Let's solve some theoretical problem in probability, specifically on convergence. The problem below is originally from Exercise 5.42 of Casella and Berger (2001). And I just want to share my solution on this. If there is an incorrect argument below, I would be happy if you could point that to me.

### Problem

Let $X_1, X_2,\cdots$ be iid (independent and identically distributed) and $X_{(n)}=\max_{1\leq i\leq n}x_i$.- If $x_i\sim$ beta(1,$\beta$), find a value of $\nu$ so that $n^{\nu}(1-X_{(n)})$ converges in distribution;
- If $x_i\sim$ exponential(1), find a sequence $a_n$ so that $X_{(n)}-a_n$ converges in distribution.

### Solution

- Let $Y_n=n^{\nu}(1-X_{(n)})$, we say that $Y_n\rightarrow Y$ in distribution. If $$\lim_{n\rightarrow \infty}F_{Y_n}(y)=F_Y(y).$$ Then, $$ \begin{aligned} \lim_{n\rightarrow\infty}F_{Y_n}(y)&=\lim_{n\rightarrow\infty}P(Y_n\leq y)=\lim_{n\rightarrow\infty}P(n^{\nu}(1-X_{(n)})\leq y)\\ &=\lim_{n\rightarrow\infty}P\left(1-X_{(n)}\leq \frac{y}{n^{\nu}}\right)\\ &=\lim_{n\rightarrow\infty}P\left(-X_{(n)}\leq \frac{y}{n^{\nu}}-1\right)=\lim_{n\rightarrow\infty}\left[1-P\left(-X_{(n)}> \frac{y}{n^{\nu}}-1\right)\right]\\ &=\lim_{n\rightarrow\infty}\left[1-P\left(\max\{X_1,X_2,\cdots,X_n\}< 1-\frac{y}{n^{\nu}}\right)\right]\\ &=\lim_{n\rightarrow\infty}\left[1-P\left(X_1< 1-\frac{y}{n^{\nu}},X_2< 1-\frac{y}{n^{\nu}},\cdots,X_n< 1-\frac{y}{n^{\nu}}\right)\right]\\ &=\lim_{n\rightarrow\infty}\left[1-P\left(X_1< 1-\frac{y}{n^{\nu}}\right)^n\right],\;\text{since}\;X_i's\;\text{are iid.} \end{aligned} $$