If $M$ is a martingale, show that $X_t = M_t − kt^2$ is a supermartingale iff $k$ is non-negative.

Question

If $M$ is a martingale, show that $X_t = M_t − kt^2$ is a supermartingale iff $k$ is non-negative.

A supermartingale is defined when $X_s \ge \mathbb E[X_t|\mathcal F_s]$ when $s\le t$. Clearly $X$ must be integrable and adaptable if $M$ is since $kt^2$ is a constant. The main thing then is that we know

$$ M_s = \mathbb E(M_t|\mathcal F_s)$$ I just don't know how to apply this to get a supermartingale for $X$?

Answer 1

Assume $k\geq0$. For $s\leq t$ we (almost surely) have $$\mathbb E[X_t|\mathcal F_s] = \mathbb E[M_t-kt^2|\mathcal F_s] = \mathbb E[M_t|\mathcal F_s] -kt^2 = M_s - kt^2 \leq M_s -ks^2 = X_s,$$ where we have used linearity of conditional expectation, the fact that $(M_t)_t$ is a Martingale and $k\geq0$. Hence, $(X_t)_t$ is a supermartingale.

For the other direction, assume $(X_t)_t$ is a supermartingale. Then, for $0<s< t$ $$X_s = M_s -ks^2 \geq\mathbb E[X_t|\mathcal F_s] = \mathbb E[M_t-kt^2|\mathcal F_s] = \mathbb E[M_t|\mathcal F_s] -kt^2 = M_s -kt^2, $$ which implies that $k\geq0$.

Answer 2

Let $t > s$; we compute $\mathbb{E}[X_t | \mathcal F_s] = \mathbb{E}[M_t - k t^2 | \mathcal F_s] = M_s - k t^2$. Since $k \ge 0$, $kt^2 \ge ks^2$ so $\mathbb{E}[X_t | \mathcal F_s] = M_s - k t^2 \le M_s - k s^2 = X_s$.