I want to find the value of the following expectation: $$E[max(B_s,B_t)^2]$$ with $$ s,t \in \mathbb{R}^{+}$$
We can also write it as $$E[B_s^2 .\mathbb{1}_{B_s>B_t}]+E[B_t^2 . \mathbb{1}_{B_t>B_s}]$$
Now we notice that $$E[\mathbb{1}_{B_s>B_t}]=E[\mathbb{1}_{B_t>B_s}]$$ since $s,t$ are interchangeable, but I do not have the independance of the random variables to separate the expectation and find $$\frac{s+t}{2}$$
In fact this result seems a bit strange to me since $$E[max(B_s,B_t)^2]+E[min(B_s,B_t)^2]=s+t$$ which would mean both $$max(B_s,B_t)^2$$ and $$min(B_s,B_t)^2$$ have the same expected value.
Do you have any hints on how to rigorously prove or disprove my heuristic result?
Okay so I am back after a bit of thinking, I think I understand why both expected values are the same and thus the result, although It it still a bit unclear to me intuitively.
Since the distributions of $(B_s)_{s\geq 0}$ and $(-B_s)_{s\geq 0}$ are the same we have
$$E[min(B_s,B_t)]=E[min(-B_s,-B_t)]$$ so $$E[max(B_s,B_t)^2]=E[(-min(-B_s,-B_t))^2]=E[min(-B_s,-B_t)^2]=E[min(B_s,B_t)^2]$$