Converging in distribution for $Z_n = n^pX_{(1)}$

Question

Hello.

The above problem is what I am working on and there is something that I would like to clarify, or at least ask for opinions regarding this.

From transformation I obtained

$$f_Z(z) = n \left( \frac{5z^4}{n^{5p}} \right) \left[ 1+\frac{-z^5}{n^{5p}} \right]^{n-1}$$

which is not too difficult to say that if $p=\frac{1}{5}$ as $n \rightarrow \infty$

$$f_Z(z) \rightarrow 5z^4e^{-z^5}$$

and since $0<x<1$ becomes $0<z<\infty$ as $n \rightarrow \infty$ I also see that indeed it is a probability distribution.

What I wonder here, though, is that is this the only $p$ that satisfies that?

From the problem it is suggestive that if $p \ne \frac{1}{5}$ it would not converge in probability anymore.

Any thoughts?

Answer 1

Your intuition is (almost) spot-on.

First a recap. Assume you have two sequences: $Y_n\overset{d}{\to}Y$ and $T_n\overset{d}{\to}0$, then we get $(Y_n,T_n)\overset{d}{\to}(Y,0)$. By the continuous mapping theorem then implies $Y_nT_n\overset{d}{\to}Y\cdot 0=0$. For our purposes, the sequence $\{T_n\}_{n\in\mathbb{N}}$ is going to be deterministic.

Your problem. Let $Z$ be the limit you obtain when $p=\frac{1}{5}$.

1. (Case $p<\frac{1}{5}$.) Set $Y_n=n^{\frac{1}{5}}X_1\overset{d}{\to}Z$ and for $p<\frac{1}{5}$, set $T_n=n^{p-\frac{1}{5}}\to0$. Then: $n^pX_1=Y_nT_n\overset{d}{\to}0$. (That is, you do get convergence in distribution, but the limit is degenerate.)
2. (Case $p>\frac{1}{5}$.) Assume $Y_n=n^pX_1\overset{d}{\to}Z'$ for some $p<\frac{1}{5}$ and some finite random variable $Z'$. Set $T_n=n^{\frac{1}{5}-p}\to0$. Then: $n^{\frac{1}{5}}X_1=Y_nT_n\overset{d}{\to}0$, but we also have $n^{\frac{1}{5}}X_1\overset{d}{\to}Z\neq 0$. This is a contradiction, so you don't have convergence in distribution. (In any case, you could argue that $n^pX_1$ diverges to "$+\infty$".)