# Find the limiting distribution of $(bn)^{-\frac{1}{\alpha}} X_{(n)}$

let $$\{X_n\}_{n\geq 1}$$ be a sequence of i.i.d random variables with common distribution $$F$$, and write

$$X_{(n)}=\max\{X_1,\cdots , X_n\}$$ , $$n=1,2,\cdots$$

(a) for $$\alpha >0$$ , $$\lim_{x\rightarrow \infty } x^{\alpha} P\{X_1 >x \}=b>0$$. Find the limiting distribution of

$$(bn)^{-\frac{1}{\alpha}} X_{(n)}$$

(b) if F satisfies

$$\lim_{x\rightarrow \infty } e^{x} [1-F(x)]=b>0$$

find the limiting distribution of

$$X_{(n)} - \log(bn)$$

(c) if $$X_i$$ is bounded above $$x_0$$ with probability 1, and for some $$\alpha>0$$

$$\lim_{x\rightarrow x_0^{-} } (x_0-x)[1-F(x)]=b>0$$

find the limiting distribution of

$$(bn)^{-\frac{1}{\alpha}} (X_{(n)}-x_0)$$

My try I got stuck in the part (a)

part (a)

$$Z=(bn)^{-\frac{1}{\alpha}} X_{(n)}$$

$$F^{(n)}_Z(z)=P(Z\leq z)=P\bigg((bn)^{-\frac{1}{\alpha}} X_{(n)} \leq z \bigg)=F_{X_{(n)}}\bigg((bn)^{\frac{1}{\alpha}} z\bigg)=\{F_{X_1}\big((bn)^{\frac{1}{\alpha}} z\big)\}^n$$

$$=\{1-P\big(X_1>(bn)^{\frac{1}{\alpha}} z\big)\}^n$$

$$=\{1-\frac{ \big( (bn)^{\frac{1}{\alpha}} z \big)^{\alpha} }{\big( (bn)^{\frac{1}{\alpha}} z \big)^{\alpha}}P\big(X_1>(bn)^{\frac{1}{\alpha}} z\big)\}^n$$

$$=\{1-\frac{ \big( (bn)^{\frac{1}{\alpha}} z \big)^{\alpha} P\big(X_1>(bn)^{\frac{1}{\alpha}} z\big) }{\big( (bn)^{\frac{1}{\alpha}} z \big)^{\alpha}}\}^n$$

$$\rightarrow \{1-\frac{ b }{\big( (bn)^{\frac{1}{\alpha}} z \big)^{\alpha}}\}^n$$ $$\hspace{1cm}$$ (1)

$$=\{1-\frac{ b }{\big( (bn) z^{\alpha} \big)}\}^n$$

$$=\{1-\frac{ 1 }{\big( n z^{\alpha} \big)}\}^n\rightarrow e^{-\frac{1}{z^{\alpha}}}$$

is this calculation is valid and what now? (if $$X_n$$ be a sequence of positive Random variables so $$\frac{1}{z}\sim weibull(\alpha,1)$$ , but $$X$$ is just continues variable).

and What is its application? any idea?

Update

in (1) I am not sure that

$$\big( (bn)^{\frac{1}{\alpha}} z \big)^{\alpha} P\big(X_1>(bn)^{\frac{1}{\alpha}} z\big)$$

can be replaced with $$b$$ so I do another way.

$$F^{(n)}_Z(z)=e^{\log F^{(n)}_Z(z)}=e^{\log F^{(n)}_Z(z)}$$

$$=e^{\log\{F_{X_1}\big((bn)^{\frac{1}{\alpha}} z\big)\}^n}$$ $$=e^{n\log\{F_{X_1}\big((bn)^{\frac{1}{\alpha}} z\big)\}}$$

define $$t=(bn)^{\frac{1}{\alpha}} z$$

$$=e^{n\log F_{X_1}(t)}=e^{n\log F(t)}=e^{n\log(1- p(X>t))}$$

$$=e^{n \bigg( - p(X>t)-\frac{1}{2} p(X>t)^2- \frac{1}{3} p(X>t)^3 \cdots \bigg) }$$ $$\hspace{1cm}$$ (2) Use Taylor series $$\log(1-p(X>t))$$

$$=e^{ \bigg( -n p(X>t)-n\frac{1}{2} p(X>t)^2- n\frac{1}{3} p(X>t)^3 \cdots \bigg) }$$

$$=e^{ \bigg( -n p(X>(bn)^{\frac{1}{\alpha}} z)-n\frac{1}{2} p(X>(bn)^{\frac{1}{\alpha}} z)^2- n\frac{1}{3} p(X>(bn)^{\frac{1}{\alpha}} z)^3 \cdots \bigg) }$$

$$\rightarrow e^{ \bigg( -n p(X>(bn)^{\frac{1}{\alpha}} z)-0 \bigg) }$$

$$=e^{ -n p(X>(bn)^{\frac{1}{\alpha}} z) }$$

$$=e^{ -\frac{\big( (bn)^{\frac{1}{\alpha}} z\big)^\alpha}{\big( (b)^{\frac{1}{\alpha}} z\big)^\alpha} p(X>(bn)^{\frac{1}{\alpha}} z) }$$

$$\rightarrow e^{ -\frac{b}{\big( (b)^{\frac{1}{\alpha}} z\big)^\alpha} }$$

$$= e^{ -\frac{1}{ z^\alpha} }$$

so

$$\lim_{n\rightarrow \infty}F_{Z_n}(z) = \left\{ \begin{array}{cc} e^{ -\frac{1}{ z^\alpha} } & z\geq 0 \\ 0 & z <0 \end{array} \right. \sim weibull(\alpha,1)$$