Asymptotic distribution of $X_{(n)}$

Question

Let $X_1,X_2,\dots,X_n \sim U(0,1)$. They are iid. Let $X_{(n)}=\max\{X_1,X_2,\dots,X_n\}$, then what is the asymptotic distribution of $X_{(n)}$.

I know the method for finding the asymptotic distribution by Delta Method but I'm unable to find the derivative here. As delta method says that $$ \sqrt{n}\Big(g(X_n)-g(\theta)\Big)\sim N\big(0,\sigma^2[g^\prime(\theta)]^2\big) $$ But here how to compute $[g^\prime(\theta)]^2$?

Answer 1

For $0<t<1$:

$$F_{X_n}(t)=P(\max\{X_1,...X_n\}\le t)=P(X_1\le t,\ldots,X_n\le t)=P(X_1\le t)\cdots P(X_n\le t)=t^n$$

So

$$F_{X_n}(t)= \begin{cases} 0 & \text{ if } t\le 0\\ t^n & \text{ if } 0<t<1\\ 1 & \text{ if } t\ge 1 \end{cases} $$

From the equation above it is clear that $X_{(n)}\xrightarrow{P} 1$ (and in distribution). However, this limit statement to a degenerate distribution does not fully reveal the asymptotic distribution of $X_{(n)}$. So we search for sequences $k_n$ and $a_n$ such that $k_n(X_{(n)}-a_n)$ has a non-degenerate limiting distribution.

Computing the distribution function of $k_n(X_{(n)}-a_n)$ directly we have:

$$F(u)=P(k_n(X_{(n)}-a_n)\le u)=P\left(X_{(n)}\le\frac{u}{k_n}+a_n\right)$$

as long as $k_n>0$. Therefore:

$$F(u)=\left(\frac{u}{k_n}+a_n\right)^n\text{ for }0<\frac{u}{k_n}+a_n<1$$

We would like this expression to tend to a limit involving only $u$ as $n\rightarrow\infty$.

We observe that if we take $a_n=1$ and $k_n=n$ so that $F(u)=\left(1+\frac{u}{n}\right)^n$ which tends to $e^u$

We further observe that the values of $u$ which make the above limit valid are $0<a_n+\frac{u}{k_n}<1$ which in this case becomes $-1<\frac{u}{n}<0$. So $u$ may be any negative real number, since for any $u<0$, $-1<\frac{u}{n}<0$ for all $n>|u|$. We conclude that the random variable $U$ has the distribution function:

$$F(u)= \begin{cases} e^u & \text{ if }u\le 0\\ 1 & \text{ if }u > 0 \end{cases} $$

then $n(X_{(n)}-1)\xrightarrow{d} U$. Since $-U$ is a standard exponential random variable, we may also write:

$$n(1-X_{(n)})\xrightarrow{d} \text{Exponential}(1)$$

Answer 2

The CDF of $X_{(n)}$ is $$F_{X_{(n)}}(t) = \mathbb P(X_{(n)} \le t) = t^n \ \text{for}\ 0 < t < 1$$ To have a nontrivial limit as $n \to \infty$, we'll want $t \sim 1 - \text{const} / n$. Thus for $0 < s < n$, $$F_{X_{(n)}}(1 - s/n) = (1-s/n)^n \to \exp(-s) $$ Since the left side can be written as $\mathbb P(n(1-X_{(n)}) \ge s$, this says $n(1-X_{(n)})$ converges in distribution to an exponential distribution with parameter $1$.

Answer 3

\begin{align} \text{right: } \sqrt{n}\Big(g(\,\overline X_n)-g(\theta)\Big) & \sim N\big(0, \sigma^2 [g^\prime(\theta)]^2\big) \\ \text{wrong: } \sqrt{n}\Big(g(X_n)-g(\theta)\Big) & \sim N\big(0, \sigma^2 [g^\prime(\theta)]^2\big) \end{align}

The first page of the pdf file to which you link, on the delta method, is about the distribution of $\bar X_n$, not about that of $X_n$ or of $X_{(n)} = \max\{ X_1, \ldots, X_n \}.$

I must dispute the answer by "Momo". Note that the weak law of large numbers tells us that in the limit, the distribution of $\overline X_n$ puts probability $1$ at the expected value, often denoted $\mu$. But the two lines displayed above speak of the distribution of $\sqrt n\left(\,\overline X_n - \mu\right)$, which is given by the central limit theorem as a normal distribution. The difference between these results is that we are scaling by multiplying by $\sqrt n$. Just as the C.L.T. gives us a normally distributed sample mean, we will get an exponentially distributed sample maximum.

We have \begin{align} \operatorname{E}(\,X_{(n)} ) & = \frac n {n+1}, \\[10pt] \operatorname{var}(\,X_{(n)}) & = \frac n {(n+2)(n+1)^2} \sim \frac 1 {(n+1)^2} \end{align} where $A\sim B$ means $A/B\to 1$ as $n\to\infty$. So the standard deviation of $X_{(n)}$ is asymptotically $1/(n+1)$.

Now we have \begin{align} \text{for } x>0, & \quad \Pr\left( \frac{1-X_{(n)}}{\sqrt{\operatorname{var}(X_{(n)})}} > x \right) = \Pr\Big( (n+1)(1 - X_{(n)}) > x\Big) \\[10pt] = {} & \Pr \left( X_{(n)} < 1 - \frac x {n+1} \right) = \left( 1 - \frac x {n+1} \right)^n \to e^{-x} \text{ as } n\to\infty. \end{align}

Thus asymptotically, $(n+1) ( 1-X_{(n)} )$ is exponentially distributed with expected value $1$.

The multiplication by $(n+1)$ plays the same kind of role as the multiplication by $\sqrt n$ in the central limit theorem.

None of this alters the fact that the limiting distribution of $X_{(n)}$ concentrates probability $1$ at $1$ (as shown in "Momo"'s answer), just as the sample mean from a distribution with finite variance concentrates probability $1$ at the population mean. But "asymptotic distribution" involves the rescaling by multiplying by $\sqrt n$ in the first two lines of this answer and in the central limit theorem, and by $n+1$ in the result shown here.