Standard Brownian Motion process

Question

I would appreciate the help in the following

$\{x(t): 0\leq t \leq 1\}$ Standard Brownian motion.

$ t^*=\inf\{t\in[0,1]:x(t)=\underset{0\leq s \leq 1}{\sup} x(s) \}$ random variable , $\{y(t): 0\leq t \leq 1\}$ Standard Brownian motion are independent, $ \mathbb{E}(t^*)<\infty $ then

$\mathbb{E}\left[(y(t^*))^2\right] =\mathbb{E}(t^*)$

I would like to know if this result is true and if it is true, how can I prove it?

Answer 1

This question is not really about Brownian motion.
You see, $t^*$ and $y$ are independent, which makes: $$\mathbb{E}( y(t^*)^2 |t^*=t )= \mathbb{E}( y(t)^2 |t^*=t)= t$$ Hence forth what follows.

P/s: $t^*$ and $y$ are independent