Let $\mu_n$ be $\text{Uniform}(\alpha_n,\beta_n)$ distribution. What condition on both sequences makes $\left\{\mu_n\right\}_n$ relatively compact?

Question

Let $\mu_{n}$ be the $\text{Uniform}(\alpha_{n},\beta_{n})$ probability distribution for $\alpha_{n}< \beta_{n}$ , $n = 1,2,3,\ldots$

What condition on these two sequences makes $\left\{\mu_{n}\right\}_{n}$ relatively compact?

I know that using Prokhorov's Theorem we can show conditions for tightness.

(If $\Omega$ is a complete separable metric space then a family of distributions is relatively compact if and only if it is tight.)

Using this theorem I will want to show,

$$\mu_{n}\{[-m,m]\} > 1-\varepsilon$$

for any $\varepsilon>0$ and $m>0$.

But I don't really know how to go about doing this. Any help would be greatly appreciated!

Answer 1

Suppose that $\left\{\mu_n,n\geqslant 1\right\}$ is relatively compact. Then there exists $M$ such that for any $n$, $a_n\lt b_n\leqslant M$. Otherwise, there would exists an increasing sequence of integers $\left(n_k\right)_{k\geqslant 1}$ such that $b_{n_k}\gt a_{n_k}\gt k$ for any $k\geqslant 1$. Then $\mu_{n_k}\left(\left[-R,R\right]\right)=0$ if $k\gt R$ and the family $\left\{\mu_{n_k},k\geqslant 1\right\}$ is not tight and cannot be relatively compact. Similarly, we can show that there exists a constant $m$ such that for any $n \geqslant 1$, $m\leqslant a_n\lt b_n$.

Conversely, if there exists $m$ and $M$ such that for any $n\geqslant 1$, $m\leqslant a_n\lt b_n\leqslant M$, then extract convergent subsequences from $\left(a_n\right)_{n\geqslant 1}$ and $\left(b_n\right)_{n\geqslant 1}$. This subsequence converges in distribution. We can also argue directed by invoking tightness.

Answer 2

A sufficient condition for the tightness is that both $(\alpha_n)_{n \in \mathbb{N}}$ and $(\beta_n)_{n \in \mathbb{N}}$ are bounded sequences. Then we can choose $M \in \mathbb{N}$ such that

$$|\alpha_n| + |\beta_n| \leq N$$

for all $n \in \mathbb{N}$. Since $\mu_n$ is supported in $[\alpha_n,\beta_n]$ we find

$$\mu_n([-M,M]) \geq \mu([\alpha_n,\beta_n]) = 1$$

for all $n \in \mathbb{N}$; this shows that $(\mu_n)_{n \in \mathbb{N}}$ is tight.

Tightness fails, for instance, if $\alpha_n$ is unbounded and $\beta_n$ is bounded. Indeed: Without loss of generality, we may assume that $\alpha_n \to - \infty$ (otherwise take a subsequence). For large $M>0$ we can choose $N \in \mathbb{N}$ such that $\alpha_n \leq -M$ and $b_n \in [-M,M]$ for all $n \geq N$. Thus,

$$\mu_n([-M,M]^c) \geq \mu_n([\alpha_n,-M)) = \frac{-M-\alpha_n}{\beta_n-\alpha_n} \xrightarrow[]{n \to \infty} 1,$$

i.e.

$$\mu_n([-M,M]) \xrightarrow[]{n \to \infty} 0.$$

This means that $(\mu_n)_{n \in \mathbb{N}}$ is not tight.

A similar thing happens if $\alpha_n \to -\infty$ and $\beta_n \to \infty$. This is not surprising since tightness means, essentially, that we do not move mass to $\pm \infty$ as $n \to \infty$.