# i.i.d. random variables uniform distribution

If I have an i.i.d. random variables $$X_i$$ from a uniform distribution on $$[0,1]$$.

How would I find scaling sequences $$a_n$$, $$b_n$$ such that $$a_n(M_n − b_n)$$ converges in distribution to a non-trivial limit function $$G$$ for

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

1. Answer to the earlier version of the question which mentioned $$M_n$$ is distributed as $$\max\{ X_1, X_2, \dots, X_n\}$$:

Note that, $$\Pr(a_n(M_n - b_n)\le x) = \Pr(M_n\le b_n +x/a_n) = \prod_{i=1}^n \Pr(Y_i\le b_n +x/a_n) = \left(b_n +x/a_n \right)^n.$$ So, one possible choice would be $$b_n=1$$ and $$a_n=n.$$ In that case $$n(M_n - 1)\Rightarrow Z$$ where $$-Z$$ follows standard exponential distribution.

1. Answer to the question which says $$M_n$$ is distributed as $$\max\{ 1/X_1, 1/X_2, \dots, 1/X_n\}$$:

Let $$Y_i = 1/X_i.$$ Note that,$$\Pr(a_n(M_n - b_n)\le x) = \prod_{i=1}^n \Pr(Y_i\le b_n +x/a_n) = \prod_{i=1}^n \Pr\left(X_i \ge \left(b_n +x/a_n\right)^{-1}\right) = \left(1-\frac{a_n}{a_n b_n + x}\right)^n.$$ Now, it is well known that if $$\displaystyle\lim_{n\to\infty} x_n = x,$$ then $$\displaystyle\lim_{n\to\infty} (1+x_n/n)^n = e^x.$$ Using this fact, we can write $$\lim_{n\to\infty} \left(1-\frac{a_n}{a_n b_n + x}\right)^n = \exp\left(\displaystyle\lim_{n\to\infty}-\frac{na_n}{a_n b_n + x}\right).$$ Thus, we can take $$a_n = 1/n$$ and $$b_n=0,$$ for which the last limit will be $$\exp(-1/x)$$. Thus, $$n^{-1}(M_n - 0)\Rightarrow Z$$ where $$Z$$ has CDF $$F(x) = \exp(-1/x) \mathbf{1}(x\ge 0).$$

Note, if $$b_n$$ is any sequence such that $$b_n/n$$ converges, say to $$b,$$ then $$n^{-1}b_n\to b$$ and $$n^{-1} M_n \Rightarrow Z$$ implies that $$n^{-1} (M_n - b_n) \Rightarrow Z-b,$$ whose CDF is $$F(x+b).$$

• thank you so so much I'm still abit confused is the bit were it says find 'an bn s.t an(mn-bn) converges to a non trivial limit function G' connected to the Yi=Xi and Mn=max(Y1,...Yn) or are they seperate things, Because I also have Ui = 1/Xi, and Mn = max{U1,...,Un} Jan 12, 2020 at 17:18
• Did you mean $Y_i = 1/X_i?$ But your statement of the problem shows $Y_i = X_i$... Jan 12, 2020 at 18:32
• Thats so amazing thank you, yes I would like to know both as you have demonstrated. I just wanted to ask one last thing is it possible to in each case to find suitable scaling sequences an, bn, where un = u/an + bn so that you get a non-trivial limit G(u) as n → ∞. Jan 14, 2020 at 18:41