# Proving an alternated Brownian Motion is a Martingale

Let $$B_t$$ denote standard Brownian motion. Show that the stochastic process $$M_t = B_t^2 - t$$ is a martingale

Ive shown the finiteness and adaptability. I am left to show that $$\mathbb{E}[M_t | \mathcal{F}_s] = B_s$$

Here is my attempt but it is very wordy and feels unrigourous:

Given $$B_s$$ we know $$B_t$$ is normally distributed with mean $$B_s$$ and variance $$t-s$$

Hence $$\mathbb{E}[B_t^2 | \mathcal{F}_s] = \mathbb{E}[B_t | \mathcal{F}_s]^2 + V[B_t | \mathcal{F}_s] = B_s^2 + (t-s)$$.

Therefore $$\mathbb{E}[M_t | \mathcal{F}_s] = \mathbb{E}[B_t^2-t | \mathcal{F}_s] = B_s^2-s = M_s$$ as desired.

Could someone show a way to make this more formal? Is there an easy trick I'm missing :) Thanks

The first equality you have, while not wrong, has not been correctly justified. Here is a slightly cleaner approach: \begin{align} \mathbb{E}[B_t^2|\mathcal{F}_s]= & \mathbb{E}[B_t B_s|\mathcal{F}_s] + \mathbb{E}[B_t (B_t-B_s)|\mathcal{F}_s] \\ =& \mathbb{E}[B_t B_s|\mathcal{F}_s] + \mathbb{E}[(B_t-B_s)^2 |\mathcal{F}_s] + \mathbb{E}[B_s(B_t-B_s)|\mathcal{F}_s] \end{align} We now use some properties of conditional expectation. Since $$B_s$$ is $$\mathcal{F}_s$$-measurable we know that $$\mathbb{E}[B_t B_s|\mathcal{F}_s]= B_s \mathbb{E}[B_t|\mathcal{F}_s]=B_s^2$$, where in the last equality we used that $$B_t$$ is an $$\mathcal{F}_t$$-martingale. For the second term note that $$B_t-B_s$$ is independent of $$\mathcal{F}_s$$ and so $$\mathbb{E}[(B_t-B_s)^2 |\mathcal{F}_s]= \mathbb{E}[(B_t-B_s)^2 ]=t-s$$. Finally, we use again that $$B_s$$ is $$\mathcal{F}_s$$-measurable and $$B_t$$ is an $$\mathcal{F}_t$$-martingale to argue that $$\mathbb{E}[B_s(B_t-B_s)|\mathcal{F}_s]=B_s \mathbb{E}[(B_t-B_s)|\mathcal{F}_s]=0$$. This leaves us with \begin{align} \mathbb{E}[B_t^2|\mathcal{F}_s]=B_s^2 + (t-s)\, , \end{align} which is what we wanted.
It is not clear as to how you got $$V[B_t | \mathcal{F}_s] = (t-s)$$.
Let $$t >s$$. Then $$E(M_t|\mathcal F_s)=E((B_t-B_s+B_s)^{2}-t|\mathcal F_s)$$. You can write this as $$E((B_t-B_s)^{2}|\mathcal F_s)+2EB_s(B_t-B_s)|\mathcal F_s)+B_s^{2}-t$$. The first term is the variance of $$B_t-B_s$$ which is $$t-s$$. The second term is $$0$$ because $$E(B_t-B_s|\mathcal F_s)=E(B_t-B_s)=0$$. So you get $$B_s^{2}-s$$