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}

Answer 2

Let $0\leqslant s<t$. We compute \begin{align} \mathbb E[B_t^2-t^2\mid\mathcal F_s] &= \mathbb E[(B_t-B_s+B_s)^2\mathcal F_s] -t^2\\ &= \mathbb E[(B_t-B_s)^2\mid\mathcal F_s] -2\mathbb E[(B_t-B_s)B_s\mid\mathcal F_s] + \mathbb E[B_s^2\mid\mathcal F_s] - t^2. \end{align} Since $B_t-B_s$ has $N(0,t-s)$ distribution and is independent of $\mathcal F_s$ and $B_s$ is $\mathcal F_s$-measurable, the above is equal to \begin{align} \mathbb E[(B_t-B_s)^2] -2\mathbb E[B_t-B_s]B_s+B_s^2-t^2 &= t-s - 0 +B_s^2-t^2\\ &= B_s^2 -s + t(1-t). \end{align} Now, the inequalities $-s + t(1-t)\geqslant -s^2$ and $-s + t(1-t)\leqslant -s^2$ vary based on $s$ and $t$, so $B_t^2-t^2$ is neither a submartingale nor a supermartingale.