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Let $X_1,X_2,\ldots,X_n$ be i.i.d and $X_{(n)}=\max_{1 \leq i \leq n}X_i$. If $X_i \sim \text{Beta}(1,\beta)$, find a value of $r$ such that $n^r(1-X_{(n)})$ converges in distribution.

So, I tried proving convergence in probability and then using a specific $\epsilon$ but I got stuck.

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Hint: write down an expression for $P(n^r ( 1 - X_{(n)}) \ge t)$, and choose $r$ so that the expression converges to some limit as $n \to \infty$.


$$P(n^r(1-X_{(n)}) \ge t) = P(X_{(n)} \le 1 - t n^{-r}) = P(X_1 \le 1 - tn^{-r})^n = ( 1 - (tn^{-r})^\beta)^n$$ where the last step uses the CDF of the $\text{Beta}(1, \beta)$ distribution. To choose the appropriate value of $r$, it may help to recall the limit $(1 + \frac{u}{n})^n \to e^u$ for real $u$.

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  • $\begingroup$ I think I have the answer r=1/$\beta$, right? $\endgroup$ – Bayesian guy Sep 1 '19 at 2:07

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