Finding Moments of an Exponential Brownian Motion

Summary

The document asks how to calculate the expectation and variance of a smooth function of Brownian motion, focusing on the exponential of the process. The key step is to use the distribution of Brownian motion at a fixed time: when it starts at zero, its value is normally distributed with mean zero and variance equal to elapsed time. Applying the normal moment-generating formula to the exponential gives the expectation of the transformed variable.

The replies also identify the exponential of a normally distributed variable as lognormally distributed, so its moments can be obtained from the lognormal distribution. These approaches explain how to proceed without deriving the full stochastic differential equation again. The stated expectation assumes a zero initial Brownian value and conditions on the starting time. Although the question asks for both expectation and variance, the response excerpt gives the expectation explicitly and points to known lognormal moments for the remaining calculation; it does not discuss more general initial conditions or other smoothing functions.

Key ideas

• At a fixed time, Brownian motion started at zero is normally distributed with mean zero and variance equal to time.
• The exponential moment formula for a normal variable yields the expectation of the exponential Brownian transform.
• Exponentiating a normal random variable produces a lognormal random variable whose moments are known.
• The stated calculation assumes a zero starting value and does not develop the variance explicitly.

Tags

# How to calculate expectation and variance of smooth function applied to brownian motion


# How to calculate expectation and variance of smooth function applied to brownian motion












I applied a smoothing function to a Brownian equation and obtained a stochastic differential equation by using Ito's lemma. The smoothing function is exp(Bt).

How do I get the expected value and variance of this function? Just looking for the required approach rather than a full fledged solution.

## Answer by Marco (score 2)

https://quant.stackexchange.com/a/61213

For a Normally distributed random variable, $X$, with mean $\mu$, and variance $V$, the following is true:

\begin{equation} \mathbb{E} \{\exp(\theta X)\} = \exp\left(\theta\mu+\frac{1}{2}\theta^2V\right) \end{equation} In your example, conditional upon time $0$, and assuming $B(0)=0$, $B(t)$ is Normally distributed variable with zero mean and variance $t$. Applying this formula with $\mu = 0$, $V=t$ and $\theta=1$, we find:

\begin{equation} \mathbb{E} \{\exp(B(t))\}=\exp \left( \frac{1}{2}t \right) \end{equation}

## Answer by tcpedersen (score 0)

https://quant.stackexchange.com/a/61218

As $B(t)$ follows a normal distribution with mean zero and variance $t$, then $\exp\{B(t)\}$ will (by definition) follow a log-normal distribution with parameters $\mu=0$ and $\sigma^2=t$. The moments of the log-normal distribution are known.

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