Let $X_0 = \frac12$, and $X_n = \text{Uniform}(0, 2X_{n-1})$ for $n \ge 1$ be a sequence of random variables. Since $E[X_{n} | X_0, \cdots, X_{n-1}] = X_{n-1}$, we have that $X_n$ is a martingale. Also, it is bounded from below by $0$.

Using (a version of) Martingale convergence theorem, we can say that $X_n \to X$ with probability $1$, where $X$ is a random variable with finite expectation.

Now, in class we argued that $X_n$ cannot converge to any value other than $0$ with positive probability. This can be seen with elementary argument by assuming that it converges to some $a > 0$ with positive probability, and considering an arbitrarily small interval around $a$ to show that that cannot be the case because of the way $X_n$ is defined.

However, we then argued that the previous argument implies that $\Pr(X_n \to 0) = 1$. This part is not clear to me, because we can consider a similar argument for $0$, by considering an interval $(0, \epsilon]$ for some $\epsilon$.

Note that this was meant as an introduction to Martingales, and we proved the Martingale convergence theorem using "basic probability" and "elementary" $\epsilon-\delta$ arguments. We do not have any background in measure theory, and we did not base our discussion of Martingales on the foundations of measure theory, which I understand is a standard way of presenting Martingales.

My question is, can we show that with probability $1$, $X_n$ converges to $0$ without using sophisticated arguments from measure theory?

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    $\begingroup$ Think of $X_n$ as ${1\over 2}V_1V_2\cdots V_n$, where the $V_k$ are i.i.d. and uniformly distributed on $(0,2)$. Take logs, observe that $\Bbb E[\log V_k] =\log 2-1<0$, and apply the strong law of large numbers to see that $\log X_n\to -\infty$ a.s. $\endgroup$ – John Dawkins Dec 15 '16 at 19:57
  • $\begingroup$ @JohnDawkins Could you expand on why $X_n$ can be written in that manner, and what is the random variable that the SLLN is being applied on? Incidentally, we also considered $X_n = \prod V_i$, where $V_i$ are Uniform$(0, 2)$ as an another example of a martingale, and argued in almost exactly the same way to conclude $X_n$ converges to $0$. $\endgroup$ – taninamdar Dec 15 '16 at 20:10

First of all, note that

$$\text{Uniform}(0,\alpha) = \alpha \cdot \text{Uniform}(0,1)$$

for any $\alpha>0$; for instance picking a random number from the interval $(0,2)$ (with uniform distribution) is the same (in distribution) as picking a random number from the unit interval $(0,1)$ (with uniform distribution) and multiplying the number by $2$.

This means that

$$X_n = \text{Uniform}(0,2X_{n-1}) = 2 X_{n-1} \text{Uniform}(0,1) .$$

If we set $\xi_n := \frac{X_n}{2X_{n-1}}$, then $\xi_n = \text{Uniform}(0,1)$ and

$$X_n = 2 X_{n-1} \xi_n. $$

Iterating the procedure, we find

$$X_n = 2X_0 \prod_{j=1}^n \xi_j \tag{1}$$

where $\xi_j = \text{Uniform}(0,1)$ for all $j=1,\ldots,n$.

From now on we assume that the random variables $\xi_1,\xi_2,\ldots$ are independent. That's an assumption we have to make; otherwise $(X_n)_{n \in \mathbb{N}}$ may fail to be a martingale. It follows from $(1)$ that

$$\log(X_n) = \log \left(2 X_0 \prod_{j=1}^n \xi_j \right) = \log(2X_0) + \sum_{j=1}^n \log(\xi_j). \tag{2}$$

The random variables $\eta_j := \log(\xi_j)$ are independent, identically distributed and integrable; therefore the law of large number shows

$$\frac{1}{n} \sum_{j=1}^n \log(\xi_j) = \frac{1}{n} \sum_{j=1}^n \eta_j \xrightarrow[]{n \to \infty} \mathbb{E}(\eta_1)$$


$$\mathbb{E}(\eta_1) = \int_0^1 \underbrace{\log(x)}_{<0} \, dx < 0$$

we find

$$\sum_{j=1}^n \log(\xi_j) = n \underbrace{\left( \frac{1}{n} \sum_{j=1}^n \log(\xi_j) \right)}_{\to \mathbb{E}(\eta_1)<0} \xrightarrow[]{n \to \infty} - \infty.$$

Letting $n \to \infty$ in $(2)$ we conclude

$$\log(X_n) \xrightarrow[]{n \to \infty} - \infty$$

which is, by the continuity of $\exp$, equivalent to saying

$$X_n = \exp(\log(X_n)) \xrightarrow[]{n \to \infty} 0.$$

  • $\begingroup$ Thank you, this is most helpful. You say "From now on we assume that the variables $\zeta_i$ are independent. That's an assumption [...]". Is this an extra assumption, or does it follow from the fact that $X_i$ are i.i.d.? $\endgroup$ – taninamdar Jan 1 '17 at 15:37
  • $\begingroup$ @taninamdar It is an extra assumption (which ensures, in particular, that $(X_n)_n$ is a martingale). Mind that the random variables $X_i$ are not independent. $\endgroup$ – saz Jan 1 '17 at 18:24
  • $\begingroup$ Right, $X_i$ are not independent. So what does the assumption that $\xi_i$ are independent translate to, in terms of the original r.v.s $X_i$? $\endgroup$ – taninamdar Jan 2 '17 at 12:31
  • $\begingroup$ @taninamdar As explained in my answer, we can write $$X_n = \text{Uniform}(0,2_X{n-1})= 2 X_{n-1} \xi_n.$$ The assumption on the independence means that we generate for each $n \in \mathbb{N}$ a random number $\xi_n(\omega)$ in the interval $(0,1)$ independently from the "history", i.e. $\xi_0,\ldots,\xi_{n-1}$. This random number we multiply by $2X_{n-1}$ to obtain $X_n$. In contrast, if we would for instance define $$Y_n := 2 Y_{n-1} \xi$$ (note that we have now one random variable $\xi$ for all $n$), then we would pick once a random number $\xi(\omega)$ in the interval $(0,1)$ ... $\endgroup$ – saz Jan 2 '17 at 12:49
  • $\begingroup$ ... and multiply in each step simply the number $2Y_{n-1}(\omega)$ by the (fixed) ration $\xi(\omega)$. The process $(X_n)_{n \in \mathbb{N}}$ is a martingale, but $(Y_n)_{n \in \mathbb{N}}$ is not.... this shows that the assumption on the independence is essential for the martingale property. $\endgroup$ – saz Jan 2 '17 at 12:50

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