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

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.

• Do you mean $\exp\{B(t)\}$ where $B(t)$ is Brownian motion? Feb 17, 2021 at 9:31
• Yes, exactly that Feb 17, 2021 at 10:49

For a Normally distributed random variable, $$X$$, with mean $$\mu$$, and variance $$V$$, the following is true:
$$$$\mathbb{E} \{\exp(\theta X)\} = \exp\left(\theta\mu+\frac{1}{2}\theta^2V\right)$$$$ In your example, conditional upon time $$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$$, we find:
$$$$\mathbb{E} \{\exp(B(t))\}=\exp \left( \frac{1}{2}t \right)$$$$
• Maybe add somewhere that the question has the case $\theta=1$ for completeness. Feb 17, 2021 at 15:51
As $$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.