WLLN tells us that if $X_1,...,X_n$ are iid, with $X_1$ having finite mean $\mu$, then their sample average converges in probability to $\mu$.

Suppose instead we know that $X_1,...,X_n$ are iid and their sample average converges in probability to a constant. Can we argue that $X_1$ has a finite mean and hence that such a constant must be the mean of $X_1$?


1 Answer 1


For counter-examples, you might want to consider cases where the weak law of large numbers applies but the strong law of large numbers does not. These cases must have $E[X_1]$ undefined.

For example, adapting the second example in Wikipedia,

  • suppose $\mathbb P\left(X_1=\frac{(-2)^n}{n} \right)=\frac1{2^n}$ for positive integer $n$
  • this does not have a mean since $\sum\limits_{n=1}^\infty \frac{(-2)^n}{n} \frac1{2^n}=\sum\limits_{n=1}^\infty \frac{(-1)^n}{n}$ does not converge absolutely
  • but the sum does converge conditionally to $-\log_e(2)\approx -0.693$ and the sample averages converge in probability to this

In this example, you need a large sample to see much convergence. For example with $10^4$ simulations each with sample sizes of $10^2$ (red), $10^3$ (green) and $10^4$ (blue), the following R code

Xbar <- function(cases){
  Y <- rgeom(cases, 1/2) + 1 # R's geometric distribution starts at 0
  X <- (-2)^Y / Y
sims4 <- replicate(10^4, Xbar(10^4))
plot(density(sims4, from=-2, to=1), col="blue")
sims3 <- replicate(10^4, Xbar(10^3))
lines(density(sims3, from=-2, to=1), col="green")
sims2 <- replicate(10^4, Xbar(10^2))
lines(density(sims2, from=-2, to=1), col="red")
abline(v = -log(2))

shows the increasing concentration of the sample mean as the sample size increases

simulated density

  • 1
    $\begingroup$ Thanks Henry, very nice answer. This gets me thinking of the restrictive definition of the mean in the Lebesgue sense, with $E[X]=E[X^{+}]-E[X^{-}].$ If, for instance, the mean was given by the conditionally convergent sum in your example, we would say the mean exists and the sample average would converge to it. What is a more general definition of the mean that would satisfy the property that a sample average (from an iid sample) that is known to converge in probability to a constant would always converge in probability to such a mean? $\endgroup$ Commented Mar 19, 2023 at 23:19
  • 2
    $\begingroup$ @Golden_Ratio - I do not know though others might. You might extend your question to whether you can have the traditional definition of mean that gives the SLLN with almost sure convergence and a weaker definition of mean which gives the WLLN with convergence in probability. $\endgroup$
    – Henry
    Commented Mar 20, 2023 at 0:00
  • $\begingroup$ For a different example illustrating a difference between WLLN and SLLN, see stats.stackexchange.com/a/205564/2958 $\endgroup$
    – Henry
    Commented Jul 30, 2023 at 8:57

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