Identifying a Gaussian Distribution from Its Characteristic Function

Summary

The document considers a stochastic integral of a deterministic volatility function against Brownian motion and asks how its characteristic function establishes the distribution of the resulting random variable. The computed characteristic function has the form of the characteristic function of a centered Gaussian, with variance equal to the integral of the squared volatility over the time interval.

The answer applies uniqueness: a characteristic function determines a probability distribution, and the displayed expression matches the known Gaussian characteristic function with that variance. Therefore the stochastic integral is normally distributed with mean zero. This conclusion relies on the deterministic integrand and Brownian-motion setup stated in the exercise; the same distributional result need not follow for a random integrand or a different driving process.

Key ideas

• A characteristic function uniquely determines a probability distribution.
• The characteristic function of a centered Gaussian has an exponential quadratic form in its argument.
• Matching the stochastic integral’s characteristic function to that form identifies its variance.
• With a deterministic integrand against Brownian motion, the integral is Gaussian with mean zero.

Tags

# Characteristic function and distribution of a random variable


# Characteristic function and distribution of a random variable












This is exercise 4.3 in Bjork, Arbitrage Theory in Continous Time. $$ X_t = \int^t_0 \sigma(s)dW_s $$ $\sigma$ is a deterministic function and $W_t$ is brownian motion.

I am asked to find the characteristic function of $X_t$ and thus showing that $X_t$ is normally distributed with mean zero and variance $\int^t_0 \sigma^2(s)ds$

I have found the characteristic function to be: $$ E[e^{iuX_t}]= \exp \left[-u^2/2 \int^t_0 \sigma^2(s)ds \right] $$ How Can I conclude that $X_t$ is normally distributed then?

## Answer by Richi Wa (score 2, accepted)

https://quant.stackexchange.com/a/42985

The characteristic function (chf) defines the distribution function in a unique correspondence. For $X$ Gaussian with mean $0$ and variance $\sigma^2$ the chf $E[e^{i u X}]$ which is given by $$ e^{-\sigma^2 u^2/2}. $$ Thus if you identify the variance term then you are done. The characteristic function is the one of the Gaussian distribution. Thus the random variable $X_t$ is Gaussian.

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