Expected value of a standard Brownian motion

Question

I'm working through a problem set on Brownian motion for a university course. The question comes in three parts. I understand the first part, but it's the second and third parts I'm completely stuck on. In particular, I'm confused when working out the expected value of the square of a standard Brownian motion and the probability of a value in a standard Brownian motion being below a certain value.

The first part is:

S​uppose $X_t$ is a standard Brownian motion. What is $E[X_t|X_s=1]$ for $t>s>0$?

Which I understand to be equal to $X_s$, which is $1$.

The second part, which I'm stuck on, asks:

W​hat is $E[X_{s+1}^2|X_s=1]$ for $s>0$?

The third part, which I'm also stuck on, asks:

G​iven $X_1=1$, what is the probability of $X_3<2$?

Can any one help with the solution to this and explain how to figure it out?

Thanks in advance.

Answer 1

You have that $\mathrm{E}[X^2_t]=\mathrm{Var}[X_t]=t$ when $X$ is an standard Brownian motion, also its easy to check that $\mathrm{E}[(X_t-X_s)^2|\mathcal{F}_s]=\mathrm{E}[X_t^2-X_s^2|\mathcal{F}_s]=t-s$ when $0\leqslant s<t$ as $X_t-X_s\sim \mathrm{N}(0,t-s)$ is independent of $\mathcal{F}_s:=\sigma (X_r, r\in[0,s])$, therefore

$$ \begin{align*} \mathrm{E}[X_{s+1}^2|X_s=1]&=\mathrm{E}[X_{s+1}^2-X_s^2+X_s^2|X_s=1]\\ &=\mathrm{E}[X_{s+1}^2-X_s^2|X_s=1]+1\\&=s+1-s+1\\&=2 \end{align*} $$ For the last part you have that

$$ \begin{align*} \Pr [X_3<2|X_1=1]&=\Pr [X_3-X_1+X_1<2|X_1=1]\\ &=\Pr [Z+1<2|X_1=1]\\ &=\Pr [Z<1]\\ &\approx 0.76 \end{align*} $$

as $Z$ is independent of $X_1$ for $Z:=X_3-X_1\sim \mathrm{N}(0,2)$.