Given a $L^2$ martingale $X_t$, for a finite stopping time $T$, is it true that $X_T$ is again $L^2$ integrable?

If $X_t$ is additionally uniform integrable, then we can use Jensen-inequality:$$E[X_T^2]=E[E[X_\infty\mid F_T]^2]\leq E[X_\infty^2]$$ to obtain the $L^2$ integrability of $X_T$ provided $X_\infty$ is square integrable.

However, I don't know if the statement is still true without the assumption on uniform integrability and $X_\infty^2$ is integrable.


No, this is, in general, not true.

Let $(B_t)_{t \geq 0}$ be a one-dimensional Brownian motion, and let $X$ be a random variable which takes values in $(0,\infty)$ and which is independent from $(B_t)_{t \geq 0}$. Clearly, $(B_t)_{t \geq 0}$ is an $L^2(\mathcal{F}_t)$-martingale with respect to the filtration

$$\mathcal{F}_t := \sigma(X, B_s; s \leq t).$$

If we set

$$T(\omega) := \inf\{t \geq 0; |B_t(\omega)| \geq X(\omega)\},$$

then it is not difficult to see that $T$ is a finite $\mathcal{F}_t$-stopping time. Moreover, it follows from the continuity of the sample paths of Brownian motion that $|B_T| = X$ almost surely. In particular,

$$\mathbb{E}(B_T^2) = \mathbb{E}(X^2),$$

and therefore $B_T \in L^2$ if, and only if, $X \in L^2$.

  • $\begingroup$ Would you mind to give an example of such $X(\omega)$ $\endgroup$ – quallenjäger Apr 12 '18 at 14:21
  • $\begingroup$ @quallenjäger What do you mean by "such $X(\omega)$"? You can pick any random variable which is not square integrable, for instance a random variable with distribution $\mu(dx) = c 1_{(0,\infty)}(x) \frac{1}{x^2+1} \, dx$ where $c>0$ is chosen such that $\mu(\mathbb{R})=1$. $\endgroup$ – saz Apr 12 '18 at 14:25
  • $\begingroup$ Thanks, I was looking for the distribution function, which is given by your comment. Everything is clear. Thanks $\endgroup$ – quallenjäger Apr 12 '18 at 14:28
  • $\begingroup$ @quallenjäger You are welcome. $\endgroup$ – saz Apr 12 '18 at 14:29

There is also Doob's example: Let $F$ be any distribution on the real line. Choose the appropriate function $g$ so that $g(B_1)$ has distribution $F$, and then define $T:=\inf\{t>1: B_t=g(B_1)\}$. Clearly $T$ is a.s. finite by the recurrence of Brownian motion, and $B_T$ has distribution $F$. And this $T$ is a stopping time of the filtration of the given Brownian motion.


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