# Subsequence density for iid sequence

Given iid sequence of $$X_{n}\in N(0,\sigma_{n})$$, from the second Borel-Cantelli we find subsequence $$\{n_{k}\}_{k\geq 1}$$ $$\sum P[X_{n}\leq c_{n}]=\infty\Rightarrow \{X_{n_{k}}\leq c_{n_{k}}\}~\forall k\geq 1~a.s.$$

for any sequence $$c_{n}$$ that doesn't converge to negative infinity. Can we optimize the density of the subsequence $$n_{k}$$ in terms of the pair $$(\sigma_{n},c_{n})$$? In particular can we figure out an optimal gauge $$b_{n}$$ s.t. for some finite $$\delta,\epsilon>0$$ we have

$$P[\delta>\frac{\{k\leq n: X_{k}\leq c_{k}\}}{b_{n}}>\epsilon]\to 1~as~n\to \infty?$$

I suppose the right object to study here is the large deviations of the counting measure $$\mu_{n}(A):=\sum_{k=1}^{n} 1_{X_{k}\in A_{k}}$$. Or maybe the Poisson-binomial rv $$S_{n}:=\sum_{k=1}^{n} 1_{X_{k}\leq c_{k}}$$ with transition jumps $$p_{k}=\Phi(c_{k})$$.

Welcome to MO! However, your post needs to be edited with care. First, the $$X_n$$'s are iid only when $$\sigma_n$$ does not depend on $$n$$. Second, you should not write $$X_n\in N(0,\sigma_n)$$, because $$s\in S$$ can only mean that $$s$$ is an element of a set $$S$$. Third, the standard notation for the normal distribution is $$N(\mu,\sigma^2)$$, rather than $$N(\mu,\sigma)$$. Fourth, the expression $$P[\delta>\frac{\{k\le n: X_k\le c_k\}}{b_n}>\epsilon]$$ does not make sense, since $$\{k\le n: X_k\le c_k\}$$ is a random set, rather than a real-valued random variable (r.v.). So, apparently your question is about a condition for
$$P_n:=P\Big(\delta>\frac{S_n}{b_n}>\epsilon\Big)\to 1$$ as $$n\to\infty$$, where $$S_n=\sum_{k=1}^n Y_k=\sum_{k=1}^n 1_{X_k\le c_k}=\#\{k\le n: X_k\le c_k\}$$ (as in your post), $$Y_k:=1_{X_k\le c_k}$$, and the $$X_k$$'s are independent normal zero-mean r.v.'s with variances $$\sigma_k^2$$. Let $$p_{k}:=EY_k=P(X_k\le c_k)=\Phi(c_k/\sigma_k)$$, where $$\Phi$$ is the standard normal cdf, so that $$ES_n=\mu_n:=\sum_1^n p_k$$ and $$\sqrt{Var\,S_n}=B_n:=\sqrt{\sum_1^n (1-p_k)p_k}$$. Then, letting $$b_n:=s\mu_n,\quad s:=\frac2{\epsilon+\delta}, \tag{1}$$ assuming $$t:=\frac{\delta-\epsilon}2>0,$$ and using Chebyshev's inequality, we have $$P_n=P\Big(\delta-\frac{\mu_n}{b_n}>\frac{S_n-\mu_n}{b_n}>\epsilon-\frac{\mu_n}{b_n}\Big) =P\Big(\Big|\frac{S_n-\mu_n}{s\mu_n}\Big| So, we will have $$P_n\to1$$ if $$b_n$$ is as in (1) and
$$\frac{B_n^2}{\mu_n^2}=\frac{\sum_1^n (1-p_k)p_k}{\big(\sum_1^n p_k\big)^2}\to0,$$ which will be the case if e.g. the $$p_k$$'s are bounded away from $$0$$ or, more generally, if $$\sum_1^n p_k\to\infty$$ as $$n\to\infty$$ (because $$\sum_1^n (1-p_k)p_k\le\sum_1^n p_k$$).