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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!

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    $\begingroup$ 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? $\endgroup$
    – whuber ♦
    Commented Nov 17, 2014 at 22:58
  • $\begingroup$ @whuber I edited my question. $\endgroup$
    – kanbhold
    Commented Nov 17, 2014 at 23:14
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    $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Commented Nov 17, 2014 at 23:18
  • $\begingroup$ 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 $X$. $\endgroup$
    – kanbhold
    Commented Nov 17, 2014 at 23:23
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    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented Nov 18, 2014 at 7:35

1 Answer 1

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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 $n \to \infty$.

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  • $\begingroup$ 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? $\endgroup$
    – kanbhold
    Commented Nov 19, 2014 at 22:32
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    $\begingroup$ 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 $1\ne 0$. $\endgroup$
    – whuber ♦
    Commented Nov 19, 2014 at 22:51

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