Determine $E[B_t^2 – t^2 | B_s]$ for $0 < s < t$ ~ Standard Brownian Motion

Question

I am trying to determine what $E[B_t^2 – t^2 | B_s]$ for $0 < s < t$ is ( Standard Brownian Motion).

This is what I tried:

$E[B_t^2 – t^2 | B_s] = E[B_t^2 | B_s]- E[t^2 | B_s]$ (using linearity)

$ = E[B_t^2 | B_s] – t^2$ (I am not sure about this step but I think it is correct)

$ = E[B_t (B_t – B_s + B_s) | B_s] – t^2$

$= E[B_t (B_t – B_s) | B_s] + E[B_t B_s | B_s] – t^2 $

$ = E[B_t | B_s]\cdot E[(B_t – B_s) | B_s] + E[B_t B_s | B_s] – t^2 $

$ = B_s \cdot 0 + E[B_t B_s | B_s] – t^2 $

I am probably doing something wrong. I am missing the fact that some increments are independent?

Answer 1

\begin{align} \mathsf{E}[B_t^2\mid B_s]&=\mathsf{E}[(B_t-B_s+B_s)^2\mid B_s] \\ &=\mathsf{E}[(B_t-B_s)^2\mid B_s]+2\mathsf{E}[(B_t-B_s)\mid B_s]B_s+B_s^2 \\ &=\mathsf{E}[(B_t-B_s)^2]+2\mathsf{E}[B_t-B_s]B_s+B_s^2 \\ &=(t-s)+0+B_s^2. \end{align}