Let $(B_t)$ be a standard Brownian motion, $n\in\mathbb{N}$ even and $t_1,\ldots,t_n\in\mathbb{N}$. I was wondering whether there is a general formula for $$\mathbb{E} \left[ B(t_1)\cdots B(t_n)\right].$$
By successively applying the property of independent increments, it is straightforward to calculate this expression for some fixed $n\in\mathbb{N}$. Now my conjecture is the following. Let $\mathcal{P}_n$ be the set of partitions of $\{1,\ldots,n\}$ into two-element subsets. Then $$\mathbb{E} \left[ B(t_1)\cdots B(t_n)\right] \stackrel{?}{=} \sum_{P\in\mathcal{P_n}} \prod_{\{i_1,i_2\}\in P} t_{i_1} \wedge t_{i_2}.$$ I confirmed this by hand for $n$ up to $6$ and tried to prove it by induction , but failed. I would be grateful for any advice on how to prove this or a known formula for $\mathbb{E} \left[ B(t_1)\cdots B(t_n)\right].$
For illustration, this is the formula for $n=4$: There are three partitions of $\{1,2,3,4\}$, namely $\{\{1,2\},\{3,4\}\}$, $\{\{1,3\},\{2,4\}\}$ and $\{\{1,4\},\{2,3\}\}$. Their contributions to the sum are $t_2 t_4$, $t_3 t_4$ and $t_4 t_3$ respectively (assuming $t_1>t_2>t_3>t_4$), yielding the correct result $2t_3 t_4 + t_2 t_4$.