# Finding scaling sequences of i.i.d. random variables

I've recently been studying i.i.d random variables and I think I'm starting to grasp the concept. I've been set this question but been given very little guidance on how to answer it and am very confused. Any help is greatly appreciated.

Consider i.i.d. random variables $(X_i)$ drawn from a uniform distribution on $[0, 1].$

In the following find scaling sequences $a_n$, $b_n$ such that $a_n(M_n − b_n)$ converges in distribution to a non-trivial limit function $G$.

(a) $Y_i = X_i,$ and $M_n = \max(Y_1, . . . , Y_n);$

(b) $U_i = \frac{1}{X_i}$, and $M_n = \max(U_1, . . . , U_n)$;

In each case find first of all the probability distribution function $P(M_n ≤ u/a_n + b_n)$ as a function of $u_n = u/a_n + b_n$. Then find suitable scaling sequences $a_n$, $b_n$ so that you get a non-trivial limit $G(u)$ as $n → ∞$.

• What have you tried so far? You should add some details about where you found problems. Jan 7, 2018 at 11:17
• I'm not really sure about the wording of the question. Also I don't really know how to find the probability distribution function in the terms given Jan 7, 2018 at 12:35
• I think that $G(u)$ transforms to one of the types in the fisher theorem Jan 7, 2018 at 12:40

Let me try to set you on your way. First of all, try to determine the distribution of $M_n$. Notice that

$$P(M_n\leq x) = P(X_i\leq x, \forall i)$$

This plus the independence and equality in distribution of the $X_i$ should lead you to the following formula:

$$P(M_n\leq x) = \begin{cases}0 &, \text{for} \; x<0 \; ,\\ x^n &, \text{for} \; 0\leq x<1 \; ,\\ 1 &, \text{for} \; 1\leq x \; .\end{cases}$$

Replace $x$ by $\frac{u}{a_n}+b_n$. You'll see that the middle term should be a familiar form, it should remind you of

$$\lim_{n\to \infty}\left(1+\frac{u}{n}\right)^n = e^u \; .$$

EDIT For part b) you'll obtain

$$P\left(M_n\leq \frac{u}{a_n}+b_n\right) = \begin{cases}0 &, \text{for} \; \frac{u}{a_n}+b_n\leq 1 \; ,\\ \left(1-\frac{1}{\frac{u}{a_n}+b_n}\right)^n &, \text{for} \; \frac{u}{a_n}+b_n > 1 \; .\end{cases}$$

Putting $a_n=\frac{1}{n}$ and $b_n=n$ this becomes

$$P(M_n\leq n(u+1)) = \begin{cases}0 &, \text{for} \; n(u+1)\leq 1 \; ,\\ (1-\frac{1}{n(u+1)})^n &, \text{for} \; n(u+1)> 1 \; .\end{cases}$$

The limiting case is then

$$\lim_{n\to +\infty}P\left(\frac{M_n}{n}-1\leq u\right) = \begin{cases}0 &, \text{for} \; u \leq -1 \; ,\\ e^{-\frac{1}{u+1}} &, \text{for} \; u>-1 \; .\end{cases}$$

• So would I be right in saying something like for a) $F_n(t)=P(M_n<t)$ =$P(X_1,...,X_n<t)$ =$P(X<t)^n$. So as $n$ tends to infinity $F_n(\frac{t}{n}+1)$ tends to $e^t$? Jan 8, 2018 at 12:19
• You're on the right track. But it's rather $P(M_n<u/a_n+b_n)$ you have to look at which will give $(b_n+u/a_n)^n$. But selecting your $a_n$ and $b_n$ properly gets you there. And you can adapt a similar reasoning for part (b). Just be careful in working out $P(U<x)$. Jan 8, 2018 at 12:27
• And for b) $F_n(t)=P(M_n<t)$ ==$P(\frac{1}{X_1},...,\frac{1}{X_n}<t)$ =$P(\frac{1}{X}<t)^n$=$P(\frac{1}{t}<X)$=$(1-P(X<\frac{1}{t}))^n$. So as $n$ tends to infinity $F_n(1-\frac{1}{nt})^n$ tends to $e^-t^-1$? I'm unsure about this second one? Jan 8, 2018 at 12:30
• Would you select $a_n=n$ and $b_n=1$ for a)? Jan 8, 2018 at 12:34
• Yep, and $a_n=1/n$, $b_n=n$ for b) . Jan 8, 2018 at 12:41