I'm wondering about a SLLN where the number of summands is also allowed to be a random variable. More specifically assume that:

  • $\{X_i\}$ are iid with $\mathbb{E}[|X_1|] < \infty $
  • $N=\{N_n\}_{n\in\mathbb{N}}$ is a sequence of random positive integers independent of $\{X_i\}$
  • $\lim_{n} N_n =\infty$ $a.s$

Can we conclude that $ \frac{1}{N_n} \sum_{i=1}^{N_n} X_i = \mathbb{E}[X_1] $ almost surely?


1 Answer 1


This holds because in fact it is an easy consequence of the strong law of large numbers.

Let $\Omega$ be the underlying probability space. Define $Y_n = \frac{1}{n}\sum_{i = 1}^n X_i\,.$ By the strong law of large numbers, if we let $\Omega_1 = \{ Y_n \to E[X_1]\}$ then $P(\Omega_1) =1$. Similarly, define $\Omega_2 = \{N_n \to \infty\}$; then $P(\Omega_2) = 1$. Then $P(\Omega_1\cap \Omega_2) = 1$ and on the set $\Omega_1 \cap \Omega_2$ we have $Y_{N_n} \to E[X_1]$. Note that in fact you don't even need $N_n$ to be independent of $\{X_i\}$ just that $N_n \to \infty$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .