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,
$$
\begin{aligned}
\lim_{n\rightarrow\infty}F_{Y_n}(y)&=\lim_{n\rightarrow\infty}P(Y_n\leq y)=\lim_{n\rightarrow\infty}P(n^{\nu}(1-X_{(n)})\leq y)\\
&=\lim_{n\rightarrow\infty}P\left(1-X_{(n)}\leq \frac{y}{n^{\nu}}\right)\\
&=\lim_{n\rightarrow\infty}P\left(-X_{(n)}\leq \frac{y}{n^{\nu}}-1\right)=\lim_{n\rightarrow\infty}\left[1-P\left(-X_{(n)}> \frac{y}{n^{\nu}}-1\right)\right]\\
&=\lim_{n\rightarrow\infty}\left[1-P\left(\max\{X_1,X_2,\cdots,X_n\}< 1-\frac{y}{n^{\nu}}\right)\right]\\
&=\lim_{n\rightarrow\infty}\left[1-P\left(X_1< 1-\frac{y}{n^{\nu}},X_2< 1-\frac{y}{n^{\nu}},\cdots,X_n< 1-\frac{y}{n^{\nu}}\right)\right]\\
&=\lim_{n\rightarrow\infty}\left[1-P\left(X_1< 1-\frac{y}{n^{\nu}}\right)^n\right],\;\text{since}\;X_i's\;\text{are iid.}
\end{aligned}
$$
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 $\nu=\frac{1}{\beta}$.
$\hspace{12.5cm}\blacksquare$ -
$$
\begin{aligned}
P(X_{(n)}-a_{n}\leq y) &= P(X_{(n)}\leq y + a_n)=P(\max\{X_1,X_2,\cdots,X_n\}\leq y+a_n)\\
&=P(X_1\leq y+a_n,X_2\leq y+a_n,\cdots,X_n\leq y+a_n)\\
&=P(X_1\leq y+a_n)^n,\;\text{since}\;x_i's\;\text{are iid}\\
&=\left[\int_{-\infty}^{y+a_n}f_{X_1}(t)\,\mathrm{d}t\right]^n
\end{aligned}
$$
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 $a_n=\log n$.
$\hspace{12.5cm}\blacksquare$
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