# Calculate $\mathbb{E}\exp \left( B_t- \int_0^t B_s \, ds \right)$ for Brownian motion $(B_t)_{t \geq 0}$

I want to calculate the following expected value: $$\mathbb{E} [e^{-\int_0^{t}B_s \ ds} \cdot e^{B_t}],$$ where $$B_t$$ is a Brownian motion, so $$B_t \sim \mathcal{N}(0, t)$$.

Moreover I know that $$\int_0^{t}B_s \ ds \sim \mathcal{N}(0, \frac{t^3}{3})$$.

The problem is that $$\int_0^{t}B_s \ dt$$ and $$B_t$$ are dependent random variables.

What I can do in this situation to calculate this expected value?

Step 1: Show that $$X_t := B_t - \int_0^t B_s \, ds$$ is Gaussian. To this end, recall that $$(B_t)_{t \geq 0}$$ is a Gaussian process which implies, in particular, that $$B_t - \sum_{j=1}^n B_{t_j} (t_{j+1}-t_j)$$ is Gaussian for any $$t>0$$ and $$t_1 < \ldots . For $$t_j := t \frac{j}{n}$$, we find that $$X_t = \lim_{n \to \infty} \left( B_t - \sum_{j=1}^n B_{tj/n} \frac{1}{n} \right)$$ is Gaussian as a pointwise limit of Gaussian random variables.
Step 2: Determine the mean and variance of $$X_t$$.
Step 3: Use the fact that $$\mathbb{E}e^Y = \exp(\sigma^2/2)$$ for any $$Y \sim N(0,\sigma^2)$$.
• Step1: I do not exactly understand it. I mean it is not clear for me that $B_t-\sum_{j=1}^n B_{t_j} (t_{j+1}-t_j)$ is Gaussian. Step 2: If I assume that $X_t$ is Gaussian, then $\mathbb{E}X_t = 0$ and $Var X_t = t+t^3/3 - t^2$. Here I used the fact that $(B_t, \int_{0}^t B_s ds)$ has two dimensional normal distribution with covariance equal to $t^2/2$. Step 3: Putting variance from Step 2 into formula you wrote me, my answer is: $\mathbb{E}[e^{B_t - \int_{0}^t B_s ds}] = e^{t/2+t^3/6-t^2/2}$. – MathMen Oct 11 '18 at 14:07
• @MathMen Fact 1: $(B_{s_1},\ldots,B_{s_m})$ is Gaussian for any choice of $s_1,\ldots,s_m>0$. Fact 2: If $(Y_1,\ldots,Y_m)$ is Gaussian then $$\sum_{j=1}^m \alpha_j Y_j$$ is Gaussian for any $\alpha_j \in \mathbb{R}$. Combining these two facts, you can easily prove Step 1. Side remark: If you already know that $(B_t,\int_0^t B_s \, ds)$ is jointly Gaussian, then you can use Fact 2 to conclude directly that $B_t - \int_0^t B_s \, ds$ is Gaussian. – saz Oct 11 '18 at 15:39
• Thanks. Is my calculation correct? I have also additional question. What if I would take $|B_t|$ or $B_t^2$ which are not Gaussian instead of $B_t$. How can I then calculate $\mathbb{E} [e^{|B_t| - \int_{0}^t B_s ds}]$ or $\mathbb{E} [e^{B_t^2 - \int_{0}^t B_s ds}]$ – MathMen Oct 11 '18 at 17:17
• @MathMen There is no general recipe... for the second expectation it is actually not even clear that it is finite for all $t$. Regarding your calculations: Yeah, I guess so. – saz Oct 11 '18 at 17:44