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Let $X_1,X_2,...$ Be independent random variables with common density:

$$f_X(x)=\alpha x^{-(\alpha+1)}. x>1$$

Where $\alpha>0$. Define a new sequence of random variables:

$$Y_n=(1/n^{1/\alpha})X_{(n)}$$

Where $X_{(n)}$ is the highest observation of n I.I.d. r.v. $X_1,…,X_n$. Show that $Y_n$ converges in distribution as $n\to \infty$ and find the limiting distribution.

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Hint: $$ \Pr(Y_n\leq y)=\Pr(\max X_n\leq yn^{\frac 1\alpha})=\left(\Pr( X_1\leq yn^{\frac 1\alpha})\right)^n\\ =\left(\int_1^{yn^{\frac 1\alpha}} \alpha x^{-\alpha-1}dx\right)^n=\left(\frac{-1}{x^\alpha}\Bigg|_1^{yn^{\frac 1\alpha}}\right)^n=\left(1-\frac{1}{ny^\alpha}\right)^n. $$

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  • $\begingroup$ Thanks Arash I ended up getting $\lim{n\to\infty}F_Y(y)=exp(-1/y^{\alpha})$ does this sound right? $\endgroup$ – user135784 Mar 16 '14 at 14:20
  • $\begingroup$ Seems correct to me! $\endgroup$ – Arash Mar 16 '14 at 14:29
  • $\begingroup$ Cheers man that one was bugging me!! $\endgroup$ – user135784 Mar 16 '14 at 21:57

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