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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!

saz
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  • note: it is meant that $X_n $ converges to a integrable random variable. –  Nov 11 '17 at 14:11
  • If you assume convergence in $L^1$ as well, it holds. That means you're looking for a martingale that converges almost surely, but not in $L^1$. – user159517 Nov 11 '17 at 14:56
  • @user159517 thanks! Actually I was working on such an example you stated. It did not work, so I missed something. Actually does not convergence in $L^1$ but convergence a.s. always imply $\sup_n E (X_{n}^+ )= \infty$? –  Nov 11 '17 at 16:32

1 Answers1

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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

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

saz
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  • wow thanks for this example! why is $ E ( \xi^+)=j^2 $? –  Nov 11 '17 at 18:16
  • @JohnDoe Well, $\xi_j^+ = j^4$ with probability $1/j^2$ and $0$ otherwise; hence $$\mathbb{E}(\xi_j) = j^4 \mathbb{P}(\xi_j = j^4) = j^4 \frac{1}{j^2}.$$ – saz Nov 11 '17 at 18:18
  • This is basically a nitpick, but you should start your sequence at $j=2$ and define $\xi_1=1$ or something. – Jason Nov 11 '17 at 18:20
  • @Jason Ah, yeah, right. – saz Nov 11 '17 at 18:23
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    wow great. How did you construct this one? I would not be able to find this one.. –  Nov 11 '17 at 18:30
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    @JohnDoe Took me some time to figure it out. Using Borel Cantelli lemma is a standard idea if you want to construct counterexample like this... at first I tried something of the form $M_n = \sum_{j=1}^n \xi_j$ but the problem was that I couldn't prove that the pointwise limit is integrable. Then I finally realized that I can get rid of this problem by considering the product instead of the sum. – saz Nov 11 '17 at 18:49
  • @saz I tried the same with $ j$ instead of $j^2$. It worked for me aswell. So actually why did you take $j^2$? –  Nov 13 '17 at 16:11
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    @JohnDoe No, it doesn't... because $\sum_j \frac{1}{j} = \infty$ (which means that you can't apply the Borel cantelli lemma the way I did). – saz Nov 13 '17 at 16:30