Showing that $B(t)^2 - t$ is a martingale

Question

I am trying to prove $B(t)^{2}-t$ is a martingale.

So far I have done:

1. By definition $E[B^{2}(t)]=t<\infty$
2. $B^{2}(t+s)= (B(t)+B(t+s)-B(t))^{2} \\ B^{2}(t)+2B(t)(B(t+s)-B(t))+(B(t+s)-B(t))^{2}$

So

$E[B^{2}(t+s)|\mathcal{F}_t]=B^{2}(t)+2E[B(t)(B(t+s)-B(t))|\mathcal{F}_t]+E[(B(t+s)-B(t))^{2}|\mathcal{F}_t]$

Im not sure what I ought to do from here to get $B^{2}(t)-t$.

I have assumed $B(t+s)-B(t)$ is $\mathcal{F}_t$ measurable and $B(t)$ is normally distributed (0,t). Is is because $2E[B(t)(B(t+s)-B(t))| \mathcal{F}_t] = 0$ and $E[(B(t+s)-B(t))^{2}|\mathcal{F}_t]= Var$ which is $t+s-t=s$...

But then I get $B^{2}(t)+s$ instead of $B^{2}(t)-t$.

Answer 1

You are supposed to show that $E(B_{t+s}^{2}-(t+s)|F_t)=B_t^{2}-t$.

$B(t+s)-B(t)$ is independent of $F_t$ (not measurable w.r.t. $F_t$). So $E(B(t+s)-B(t)^{2}|F_t)=E(B(t+s)-B(t)^{2})=s$. You can now finish by subtracting $t+s$ from both sides.

Answer 2

Using simpler notation might help. For simplicity define your Brownian motion as $(Y(t) = B(t)^{2}-t,t\geq0)$.You have to show that for $s<t$,$$E[Y(t)|\mathcal{F}_{s}] = Y(s) $$ Note that $$E[Y(t)|\mathcal{F}_{s}]= E[B(t)^2|\mathcal{F}_{s}]-t$$ Now $$E[B(t)^2|\mathcal{F}_{s}] = E[(B(t) - B(s) + B(s))^{2}|\mathcal{F}_{s}]$$ $$= E[(B(t)-B(s))^{2}|\mathcal{F}_{s}] + E[B(s)^{2}|\mathcal{F}_{s}] +E[2(B(t)-B(s))B(s)|\mathcal{F}_{s}]$$ $$=E[(B(t)-B(s))^{2}] + B(s)^{2} + 0=(t-s) + B(s)^{2} $$ Thus $$E[Y(t)|\mathcal{F}_{s}]= (t-s) + B(s)^{2} -t = Y(s) $$ I hope it helps.

Answer 3

In general $f(B,\,t)$ is a martingale iff $\frac{\partial f}{\partial t}=-\frac12\frac{\partial^2f}{\partial B^2}$ (both sides are $-1$ for $f=B^2-t$), since then by Itô's lemma $df=\frac{\partial f}{\partial B}dB$ has no $dt$ term.