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$X_1,X_2,\dotsc ,X_n$ are independent, uniformly distributed random variables on the interval $[0,1]$

The question is the convergence of the sequence: $X_{{\lfloor n/3 \rfloor}}^ \space\small{(n)}$. It denotes the $\lfloor n/3 \rfloor$-th smallest value of a sample size $n$.

I know the k-th order statistic of a uniform distribution follows a beta distribution.

In my case $U_{\lfloor n/3\rfloor}\sim B(\lfloor n/3\rfloor,n+1-\lfloor n/3\rfloor)$

I think the answer should be the expected value of this beta distribution, so: $\frac{1}{3}$

Is that correct?

I'm studying probability, and tihs was a question on an exam years ago.

Can someone help me?

Thanks in advance!

kanbhold
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    Although we can make educated guesses about what the notation might mean, please supply some context to help us out. Where does this question come from? Why is it interesting or important? – whuber Nov 17 '14 at 22:58
  • @whuber I edited my question. – kanbhold Nov 17 '14 at 23:14
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    Thanks, but the edits don't add any more information. As it is written, the question does not lead to any definite convergence. We are practically forced to view "$X_{\lfloor n/3\rfloor}$" as being an *order statistic* (either ascending or descending won't matter)--but then the convergence to $0$ is such a trivial result that the problem hardly looks worthy of an examination. – whuber Nov 17 '14 at 23:18
  • It doesn't really make sense to me either. Actually the notation was: $X_{\lfloor n/3 \rfloor}^{(n)}$. I interpreted it as the n-th power of $X$. – kanbhold Nov 17 '14 at 23:23
  • Nobody puts parentheses around an exponent unnecessarily, so our first efforts at interpretation should assume this is *not* a power. I suspect it might be a rather fussy (but ambiguous) notation for an order statistic, in which case there is a bit more content to the question--and the convergence is not to zero. (What it actually converges to depends on whether the values are sorted in ascending or descending order.) Another problem with the notation is that it won't even make sense for $n\lt 3$ for then $\lfloor n/3\rfloor=0$! – whuber Nov 17 '14 at 23:27
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    I agree with @whuber - with the $^{(n)}$ superscript I'd take that to be the $\lfloor n/3 \rfloor$ th largest observation of $n$ (but we have to assume $n\geq 3$). It's best not to alter notation you're not certain of. – Glen_b Nov 18 '14 at 07:35
  • @Glen_b Can you tell me how to proceed in that case? – kanbhold Nov 18 '14 at 07:42
  • Perhaps start [here](http://en.wikipedia.org/wiki/Order_statistic#Probability_distributions_of_order_statistics). A number of questions here deal with uniform order statistics. – Glen_b Nov 18 '14 at 09:31
  • That information might give you some ideas about how to proceed to think about convergence, perhaps. – Glen_b Nov 18 '14 at 09:38
  • See the analysis at http://stats.stackexchange.com/questions/45124/central-limit-theorem-for-sample-medians/86804#86804. – whuber Nov 18 '14 at 16:03
  • @whuber thanks, but I don't see how that helps me. – kanbhold Nov 18 '14 at 16:45
  • I think it would be beneficial if the OP posted the _exact_ and _full_ exam question. It appears that the issue here is: We have a sequence of i.i.d. uniforms, and we consider the $\lfloor (n/3) \rfloor$ order statistic. E.g. if $n=30$ or $31$, then $\lfloor (n/3) \rfloor = (10)$ i.e. the $10$-th rder statistic -and the "tricky part" here is to determine whether, as $n \rightarrow \infty$, $\lfloor (n/3) \rfloor$ goes also to infinity or not. But since the notation is not $\lfloor (n/3) \rfloor$, but $(n)$ and $\lfloor n/3 \rfloor$ are kept separate, it might be something else, after all. – Alecos Papadopoulos Nov 18 '14 at 17:28
  • @Alecos There must be a typo in your comments, because obviously $\lfloor n/3 \rfloor \to \infty$ as $n\to\infty$. The question, as we are interpreting it, is resolved by studying the sampling distribution of this particular tercile, which is going to converge to the tercile of the parent distribution. (There are other ways to solve the problem, too: very little information is actually needed about the sampling distribution.) Since we have both exact and asymptotic formulas available for the sampling distribution--courtesy of the thread I have linked to--all the analysis needed has been done. – whuber Nov 18 '14 at 18:19
  • @whuber No typo -it may be the case that this "obviously" part is just what the exam exercise wanted to test the students about -whether fruitful or not (we don't even know what level the course is). But exactly this is the reason why I asked for the whole original exam question. The meaning of $X_{\lfloor n/3 \rfloor}^{(n)}$ is still unclear. – Alecos Papadopoulos Nov 18 '14 at 18:26
  • @AlecosPapadopoulos I can't post the exact question here because it is written in my native language. But the notation looks like just as I posted, and there is no additional information. – kanbhold Nov 18 '14 at 18:37
  • Re the edit: What is the variance of those Beta distributions? What happens to it as $n$ increases? (To turn these observations into a formal demonstration, use Chebyshev's Inequality.) – whuber Nov 19 '14 at 19:28
  • @whuber So $1/3$ is not the right answer? Why do I need the variances? – kanbhold Nov 19 '14 at 19:34
  • You have to prove that this sequence of random variables converges--and you have to be clear about the sense in which it converges. Most graders of an exam like this would give you no, or almost no, partial credit simply for asserting the "answer" is $1/3$. – whuber Nov 19 '14 at 19:35
  • And how someone proves it? I have never done such thing before. – kanbhold Nov 19 '14 at 19:47

1 Answers1

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In general, one can show that if $X_1, \ldots, X_n \text{ i.i.d. } \sim F$, then: $$\sqrt{n}(X_{(np)} - p)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{p(1 - p)}{f^{2}(\xi_p)}\right)$$ where $f$ is the density of $F$. This result is classical, and can be proven in many ways, for one possible reference, see Example 2.4.9 in Lehmann's book: Elements of large-sample theory. In your case, based on this result, you can show that: $$\sqrt{n}(X_{(n/3)} - 1/3)\overset{d}\rightarrow \mathrm{N}\left(0, \frac{2}{9}\right).$$ The above statement is about convergence in distribution, using Slutsky's theorem, your educated case can be confirmed --- but you have to point out the convergence mode: it converges in probability. It is incorrect to just say "it converges to 1/3". To show it converges in probability to 1/3, it is more straightforward, you can just show $\mathrm{Var}(X_{(n/3)}) \rightarrow 0$ as $n \to \infty$.

amoeba
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Zhanxiong
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  • I have one more probably silly question. If a series of r.v.s converge to a constant does it imply that the variance converges to zero, or only the opposite is true? – kanbhold Nov 19 '14 at 22:32
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    Consider a sequence of independent $X_n$ for which $X_n/\sqrt{n}$ is a Bernoulli$(1/n)$ variable, $n=1, 2, 3, \ldots$. Although they converge to $0$ (in probability and distribution), their variances are equal to $1-1/n$ which converges to $1\ne 0$. – whuber Nov 19 '14 at 22:51