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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

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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.

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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 $B_s^{2}-s$

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