# Expectation and Variance of $e^{B_T}$ for Brownian motion $(B_t)_{t \geq 0}$

I'm working through the following problem, and I need a nudge on the variance of the process. I'm almost certain the expectation is correct, but I'm struggling a lot on applying the isometry property and deriving variances for these types of problems.

Let $B=\{B_t,t \in [0,T]\}$ be a Brownian Motion. Consider $X_T=e^{B_T}$. Find the expectation and variance of the process. (Hint: consider $Y_t=e^{B_t-\frac{t}{2}}$)

So, applying Ito to the process $Y_t$, we get

\begin{align*} dY_t=e^{B_t-\frac{t}{2}} \, dB_t &\iff Y_T-Y_0=\int_0^T e^{B_s-\frac{s}{2}} \, dB_s \\ &\iff e^{B_T-\frac{T}{2}}-1=\int_0^T e^{B_s-\frac{s}{2}} \, dB_s \\ &\iff e^{B_T}=e^{\frac{T}{2}}+e^{\frac{T}{2}}\int_0^T e^{B_s-\frac{s}{2}}\, dB_s \end{align*}

Making the expectation simply $e^{\frac{T}{2}}$

Now for the variance, we simply need to focus on the stochastic integral, since the drift has no variance. This is also easily proved by looking at $E[\{X_T-E(X_T)\}^2]$ and seeing that you're just left with $E[\{e^{\frac{T}{2}}\int_0^T e^{B_s-\frac{s}{2}} \, dB_s\}^2$. I make an attempt to apply the isometry formula and just end up with $4e^{2T}-8e^\frac{3T}{2}+4e^T$, which just feels wrong.

Applying Itô's formula to solve this problem is kind of overkill. It is well-known that for any Gaussian random variable $X$ with mean $0$ and variance $\sigma^2$ it holds that

$$\mathbb{E}e^{\xi X} = \exp \left( \frac{1}{2} \sigma^2 \xi^2 \right) \qquad \text{for all \xi \in \mathbb{R}.}$$

Since $B_T \sim N(0,T)$ this implies

$$\mathbb{E}e^{B_T} = e^{T/2} \qquad \text{and} \qquad \mathbb{E}e^{2 B_T} = e^{2T}.$$

Using that $\text{var}(X) = \mathbb{E}(X^2)-(\mathbb{E}(X))^2$ we get

$$\text{var}(e^{B_T}) = \mathbb{E}(e^{2B_T}) - (\mathbb{E}e^{B_T})^2 = e^{2T}-e^{T}.$$

$$e^{B_T} = e^{T/2} + e^{T/2} + e^{T/2} \int_0^T e^{B_s-s/2} \, dB_s$$
\begin{align*} \text{var}(e^{B_T}) &= \mathbb{E} \big[ (e^{B_T} - e^{T/2})^2 \big] \\ &= e^T \mathbb{E} \left| \int_0^T e^{B_s-s/2} \, dB_s \right|^2 \\ &= e^T \mathbb{E} \left( \int_0^T e^{2B_s-s} \, ds \right) \\ &= \int_0^T e^{-s} \mathbb{E}e^{2B_s} \, ds. \end{align*}
In order to calculate the remain integral you have to calculate $\mathbb{E}e^{2B_s}$. Since we have already seen in the first part of my answer that we can easily calculate the variance of $e^{B_t}$ if we know $\mathbb{E}e^{2B_t}$, this approach doesn't make any sense, from my point of view.
• Thanks. The variance is $e^{2T}-e^T$ though, based on what you derived before no? – Archetupon Apr 29 '17 at 16:36