$(X_n) $ a $\{F_n \}$ martingale that a.s. converges. Does this imply $\sup_n E (X_n^+) < \infty$?

Question

Let $(X_n) $ a $\{F_n \}$ martingale that a.s. converges. Does this imply $\sup_n E (X_n^+) < \infty$? The other way round this statement is true (MCT). I tried to construct a counter example like symmetric random walk in combination with a well constructed stopping time, but that does not work. An example is highly appreciated!

Answer 1

Let $(\xi_j)_{j \geq 2}$ be a sequence of independent random variables such that

$$\begin{align*} \mathbb{P}(\xi_j = j^4) &= \frac{1}{j^2} \\ \mathbb{P}\big(\xi_j = -j^2 (j^2-1)\big)& = \frac{1}{j^2} \\ \mathbb{P}(\xi_j=0) &= 1- \frac{2}{j^2}. \end{align*}$$

Since $\mathbb{E}(\xi_j)=1$ for all $j \geq 2$, it follows from the independence of the random variables that

$$M_n := \prod_{j=2}^n \xi_j$$

defines a martingale. Moreover, by the Borel-Cantelli lemma, $\mathbb{P}(\xi_j \neq 0$ infinitely often)$=0$, and so $M_n \to M_{\infty} := 0$ almost surely. On the other hand, we have

$$M_n^+ \geq \prod_{j=2}^n \xi_j^+$$

and so

$$\mathbb{E}(M_n^+) \geq \prod_{j=2}^n \mathbb{E}(\xi_j^+) = \prod_{j=2}^n j^2 \xrightarrow[]{n \to \infty} \infty.$$