Consider the classical triangular sequence $\{(X_{n,i})_{i=1}^n\}_{n \in \mathbb{N}}$. Assume $X_{n,1},\ldots,X_{n,n}$ are i.i.d random variables with mean $\mu_n$. Then under what conditio(s) can we guarantee the strong law of large number: $$ {1 \over n}\sum_{i=1}^n X_{n,i}-\mu_n \to_{a.s.} 0?$$


1 Answer 1


I'm not confident in the case where the distributions of the $X_{n,i}$ do not depend on $n$.

Assume that the variances $\rm{Var}(X_{n,i}) = \sigma_n^2$ tend towards zero sufficiently fast that $$\sum_n \frac{\sigma_n^2}{n} < \infty.$$ In particular, it suffices to make the $\sigma_n$ go like any negative power of $n$. Then $$\rm{Var}\left(\sum_{I=1}^n \frac{X_{n,i} - \mu_n}{n}\right) = \frac{\sigma_n^2}{n}.$$ Thus, by Chebyshev's inequality, $$\mathbf{P}\left(\left|\sum_{I=1}^n \frac{X_{n,i} - \mu_n}{n}\right| > c\right) \leq \frac{(\sigma_n^2/n)}{c^2}.$$ By our earlier assumption, these probabilities are summable in $n$, so by Borel-Cantelli, the sums converge a.s. to 0.

  • $\begingroup$ +1. By the way, if the 4th moments are uniformly bounded, so $E[(X_{n,i} - \mu_n)^4] \leq b$ for all $i,n$, then we also get almost sure convergence. This is nice because then the variances do not need to go to zero. So the distributions $X_{n,i}$ could even be the same for all $n,i$. $\endgroup$
    – Michael
    May 29, 2020 at 0:29

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