Square integrability and continuity for local martingales

Question

For local martingales (as in Pro05)

$\text{[Local/general square integrability]} \overset{?}{\leftrightarrow} \text{[continuity]}$

That is, does one imply the other? I believe $ [\text{continuity}] \rightarrow [\text{local square integrability}]$

Can anyone say one way or another? Any counterexamples to help develop intuition? (I have very little here)

Answer 1

Neither implies the other.

Let $X$ be any integrable random variable with $EX = 0$ and $E X^2 < \infty$, and let $X_t = X 1_{(1, \infty)}(t)$. Then $X_t$ is a square integrable martingale which is not continuous.

Let $Y$ be any integrable random variable with $E Y^2 = \infty$. Let $Y_t = Y$ for all $t$. Then $Y_t$ is a continuous martingale which is not even locally square integrable.

Answer 2

If $(X_t)_{t\geq 0}$ is a continuous local martingale and $(\sigma_n)$ is a localizing sequence such that $(X^{\sigma_n}_t)_{t\geq 0}$ is a martingale, then you can use the localizing sequence $$ \tau_n=\inf \{t>0\mid |X_t|>n\},\quad n\geq 1, $$ and put $\rho_n=\tau_n\wedge \sigma_n$. Then by continuity we have $|X^{\rho_n}_t|\leq n$ for all $n\geq 1$ and hence $(X^{\rho_n}_t)$ is a bounded martingale for every $n$.