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let $Y_1,Y_2,..$ be a sequence of equally distributed, independent and positive random variables.

Consider $X_n = Y_1…Y_n$. Under which condition is $X_n$ a (super)-martingale? Show that neglecting the case $Y_n=1$ we have $X_n -> 0$.

$E[X_n | Y_1,…,Y_{n-1}] = Y_1 … Y_{n-1} E[Y_n | Y_1 ,…, Y_{n-1}] = Y_1…Y_{n-1} E[Y_n]$

This is a (super)-martingale if $E[Y_n] \le 1$. Since the variables are equally distributed we have $E[Y_i]=E[Y_j]$.

Since $X_n \ge 0$ and $E[X_n] = E[Y_1]…E[Y_n] \le 1$ the martingale is converging $X_n -> X$.

If $E[Y_i] < 1$ we have $E[X_n] = E[Y_1] … E[Y_n] -> 0$ hence with fatuous lemma and since $X_n \ge 0$ we have $X=0$.

But what is with the case $E[Y_i] =1$? Here the result is remarkable, since $E[X_n] =1$ for all n but $E[X] = 0$. So somehow the probability of the events, where $X_n \ge 1$ have to go to 0, while $X_n -> \infty$ there. Any ideas?

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  • $\begingroup$ Why is $E[X]=0$? $\endgroup$ – zoli May 4 '15 at 8:26
  • $\begingroup$ Since $X=0$ (this what i have to show) $\endgroup$ – crankk May 4 '15 at 9:03
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In order to identify the limit of $(X_n)$, one can use the strong law of large numbers: we have $$\lim_{n\to \infty}\frac{\log X_n}n=\mathbb E\left[ \log(Y_1)\right].$$ Here we used the not necessarily integrable version of the law of large numbers, since the positive part of $\log (Y_1)$ is integrable.

Using Jensen's inequality, note that $\mathbb E\left[ \log(Y_1)\right]\leqslant 0$, and the inequality is strict if $\log Y_1$ is non-degenerated.

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  • $\begingroup$ I don't understand…$ \frac{log X_n}{n} = \frac{1}{n} (log Y_1 + … + log Y_n)$ … why this is $E[log X_1]$ ? And by $X_1$, do you speak about the limit of the series or $X_1 = Y_1$? $\endgroup$ – crankk May 4 '15 at 9:21
  • $\begingroup$ It is not equal: the limit as $n\to\infty$ is equal to $E[ \log X_1]$. I should write instead $Y_1$, but it is the same. $\endgroup$ – Davide Giraudo May 4 '15 at 9:23
  • $\begingroup$ Ah okay … so if we exclude $Y=1$ a.s. we have lim $log X_n = n c$ with $c < 0$. So $log X_n -> - \infty$ => $X_n -> 0$. Thank you! $\endgroup$ – crankk May 4 '15 at 10:09
  • $\begingroup$ one more thing: first of all we need $Y>0$ a.s. to make the logarithm be defined (the other case i already verified). But still, one can have Y where the expectation value is unbounded (since the logarithm is not integrable on $(0,\infty)$), hence we can't use the law of strong numbers. $\endgroup$ – crankk May 5 '15 at 15:51
  • $\begingroup$ @crankk I see your point. Actually, there are version of the law of large numbers when the expectation is infinite. I should have been more careful. $\endgroup$ – Davide Giraudo May 5 '15 at 18:32
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Here is the case $P( Y = 0) = d > 0$:

one has $P(X > 0) \le P(X_n > 0) = P ( \cap Y_i > 0 , 1 \le i \le n) = P(Y_0 > 0) … P( Y_n > 0) = (1-d)^n$ for all n.

So P(X > 0) = 0 a.s.

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