Let $X_i, i \geq 1$ be i.i.d. r.v.s where density $g(x)$ is $$ \lim_\limits{x\to\infty} x^3g(x)=1 $$
- calculate $\lim_\limits{x\to\infty} x^2\mathbb{P}(X_1>x)$
- use that do prove that there exists sequence $a_n$ such that $\frac{X_n^*}{a_n}$ converges weakly to non generate distribution. What is its density?
Now, I start with:
$$ \lim_\limits{x\to\infty} x^2\mathbb{P}(X_1>x) = \lim_\limits{x\to\infty} x^2\int_\limits{x}^{\infty}g(s)ds = \dots $$ here I assumed $\lim_\limits{x\to\infty} x^3g(x)=1 \implies \lim_\limits{x\to\infty} g(x)=\frac{1}{x^3}$ so $\lim_\limits{x\to\infty} x^2\int_\limits{x}^{\infty}g(s)ds$ is like $\lim_\limits{x\to\infty} x^2\int_\limits{x}^{\infty}\frac{1}{s^3}ds $: $$ \cong \frac{1}{2x^2}x^2=\frac{1}{2} $$ Is this correct so far?
Now, to show $1-F$ is regulary-varying in infinity we do as follows:
$$ \lim_\limits{t\to\infty} \frac{\mathbb{P}(X_1>xt)}{\mathbb{P}(X_1>t)} = \lim_\limits{t\to\infty} \frac{\mathbb{P}(X_1>xt)x^2t^2}{\mathbb{P}(X_1>t)x^2t^2} = \frac{\frac{1}{2}}{\frac{1}{2}x^2} = x^{-2} \implies \gamma=-\frac{1}{-2} = \frac{1}{2} $$
Anyway, using Fisher–Tippett–Gnedenko and having proved $1-F$ is regulary-varying in infinity, that means $X_n^*$ indeed weakly converges to non-generate distribution. How does one find the limiting distribution though?
FTG says: $$ G_{\gamma}(s) = \exp{\left(-(1+\gamma s)^{\frac{-1}{\gamma}}\right)} $$ our gamma is $\frac{-1}{\alpha} = \frac{1}{2}$, so is the limiting distribution CDF $$ G_{\gamma}(s) = \exp{\left(-(1+\frac{1}{2}s)^{-2}\right)} $$ and the density is just $\frac{\partial d}{\partial s}\left(\exp{\left(-(1+\frac{1}{2}s)^{-2}\right)}\right)$?