# Question about asymptotic distribution of order statistics.

Let $$U_i$$'s be i.i.d. Uniform$$(0,1)$$ random variables, and let $$X_n = min\{U_1, \dots , U_n \}, Y_n = max\{ U_1 , \dots , U_n \}, Z_n = Y_n - X_n$$. Then find asymptotic distribution of $$n(1-Z_n)$$.

I understand how to find asymptotic distribution of $$X_n , Y_n$$. But this problem makes me confused. Also I tried to use this method($$X_{(r)} \sim Beta(r,n-r+1)$$), but it failed.

Please help!

I know $$Z_n=Y_n-X_n \sim Beta(n-1,2)$$. So, cdf of $$Z_n$$ is

$$P(Z_n \leq t) = n(n-1)\int_{0}^{t} x^{n-2}(1-x)dx$$

$$P(1-Z_n \leq t) = 1-n(n-1) \int_{0}^{1-t}x^{n-2}(1-x)dx$$

$$P(n(1-Z_n) \leq t) = 1-n(n-1)\int_{0}^{1-t/n}x^{n-2}(1-x)dx$$

I don't know what to do after this.

• did you mean $U_1, \dots, U_n$ instead of $X_1, \dots, X_n$? – tortue 2 days ago
• Find the distribution of $Z_n$ (see this or this) and hence cdf of $n(1-Z_n)$. – StubbornAtom 2 days ago
• @tortue Oh, sorry. That's typo. – voidcome 2 days ago
• @StubbornAtom I understand your comment. Thx! – voidcome 2 days ago
• @StubbornAtom I'm sorry to bother you, but can you explain how to find cdf of that one? – voidcome 2 days ago

## 1 Answer

starting from your Range density that is a $$Z_n\sim Beta(n-1;2)$$ transforming your

$$Y_n=n(1-Z_n)$$

with usual technics:

$$f_Y(y)=f_X(g^{-1}(y))|\frac{d}{dy}g^{-1}(y)|$$

you get

$$f_{Y_n}(y)=\frac{n-1}{n}\cdot y\cdot \left(1-\frac{y}{n}\right)^{n-2}$$

whose limit results to me

$$\lim\limits_{n \to \infty}f_{Y_n}(y)=ye^{-y}$$

which is the density of a $$Gamma(2;1)$$

• Did you mean that pdf of $Y = n(n-1)(y/n)(1-y/n)^{n-2} (1/n)$ because jacobian is 1/n? – voidcome 2 days ago
• @voidcome : no, it is not a $2\rightarrow 2$ transformation , Just use the univariate transformation theorem. See my edits. If my answer has been useful you can mark it as accepted – tommik 2 days ago