormal distribution.
• The mean of the log value includes drift minus one half of integrated variance.
• The log variance equals the integral of squared time-varying volatility.
• Threshold probabilities can be calculated through the corresponding normal distribution.

Tags

# Distribution of Geometric Brownian with time-dependant volatility


# Distribution of Geometric Brownian with time-dependant volatility












The process $S(t) =\exp\left(\mu.t + \int_0^t\sigma(s) \text{d}W(s) - \int_0^t \frac{1}{2}\sigma^2(s)\text{d}s\right)$ where $\sigma(s) = 0.03s$ is log-normally distributed, but i'm not sure about the parameters of this log-normal distribution. Can someone help me to find them ?

Based on this, how can I calculate some probabilities like the following one, for instance : $P[S(15) < 1]$

Thanks for your contributions !

## Answer by Valter (score 1)

https://quant.stackexchange.com/a/77694

Recognize that the process $S(t)$ is log-normally distributed because it is an exponential of a Gaussian process $X(t)$, in this case $S(t)=e^{X(t)}$.

To find the parameters of the log-normal distribution, you need to calculate the mean and variance of $X(t)$,as the mean and variance of $ln(S(t))$ are those of $X(t)$. We have $$ E(X(t)) = \mu t - \frac{1}{2}\int_0^t \sigma^2(s)ds, \quad Var(X(t)) = \int_0^t\sigma^2(s)ds $$ Substitute $\sigma(s)=0.03s$ and solve.

Now that you know the mean and variance, we can find the probabilities using the log-normal distribution with these parameters. This is done by transforming the problem into a standard normal distribution problem, as the log-normal distribution of $S(t)$ is defined by the normal distribution of $X(t)$.

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