# Condition of Existence of Asymptotic Distribution

Let $X_1... X_n$ be random samples from $Beta(\alpha, 1), \alpha\gt0$ and

Let $Y_n = \min X_{i\in\{1,2,..n\}}$

Then evaluate the condition of $r$ for the asymptotic distribution $n^rY_n$ to exist and specify the asymptotic distribution upon $r$.

\lim\limits_{n\rightarrow \infty}P(n^rY_n \le y) = \begin{align}1 \;\;\;\text{when}\;\ 0\lt r\alpha\lt 1\\1-\exp(-y^\alpha) \;\;\;\text{when}\;r\alpha=1\\0\;\;\;\text{when}\;\;r\alpha\gt1\end{align}

Then now, How I need to specify another distribution $Z$ s.t. above given limit probability equal to $P(Z\le z)$ where z denotes every continuous point of cdf of $z$ and $Z = n^rY_n$

1) My first question is : How to specify the range of $Z$ about given $r$?

If $\;0\lt r\lt1/\alpha$, since $0\lt y \lt 1$, range of Z depends on $n^r$, however, I am confused with when it comes to deal with the case of form $\infty^r$.

2) I presumably guess for the second case of $r\alpha$ = 1, I can easily derive Z by just differentiating given $1-exp(-y^\alpha)$, however, what about $1$ and $0$ case? 1 corresponds to $0$ when differentiated so is it enough for me to say there exist no asymptotic distribution in case of $0<r\alpha<1$ and same conclusion with same reasoning to $r\alpha>1$ case?

You are almost done, just reflect what the question is about. It should be clear, that $Y_n$ converges to $0$ a.s.. In order to "see" something more, e.g. the asymptotics (how converges $Y_n$ to zero) you have to find a suitable normalization (Think of the standard central limit theorem: For integrable RVs, you know that $1/n\sum_{i\le n} X_i$ converges to the expectated value a.s.. The central limit theorem tells you, "how"). They already hinted you, that the right normalization should be of the form $n^r$ for some $r$. In fact you have done all the work. If you choose $r<\alpha^{-1}$, then your normalization is not "strong enough" and $n^rY_n\to0$, hence $Z=\delta_0$. (degenerated). For $r>\alpha^{-1}$ your normalization is "too strong", and $n^rY_n \to \infty$ (you could write $Z={\delta}_{\infty}$). The right normalization is $r=\alpha^{-1}$, then $n^rY_n$ converges in distribution to a nondegenerated $Z$, with distribution $1-\exp(-y^{\alpha})$ on $[0,1]$. in the other cases there are existing only degenerated distributions.