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Suppose $X_i \sim\operatorname{uniform}[0,1]$ and that they are iid. Does $n/\Sigma_{i=0}^n(1/X_i)$ converge to $0$ in probability? A simulation seems to indicate that it does. But as the expected value of $1/X_i$ does not exist, I cannot use LLN. Any hints?

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  • $\begingroup$ sry guys, it's my fault, I missed one n here.... $\endgroup$
    – abc1m2x3c
    Feb 14, 2016 at 4:30
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    $\begingroup$ Recall that the law of large numbers, both weak and strong, continues to hold when the expectation is either $+\infty$ or $-\infty$ by truncation argument. $\endgroup$ Feb 14, 2016 at 5:53
  • $\begingroup$ The only LLN without assuming first moment I can find is called Feller's WLLN, however it requires $xP[X>x] \rightarrow 0$... $\endgroup$
    – abc1m2x3c
    Feb 15, 2016 at 0:01
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    $\begingroup$ You can, for example, check Theorem 2.4.5 of Durrett (4.1st ed). This is really a straightforward corollary of the standard SLLN. $\endgroup$ Feb 15, 2016 at 0:10
  • $\begingroup$ Thanks so much! It's more important to understand a theorem than just to memorize it! $\endgroup$
    – abc1m2x3c
    Feb 15, 2016 at 0:35

3 Answers 3

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Let $$Y_n=\frac{1}{\sum_{i=0}^n(1/X_i)}$$

Then, for any $0<a<1$, $$P(Y_n>a)=P(\sum_{i=0}^n X_i^{-1}<a^{-1})\le P( X_i^{-1}<a^{-1}; \forall i)=\prod_{i=1}^n P(X_i>a)=(1-a)^n$$

This tends to $0$ as $n\to \infty$, so...

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  • $\begingroup$ sry, I missed an n in my question, I have revised..... Thanks for your answer, could you have a look on it again? $\endgroup$
    – abc1m2x3c
    Feb 14, 2016 at 4:32
  • $\begingroup$ see Bombyx mori's answer $\endgroup$
    – leonbloy
    Feb 14, 2016 at 14:42
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Let $$ Z =\frac{1}{ \sum_{i=0}^n \frac{1}{X_i}} $$ Observe that $$ P\left(\sum_{i=0}^n \frac{1}{X_i}\geq n\right) =1 $$ Then, $$ P\left(Z\leq \frac{1}{n}\right) = 1 $$ Moreover, $P(Z<0)=1$.

Now let $k> 0$, $\varepsilon>0$ and Notice that $\lim_{n\rightarrow\infty} P(|Z-k|\geq \varepsilon) = 0$ because as $n$ gets larger, $Z-k$ gets arbitrarily close to $0$.

As a result, $Z$ converges in probability to $0$.

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  • $\begingroup$ sry, I missed an n in my question, I have revised..... Thanks for your answer, could you have a look on it again? $\endgroup$
    – abc1m2x3c
    Feb 14, 2016 at 4:32
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(does not work and should be discarded)

For the modified problem: Let $Z_{n}=\sum^{n}_{i=0}\frac{1}{X_{i}}, X_{i}\sim U(0,1)$. Then we have the following inequalities using pigeon hole principle:

$$ P(X_{(1)}\ge \frac{1}{x})=P(\frac{1}{X_{(1)}}\le x)\le P(Z_{n}/n\le x)\le P(\frac{1}{X_{(n)}}\le x)=P(X_{(n)}\ge \frac{1}{x}) $$ which gives us $$ (1-\frac{1}{x})^{n}\le P(\frac{Z_{n}}{n}\le x)\le 1-x^{-n} $$ This does not give us any useful info when $x>1$, since then we get a useless $(0,1)$ bound. A direct $R$-simulation showed the mean did not diminish when $n$ goes from $10^{3}$ to $10^{5}$, and the histogram is tilted towards the right. But it could also be because the $n$ is still relatively small.

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    $\begingroup$ if $x >1$, I think it should be $1-\frac{1}{x^n}$, is it? $\endgroup$
    – abc1m2x3c
    Feb 14, 2016 at 22:19
  • $\begingroup$ @abc1m2x3c: I think you are right. I am thinking how to amend the proof. $\endgroup$ Feb 14, 2016 at 23:34
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    $\begingroup$ I think Sangchul Lee's comments (which is just below my question) are right. Just use an "extended" Law of Large Numbers, $Z_n/n$ goes to infinity and thus its inverse goes to 0... $\endgroup$
    – abc1m2x3c
    Feb 15, 2016 at 0:54
  • $\begingroup$ @abc1m2x3c: I see. I think he is totally right. Someone should make this into answer. It is kind of embarrassing to see I got up-voted for a wrong proof at here. $\endgroup$ Feb 15, 2016 at 0:58
  • $\begingroup$ @ Bombyx mori , haha, it's OK! Thank you all the same! $\endgroup$
    – abc1m2x3c
    Feb 15, 2016 at 1:58

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