Law of large numbers on a random number of samples

Question

Let $X_1,\dots,X_n$ be $n$ iid random variables, each with expectation $1$. From the law of large numbers one has $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n X_i = 1 \text{ a.s.}$$

Now let $Y_n$ be a random variable such that $\lim_{n\to\infty}Y_n = y > 0 \text{ a.s.}$

Let $a_n\in\mathbb{R}$ be an increasing sequence such that $\lim_{n\to\infty}a_n=\infty$ and let $K_n = Y_n a_n$

How may I prove that $$\lim_{n\to\infty}\frac{1}{K_n}\sum_{i=1}^{K_n} X_i = 1 \text{ a.s.}$$

Answer 1

We claim that $A:=(\frac{1}{n}\sum_{i=1}^n X_i\to 1)\cap (K_n\to\infty)\subseteq (\frac{1}{K_n} \sum_{i=1}^{K_n} X_i\to 1).$ As the first set is a finite intersection of almost sure sets, it is almost sure, and thus, this would prove your statement.

Now, let $\omega\in A$ and let $\varepsilon>0$. Then, there exists $N(\omega)$ such that $|\frac{1}{n}\sum_{i=1}^n X_i(\omega)-1|\leq \varepsilon$ for all $n\geq N(\omega)$. Now, by assumption there exists some $M(\omega)$ such that $K_n(\omega)\geq N(\omega)$ for all $n\geq M(\omega)$. Adding these two statements, we get that for all $n\geq M(\omega)$,

$$ \left|\frac{1}{K_n(\omega)}\sum_{i=1}^{K_n(\omega)} X_i(\omega) -1\right|\leq \varepsilon, $$ which was what we wanted.

Answer 2

For simplicity, assume that $X_i$ have expectation $0$. Then for all $N$, $$\tag{*} \left\lvert \frac 1{K_n}\sum_{i=1}^{K_n}X_i\right\rvert \leqslant \sup_{\ell\geqslant N}\frac 1{ \ell}\left\lvert\sum_{i=1}^{\ell}X_i\right\rvert +\mathbf 1\left\{K_n\leqslant N\right\}\left\lvert \frac 1{K_n}\sum_{i=1}^{K_n}X_i\right\rvert. $$ Take the $\limsup_{n\to +\infty}$ to see that the inequality $$ \limsup_{n\to +\infty}\left\lvert \frac 1{K_n}\sum_{i=1}^{K_n}X_i\right\rvert \leqslant \sup_{\ell\geqslant N}\frac 1{ \ell}\left\lvert\sum_{i=1}^{\ell}X_i\right\rvert $$ holds almost surely (the term in the right hand side of (*) is zero for almost all $\omega$ and for $n\geqslant N(\omega)$)). Then conclude by the classical law of large numbers.