6
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ 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
2
+50
$\begingroup$

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

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

Since

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

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

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

$\endgroup$
9
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.