Partition of Brownian motion I would like to see a solution to the following problem if possible at all. Unfortunately I have no idea on how to proceed on solving it. Any help is appreciated.
Let $\{B(t); t\geq 0\}$ represent standard Brownian motion. Assume all paths of $\{B(t); t\geq 0\}$ are continuous on $[0, \infty)$. 
a. Determine, given a fixed t>0 and a partition $0=t_0 < t_1< t_2< ...< t_n=t$ of $[0, \infty)$, the distribution of the random variable $\sum_{k=1}^n B(t_{k-1})(t_k-t_{k-1}))$.
b. Given a fixed t>0, determine the distribution of $\int_0^t B(s)ds$. 
Hint: This integral is well -defined for each possible path. Furthermore, because the path of $\{B(t); t\geq 0\}$ is continuous on [o,t], we have $\int_0^t B(s) ds = lim_{n \rightarrow \infty}\sum_{k=1}^{2^n}B\left(\frac{(k-1)t}{2^n}\right)\left(\frac{kt}{2^n}-\frac{(k-1)t}{2^n}\right)$. This also implies that the limiting cdf of the random variable on the right hand side coincides with the cdf of the left hand side.
 A: You create a subdivision of size $2^n$ $[0,t]$ where the $k^{th}$ point is $\frac{kt}{2^n}$.
The Brownian motion is continuous, therefore you can write it as the limit of a Riemann sum.
It is $$\int_0^t B(s) ds = lim_{n \rightarrow \infty}\sum_{k=1}^{2^n}B\left(\frac{(k-1)t}{2^n}\right)\left(\frac{kt}{2^n}-\frac{(k-1)t}{2^n}\right)$$
The sequence $\sum_{k=1}^{2^n}B\left(\frac{(k-1)t}{2^n}\right)\left(\frac{kt}{2^n}-\frac{(k-1)t}{2^n}\right)$ is normally distributed by definition of the Brownian motion, so is the limit. Therefore, $\int_0^t B(s) ds$ is normally distributed.
We now have to define the mean and the variance to entirely define the distribution of $\int_0^t B(s) ds$
The mean and variance are easily calculated , you can use Fubini theorem and write
$$E\left(\int_0^t B(s) ds\right)=\int_0^t E(B(s)) ds=0$$ 
For the variance,
$$E\left[\left(\int_0^t B(s) ds\right)^2\right]=E\left[\left(\int_0^t B(s) ds\right)\left(\int_0^t B(u) du\right)\right]$$
$$=\int_0^t\int_0^t{E\left[B(u)B(s)\right]duds}=\int_0^t\int_0^t{min(u,s)duds}=\frac{t^3}{3}$$ 
EDIT :
As for the first question, notice that 
$$\sum_{k=1}^n B(t_{k-1})(t_k-t_{k-1})=\sum_{k=1}^n B(t_{k-1})t_k-\sum_{k=1}^n B(t_{k-1})t_{k-1}$$
$$=\sum_{k=1}^n B(t_{k-1})t_k-\sum_{k=0}^{n-1} B(t_{k})t_{k}=\sum_{k=1}^n\left(B(t_{k-1})- B(t_{k})\right)t_k+t_nB_{t_n}$$
Notice that 
$$t_nB_{t_n}=-t_n\sum_{k=1}^n\left(B(t_{k-1})- B(t_{k})\right)$$
Thus,
$$\sum_{k=1}^n B(t_{k-1})(t_k-t_{k-1})=\sum_{k=1}^n\left(B(t_{k})- B(t_{k-1})\right)(t_n-t_k)$$
Using the property of the Brownian increments, you can finish the proof
