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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.

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

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  • $\begingroup$ Did you mean that pdf of $Y = n(n-1)(y/n)(1-y/n)^{n-2} (1/n)$ because jacobian is 1/n? $\endgroup$ – voidcome 2 days ago
  • $\begingroup$ @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 $\endgroup$ – tommik 2 days ago

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