### Probability Theory: Convergence in Distribution Problem

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 $X_{(n)}=\max_{1\leq i\leq n}x_i$.
1. If $x_i\sim$ beta(1,$\beta$), find a value of $\nu$ so that $n^{\nu}(1-X_{(n)})$ converges in distribution;
2. If $x_i\sim$ exponential(1), find a sequence $a_n$ so that $X_{(n)}-a_n$ converges in distribution.

### Solution

1. 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} And because $x_i\sim$ beta(1,$\beta$), the density is $$f_{X_1}(x)=\begin{cases} \beta(1-x)^{\beta - 1}&\beta>0, 0\leq x\leq 1\\ 0,&\mathrm{Otherwise} \end{cases}$$ Implies, \begin{aligned} \lim_{n\to \infty}P(Y_n\leq y)&=\lim_{n\to \infty}\left\{1-\left[\int_0^{1-\frac{y}{n^{\nu}}}\beta(1-t)^{\beta-1}\,\mathrm{d}t\right]^n\right\}\\ &=\lim_{n\to \infty}\left\{1-\left[-\int_1^{\frac{y}{n^{\nu}}}\beta u^{\beta-1}\,\mathrm{d}u\right]^{n}\right\}\\ &=\lim_{n\to \infty}\left\{1-\left[-\beta\frac{u^{\beta}}{\beta}\bigg|_{u=1}^{u=\frac{y}{n^{\nu}}}\right]^{n}\right\}\\ &=1-\lim_{n\to \infty}\left[1-\left(\frac{y}{n^{\nu}}\right)^{\beta}\right]^{n} \end{aligned} We can simplify the limit if $\nu=\frac{1}{\beta}$, that is $$\lim_{n\to\infty}P(Y_n\leq y)=1-\lim_{n\to\infty}\left[1-\frac{y^{\beta}}{n}\right]^{n}=1-e^{-y^{\beta}}$$ To confirm this in Python, run the following code using the sympy module

Therefore, if $1-e^{-y^{\beta}}$ is a distribution function of $Y$, then $Y_n=n^{\nu}(1-X_{(n)})$ converges in distribution to $Y$ for $\nu=\frac{1}{\beta}$.
$\hspace{12.5cm}\blacksquare$
2. \begin{aligned} P(X_{(n)}-a_{n}\leq y) &= P(X_{(n)}\leq y + a_n)=P(\max\{X_1,X_2,\cdots,X_n\}\leq y+a_n)\\ &=P(X_1\leq y+a_n,X_2\leq y+a_n,\cdots,X_n\leq y+a_n)\\ &=P(X_1\leq y+a_n)^n,\;\text{since}\;x_i's\;\text{are iid}\\ &=\left[\int_{-\infty}^{y+a_n}f_{X_1}(t)\,\mathrm{d}t\right]^n \end{aligned} Since $X_i\sim$ exponential(1), then the density is $$f_{X_1}=\begin{cases} e^{-x},&0\leq x\leq \infty\\ 0,&\mathrm{otherwise} \end{cases}$$ So that, \begin{aligned} P(X_{(n)}-a_{n}\leq y)&=\left[\int_{0}^{y+a_n}e^{-t}\,\mathrm{d}t\right]=\left\{-\left[e^{-(y+a_n)}-1\right]\right\}^n\\ &=\left[1-e^{-(y+a_n)}\right]^n \end{aligned} If we let $Y_n=X_{(n)}-a_n$, then we say that $Y_n\rightarrow Y$ in distribution if $$\lim_{n\to\infty}P(Y_n\leq y)=P(Y\leq y)$$ Therefore, \begin{aligned} \lim_{n\to\infty}P(Y_n\leq y) &= \lim_{n\to\infty}P(X_{(n)}-a_n\leq y)=\lim_{n\to \infty}\left[1-e^{-y-a_n}\right]^n\\ &=\lim_{n\to\infty}\left[1-\frac{e^{-y}}{e^{a_n}}\right]^n \end{aligned} We can simplify the limit if $a_n=\log n$, that is $$\lim_{n\to\infty}\left[1-\frac{e^{-y}}{e^{\log n}}\right]^n=\lim_{n\to\infty}\left[1-\frac{e^{-y}}{n}\right]^n=e^{-e^{-y}}$$ Check this in Python by running the following code,

In conclusion, if $e^{-e^{-y}}$ is a distribution function of Y, then $Y_n=X_{(n)}-a_n$ converges in distribution to $Y$ for sequence $a_n=\log n$.
$\hspace{12.5cm}\blacksquare$