Expected value of Brownian motion when it is less that a given number:$E[W_t\mathbb{1}_{(W_t \leq a)}] $

Question

I want to find $E[W_t\mathbb{1}_{(W_t \leq a)}] $, where $W_t$ is Brownian motion and $a \in \mathbb{R}$. I thought that since $W_t \sim N(0,t)$, that its pdf would be $\frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}$, and tried to use this to obtain: $$\begin{aligned} E[W_t\mathbb{1}_{(W_t \leq a)}] & = \int_{-\infty}^a\frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}\\ & = \int_{-\infty}^{a/\sqrt{t}}n(y) dy\\ & = N(a/\sqrt{t}) \end{aligned}$$ Where I've used the change of variable $y = x/\sqrt{t}$, and $n(x)$ is standard normal distribution, and $N(x)$ is its CDF. I have been told that this is wrong but I'm not sure why, or how to do it correctly. Thanks!

Answer 1

What you have calculated is $EI_{{W_t} \leq a}$ and not $EW_tI_{{W_t} \leq a}$. Multiply the integrand by $x$ and then integrate.

Answer 2

For the sake of completeness, I posted my solution below:

\begin{align*} \mathbb{E}[W_t 1_{\{W_t\leq a\}}] & = \int_{-\infty}^a \frac{x}{\sqrt{2\pi t}} e^{-\frac{x^2}{2t}}dx \\ & = \frac{-\sqrt{t}}{\sqrt{2\pi}} \int_{-\infty}^a \frac{-x}{t} e^{-\frac{x^2}{2t}}dx \\ & = -\sqrt{\frac{t}{2\pi}} e^{-\frac{a^2}{2t}} \end{align*}