# Understanding proof of convergence

I'm going through section 6.1 in this paper for the proof of Theorem 2.1. However, I can't seem to get the result, which I will explain below.

Setup

Let $T_i$, $1\leq i\leq m$ be independent Student's $t$ test statistics which are constructed as $$T_i=\frac{\bar{X}_i}{\hat{s}_{ni}/\sqrt{n}}$$ where $$\bar{X}_i=\frac{1}{n}\sum_{k=1}^n X_{ki},\hspace{5mm}\hat{s}^2_{ni}=\frac{1}{n-1}\sum_{k=1}^n (X_{ki}-\bar{X}_i)^2$$ and $X_{ki}\stackrel{iid}{\sim} \mathcal{N}(\mu_i,\sigma_i^2)$, $1\leq k\leq n$, $1\leq i\leq m$.

Problem Assumptions

1. Suppose $\log m=o(n^{1/2})$. Assume that $\max_{1\leq i\leq m}EY_i^4\leq b_0$ for some constant $b_0>0$ and $$\text{Card}\left\{i: |\mu_i/\sigma_i|\geq 4\sqrt{\log m/n} \right\}\to\infty$$

1. Further suppose that for $0\leq t\leq o(n^{1/4})$, $$P(|T_i-\sqrt{n}\mu_i/\hat{s}_{ni}|\geq t)=\frac{1}{2}G(t)\left[\exp\left( -\frac{t^3}{3\sqrt{n}}\kappa_i\right)+\exp\left(\frac{t^3}{3\sqrt{n}}\kappa_i \right) \right](1+o(1))$$ where $o(1)$ is uniform in $1\leq i\leq m$, $G(t)=2-2\Phi(t)$, $\Phi(t)$ is the normal cdf, and $\kappa_i=EY_i^3$.

1. Finally, let $\mathcal{M}\subset\{1,2,\dots,m\}$ satisfying $\mathcal{M}\subset\{i:|\mu_i/\sigma_i|\geq 4\sqrt{\log m/n}\}$ and $\text{Card}(\mathcal{M})\leq\sqrt{n}$. Also for any $\epsilon>0$, $$P(\max_{i\in\mathcal{M}}|\hat{s}^2_{ni}/\sigma_i^2-1|\geq \epsilon)=O(1/\sqrt{n})$$

Question I want to show equation (15) in section 6.1 which says that

for some $c>\sqrt{2}$ and some $b_m\to \infty$ (the subscript means that the constant depends on $m$), $$P\left(\sum_{i=1}^m I\{|T_i|\geq c\sqrt{\log m} \}\geq b_{m} \right)\to 1$$

I believe the idea is to split up the quantity for $i\in\mathcal{M}$ and $i\not\in\mathcal{M}$ and use the fact that $$P\left(\sum_{i\in\mathcal{M}} I\{|T_i|\geq c\sqrt{\log m} \}\geq b_{m} \right)\leq P\left(\sum_{i=1}^m I\{|T_i|\geq c\sqrt{\log m} \}\geq b_{m} \right)$$

Then if I show that the LHS goes to 1, I have the claim.

However, I can't seem to get the inequalities needed to get the convergence to 1. Any hints or insights?

What I have so far

\begin{eqnarray} P\left(\sum_{i\in\mathcal{M}} I\{|T_i|> c\sqrt{\log m} \}\geq b_m\right) &=& 1- P\left( \sum_{i\in\mathcal{M}} I\{|T_i|> c\sqrt{\log m} \}< b_m \right)\\ &\geq& 1-P\left( \sup_{i\in\mathcal{M}} I\{|T_i|> c\sqrt{\log m} \}< b_m/\text{Card}(\mathcal{M}) \right)\\ &\geq& 1- \sum_{i\in\mathcal{M}} P\left(I\{|T_i|> c\sqrt{\log m} \}< b_m/\text{Card}(\mathcal{M}) \right)\\ &\geq& 1- \sum_{i\in\mathcal{M}} [1-P\left(I\{|T_i|> c\sqrt{\log m} \}\geq b_m/\sqrt{n} \right)]\\ \end{eqnarray}

Now it suffices to show that $P\left(I\{|T_i|> c\sqrt{\log m} \}\geq b_m/\sqrt{n}\right)\to 1$, which I'm having trouble showing.

First we look at the magnitude of some basic quantities.

$$\log m = O(n^{1/2})\Rightarrow \sqrt{\log m} =O(n^{1/4}).$$

By assumption 1 and 3, we have

$$\sqrt{n}\geq {\rm Card}\{i:|\frac{\sqrt{n}\mu_i}{\sigma_i}|\geq 4\sqrt{\log m}\}\rightarrow \infty.$$

Now we get the intuition. The law of large numbers and Slusky's theorem tell us that $T_i/\sqrt{n}\overset{p}{\rightarrow} {\mu_i}/{\sigma_i},$ which roughly means $T_i\approx \frac{\sqrt{n}\mu_i}{\sigma_i}$ for large $n$. Therefore roughly we may expect that ${\rm Card}\{i:|T_i|\geq 4\sqrt{\log m}\}\rightarrow \infty$ with probability tending to $1$. This is exactly the result we want for $c=4$ and some $b_m$.

To prove this intuition, we will use a commonly used technique in statistics: to prove $P(A_n)\rightarrow 1$, we try to find some nice $B_n$ such that $P(B_n)\rightarrow 0$ and $P(A_n| B_n^c)\rightarrow 1$, where $P(A_n\cap B_n^c)$ will be easier to compute than $P(A_n)$ itself. And use $P(A_n)\geq P(A_n| B_n^c)P(B_n^c)\rightarrow 1$.

In our problem, we shall use two such nice sets. one of them is

$$B_{ni} = \{\max_{i\in \mathcal{M}}|\frac{\hat{s}^2_{ni}}{\sigma_i^2}-1|\geq \epsilon\},\ {\rm with}\ P(B_n) = O(1/\sqrt{n})$$

and the other is

$$C_{ni} = \{|T_i-\sqrt{n}\mu_i/\hat{s}_{ni}|\geq \delta \sqrt{\log m}\},\ {\rm with}\ P(C_n) = O(1/\log m),$$

where $\delta>0$ is some constant. To get $P(C_n) = O(1/\log m)$, we can let $t = \delta\sqrt{\log m} = o(1/n^{1/4})$ in assumption $2$ and use the trivial fact that

$$1-\Phi(\delta\sqrt{\log m})\leq c\exp(-\delta^2\log m)\leq c/(\delta^2\log m).$$

Now on event $B_{ni}^c$, $\{|\frac{\sqrt{n}\mu_i}{\sigma_i}|\geq 4\sqrt{\log m}\}\Rightarrow \{|\frac{\sqrt{n}\mu_i}{\hat{s}_{ni}}|\geq a\sqrt{\log m}\}$, for some constant $a>3$ (we can pick $\epsilon$ small enough to make $a>3$).

On event $C_{ni}^c$, $\{|T_i-\sqrt{n}\mu_i/\hat{s}_{ni}|<\delta \sqrt{\log m}\}\Rightarrow |T_i|>|\sqrt{n}\mu_i/\hat{s}_{ni}|-\delta \sqrt{\log m}$.

Therefore on $B_{ni}^c\cap C_{ni}^c$, we have $|T_i|>(a-\delta) \sqrt{\log m}$.

Let $c = (a-\delta)>\sqrt{2}$ (again we can pick $\delta$ small enough), $b_m = [{\log m}]\rightarrow \infty$ (the smallest integer that no less than ${\log m}$), $A_m = \left\lbrace\sum_{i=1}^m I\{|T_i|\geq c\sqrt{\log m} \}\geq b_{m} \right\rbrace$. Let $B_m = \bigcup_{i=1}^{b_m}(B_{ni}\cap C_{ni})$, then

$$P(B_m)\leq \sum_{i=1}^{b_m} (P(B_{ni}) +P(C_{ni})) = \log m\cdot (O(1/\sqrt{n}) + O(1/\log m))\rightarrow 0,$$

and over set $B_m^c$, $|T_i|\geq c\sqrt{\log m}$ for $i=1,\cdots, b_m$, which implies

$$P\left(A_m|B_m^c \right)= 1.$$

Therefore we finally get $P(A_m)\geq P(A_m|B_m^c)P(B_m^c)\rightarrow 1$.