I have a random sample of size $n$ from a uniform distribution $U(0, \theta).$

And I've proven that the pdf of $Y_n$, the n-th order statistic of the sample is:

$$ f_{Y_n}(y) = \begin{cases}\frac{n}{\theta^n} y^{n-1} & 0<y<\theta ,\\ 0 &\text{elsewhere}\end{cases} $$

Now, what I'm trying to do next is calculating the limiting distribution of $Y_n$, and I'm not sure how to do that.

Am I supposed to calculate the limit of the pdf as $ n \rightarrow \infty $ ? or the cdf?

  • 1
    $\begingroup$ (a) The $n!$ in your pmf is not the right constant; (b) it is easier to use the cdf than the pmf and you have to find the right $\alpha$ power for $$n^\alpha\{\theta-Y_n\}\stackrel{n\to\infty}{\longrightarrow} \mathcal{L}$$ where $\mathcal{L}$ is your limiting distribution. $\endgroup$
    – Xi'an
    Dec 27, 2014 at 15:13
  • $\begingroup$ @Xi'an (a) You are right. It should be $n$. I edited the question. (b) What is the notion/formula you are using here? $\endgroup$
    – iTurki
    Dec 27, 2014 at 15:16
  • $\begingroup$ stats.stackexchange.com/q/96216/119261 $\endgroup$ Jun 4, 2020 at 6:24

2 Answers 2


If you look at the cdf of $Y_n$, $$F_n(\delta)=\mathbb{P}(Y_n\le\delta)=(\delta/\theta)^n\qquad0\le\delta\le\theta\,,$$ you get that $F_n(\delta)$ converges to zero when $0\le\delta<\theta$, which means that $Y_n$ converges in probability to $\theta$: $$Y_n\stackrel{\text{prob}}{\longrightarrow} \theta\,.$$ This implies that $Y_n$ also converges in distribution to the constant value random variable $\theta$: $$Y_n\stackrel{\text{dist}}{\longrightarrow} \theta\,.$$ To get a more precise description of the asymptotic behaviour of $Y_n$, you need to zoom around $\theta$, i.e., to consider $(\theta-Y_n)$ scaled by a power of $n$, $n^\alpha$, so that, while $(\theta-Y_n)$ converges to zero in distribution and $n^\alpha$ goes to infinity, the product $$n^\alpha(\theta-Y_n)$$ converges to a standard distribution (in distribution).

This is e.g. the case for the Central Limit theorem: if the mean of $X$ is well-defined, $\bar{X}_n-\mathbb{E}[X]$ converges to zero in distribution, while $$\sqrt{n}(\bar{X}_n-\mathbb{E}[X])=n^{1/2}(\bar{X}_n-\mathbb{E}[X])$$ converges to a Normal distribution (in distribution).

To answer your question you thus have to find the right scale $n^\alpha$ (there is only one!) and then deduce the associated limiting distribution. Hint: Remember that $$\lim_{n\to\infty} \left\{1-\frac{\beta}{n} \right\}^n = \exp\{-\beta\}\,.$$


As suggested by Xi'an, work with the CDF $$ F_n(t) = \Pr(Y_n\leq t)=\bigcap_{i=1}^n \Pr(X_i\leq t) = \prod_{i=1}^n \Pr(X_1\leq t) \, . $$ Hence, $$ F_n(t) = \begin{cases} 0 &,& t<0\,; \\ \left(t/\theta\right)^n &,& 0\leq t<\theta\,; \\ 1 &,& t\geq\theta \, . \end{cases} $$ Plot a graph of this $F_n$.

For $t<0$ and $t\geq\theta$ you're done. Now, what is $\lim_{n\to\infty}F_n(t)$ for $0\leq t<\theta$?

Put everything together to find $F(t)$ such that $F_n(t)\to F(t)$ for every $t$ when $n\to\infty$.

Graph this $F$. What is the interpretation of $F$?

Is this result intuitive? Imagine yourself drawing huge samples from a $U[0,\theta]$ distribution and computing the sample maximum every time.

Can you code a short simulation in R which confirms your results?

  • $\begingroup$ I am not sure I understand your solution, Paulo: when $n$ goes to $+\infty$, the plot of $F_n$ will spike from $0$ to $1$ at $\theta$, which does not help in finding the convergence speed of $Y_n$. $\endgroup$
    – Xi'an
    Dec 27, 2014 at 20:32
  • $\begingroup$ Hi @Xi'an! I thought he just wanted the limiting distribution, which is a point mass at $\theta$. Did he also ask for the speed of convergence? $\endgroup$
    – Zen
    Dec 28, 2014 at 0:26
  • $\begingroup$ I would presume such a question about convergence in distribution implicitly calls for the speed $n^\alpha$ that produces a non-degenerate limit. Which leads to an interesting query: if $Y_n$ converges in distribution to $0$, when does it exist an $\alpha$ such that $n^\alpha Y_n$ converges in distribution to a non-degenerate limit? $\endgroup$
    – Xi'an
    Dec 28, 2014 at 7:17

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