 Centered Brownian powers need not be martingales

The process $B_t^k-\mathbb{E}{B}_t^k$ is a martingale only for $k=1,2$. For higher powers its finite-variation drift is $k(k-1)(B_t^{k-2}-\mathbb{E}{B}_t^{k-2})/2$, which is not identically zero. Subtracting the random compensator $k(k-1){\int }_0^t{B}_s^{k-2}ds/2$ instead produces a true martingale on each finite horizon.

// Target: probability-and-statistics.bigb