Showing this is a martingale

Question

$$X_t = t^2 B_t - 2 \int_0^t s B_s ds$$

where $B_t$ is standard Brownian motion.

My textbook says this is a martingale, but I'm not sure how to prove the expectation condition.

Taking conditional expectation,

$$E[X_t | \mathcal{F}_s ] = E[t^2 B_t |\mathcal{F}_s] -2 E\left[\int_0^t s B_s ds|\mathcal{F}_s \right] = t^2 B_s -2 \int_0^s u B_u du -2 E\left[\int_s^t u B_u du|\mathcal{F}_s \right]$$

So presumably that last expectation should equal

$$s^s B_s - t^2 B_s$$

in order for it to be a martingale. But I don't see how to take the expectation inside, nor why it should equal $s^2 B_s - t^2 B_s$. Can someone explain how to evaluate that last expecation?

edit: Misread the question. Corrected

Answer 1

I guess $X_t$ should be $$X_t = t^2 B_t-2 \int_0^t s B_s ds$$ to be a martingale. We can see that $$ E\left[\int_s^t r B_r dr\;|\mathcal{F}_s\right] = E\left[\int_s^t r B_r dr\;|\mathcal{F}_s\right]=\int_s^t r E[B_r |\mathcal{F}_s] dr = \frac{t^2-s^2}{2}B_s. $$ Hence, we have that$$ E\left[X_t - X_s |\mathcal{F}_s\right]= t^2B_s -s^2B_s -2\cdot\frac{t^2-s^2}{2}B_s=0, $$ showing that $\{X_t,\mathcal{F}_t\}_{t\geq 0}$ is a martingale.