Standard Brownian Motion problem. How can I find the distribution of $X_{t}|X_{s}=1$? 
Let be $X_t$ a Standard Brownian Motion. Then compute:

*

*$\mathbb{E}\left[X_t | X_{s} =1\right]$, with $t>s>0$

*$\mathbb{E}\left[X^{2}_{s+1} | X_{s}=1\right]$

*$\mathbb{P}(X_{3}<3 | X_{1}=1)= \mathbb{F}_{X_{3}|X_{1}=1}(2)$

My Attempt:

*

*For the first bullet point
We know that $X_t \sim N(0,t)$, since $X_t$ is a Standard Brownian Motion. Thus, $\mathbb{E}\left[X_t \right]=0$.
Also, we know that if we define $t=s+r$, then
\begin{align*}
X_{r+s}-X_{s} \sim N(0,s)
\end{align*}
Do you have any hint for me to continue?

*

*For the second bullet point
I think, that we can use this:
\begin{align}
\mathbb{E}\left[X^{2}_{s+1}|X_{s}=1\right]=\underbrace{\mathbb{V}\left[X_{s+1}|X_{s}=1 \right]}_{\text{(1)}}+\underbrace{\mathbb{E}^{2}\left[X_{s+1} | X_{s}=1\right]}_{\text{(2)}}
\end{align}
I could compute (1), only if I knew the distribution of $X_{s+1}|X_{s}=1$. But I don't know how to use the fact that I mentioned in the first bullet. And (2) is direct from the first part of the problem.
How can I know the distribution of $X_{s+1}|X_{s}=1$? Is it correct that equality?

*

*By independent increments, $X_t-X_s \sim N(0, t-s)$ and is independent of $X_s$. So write $X_t = (X_t - X_s) + X_s$ to help compute the conditional expectation.


*Similarly, note $X_{s+1}-X_s \sim N(0, 1)$ is independent of $X_s$. Further, $(X_{s+1}-X_s)^2 = X_{s+1}^2 - 2 X_{s+1} X_s + X_s^2$. This can help you compute the conditional expectation.


*Again, try to write the result in terms of $X_3-X_1$.
