Method to evaluate $\mathbf{E}[B(t)\cdot e^{B(t)}]$, where $B(t)$ is a standard Brownian motion

Question

Let $\xi$ be a normally distributed random variable with mean $0$ and variance $1$, and that we have the formula

$$\begin{equation} \mathbf{E}[e^{\lambda \xi}]=e^{\frac{1}{2} \lambda^2}, \hspace{3mm} -\infty<\lambda <\infty \end{equation} $$

I will like to evaluate $\mathbf{E}[B(t)\cdot e^{B(t)}]$, where $B(t)$ is a standard Brownian motion.

Am I allowed to do the following:

$$\begin{equation} \mathbf{E}[B(t)\cdot e^{B(t)}] = \mathbf{E}[B(t)]\cdot \mathbf{E}[e^{B(t)}]? \end{equation} $$

I'm still new to the topics on Brownian motion, so some clarification will be deeply appreciated.

Answer 1

You are right, $B(t)\sim N(0,t)$. But no, you are not allowed to pretend that $\mathbf E(B(t)\exp(B(t))=\mathbf E(B(t))\mathbf E(\exp(B(t))$ because the random variables $B(t)$ and $\exp(B(t))$ are not independent. (In fact, they are positively correlated, so your proposal would give the wrong answer.)

Instead use the fact that $\mathbf E\exp(\lambda B(t))=\exp(t\lambda^2/2)$, or rather, that $$\mathbf E(\frac d{d\lambda} \exp(\lambda B(t)) = \frac d{d\lambda} \exp(t\lambda^2/2)=t\lambda\exp(t\lambda^2/2).$$ When evaluated at $\lambda=1,$ you get the desired result.

The justification of the interchange of expectation and differentiation with respect to $\lambda$ is left to you. It does not involve any Brownian-motion specific reasoning, and hardly any specific facts about the normal distribution.

Answer 2

You cannot do what you want since $B(t)\ and\ e^{B(t)}$ are dependent. The basic formula you need is $E=\int_{-\infty}^{\infty} xe^xf(x)dx$, where f(x) is the probability density function for B(t).