I recently solved a problem on the distribution of the sum of two Brownian motions.
That is, if $\{B(t); \,t\geq 0\}$ is the standard brownian motion, what is the distribution of $B(s)+B(t)$ if $0\leq s \leq t$?
The idea in solving this problem is to represent the sum $B(s)+B(t)$ as the sum of an increment. That is, $$ B(s) + B(t) = 2B(s) + B(t) - B(s) $$ and since we know incrememnts of a brownian motion are independent, then $2B(s)$ is independent of $B(t) - B(s)$. Thus, we can easily get that $\mathbb{E}[B(s)+B(t)] = 0$ & $\mathbb{V}\mathrm{ar}[B(s)+B(t)] = 4s+t-s = 3s+t$. So, $B(t) + B(s) \sim \mathcal{N}(0,3s+t)$.
Now my question is, can we exploit this same idea to determine the sum of $n$ brownian motions? That is, what is the distribution of $$ \sum_{i=1}^n B(t_i) $$ where $0\leq t_1 \leq \cdots \leq t_n?$
If I am allowed to explot this idea, would it mean that $$ B(t_1) + B(t_2) + \cdots + B(t_n) = \left[2B(t_1) + B(t_2) - B(t_1)\right] + \left[2B(t_3) + B(t_4) - B(t_3)\right] + \cdots + [2B(t_{n-1}) + B(t_n) - B(t_{n-1})] $$
giving us, $$ \mathbb{E}\left[\sum_{i=1}^n B(t_i)\right] = 0,\qquad \mathbb{V}\mathrm{ar}\left[\sum_{i=1}^n B(t_i)\right] = 3\sum_{\text{odd}}t_i + \sum_{\text{even}}t_i$$
and so $$ \sum_{i=1}^n B(t_i) \sim \mathcal{N}\left(0,3\sum_{\text{odd}}t_i + \sum_{\text{even}}t_i \right) ?$$
Although this is a pretty faithful use of the trick in the $n=2$ case, I feel as if this looks a bit silly. I'm mainly concerned with my derivation of the variance, because I think I have assumed that any two pairs $2B(t_{i})+B(t_{i+1})-B(t_i)$ and $2B(t_{j})+B(t_{j+1})-B(t_j)$ where $i\neq j$ are independent of one another. Which may or may not be true, I'm not exactly sure.
If they are not independent, how can I reuse this trick to get the distribution of the sum? Would it mean I have to take $2[B(t_1)+B(t_3)+B(t_5)+\cdots]$ and reapply the same trick? If so, would I have to keep on redoing that trick, until everything is an increment except some multiple of $B(t_n)$?
Any help/illumination on how to use this trick for deriving the distribution of the sum of $n$ brownian motions would be greatly appreciated. Thank you.
$E(B(t)B(s))=t\wedge s$ for amny $s,t \geq 0$. So $var ( \sum\limits_{i=1}^{n}B(t_i))=\sum\limits_{i,j=1}^{n}t_i\wedge t_j$.
If you prefer you can write this as $2 \sum\limits_{i=1}^{n-1}(n-i)t_i+\sum\limits_{i=1}^{n} t_i$.