When does the average of $n$ independent random variables concentrate around the average of their means as $n$ tends towards infinity?

I'm having trouble understanding Theorem 11 in these class notes from Wade Trappe's class. It gives the following sufficient condition:

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However, I think this might be incorrect. Consider $X_2,X_3,X_4,\ldots=0$. This scenario does fulfill the theorem's condition and it also implies that the probability of $X_1$ being equal to its expected value is 1, which is applicable to any random variable $X_1$ that has a finite variance. This does not seem right.

So, what is the accurate condition for this convergence to occur?

This related post provides a formula for the variance of a linear transformation of a 'd'-dimensional Dirichlet random variable. This might suggest that the sum of the squares of the differences between the variances of all pairs of these random variables (i.e., $\sum_{i<j}^n (\operatorname{var}(X_i)-\operatorname{var}(X_j))^2$) needs to grow at a rate slower than $O(n)$ for this convergence to happen.


1 Answer 1


In your example, $M_n=\frac {X_1} n$ so $\lim M_n=0$. Also, $\lim \frac 1n \sum\limits_{i=1}^{n}EX_i=\lim \frac {EX_1} n=0$. So there is no contradiction.


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