1
$\begingroup$

SLLN tells us that if $X_1,...,X_n$ are iid, with $X_1$ having finite mean $\mu$, then their sample average converges almost surely to $\mu$.

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

Note this is a follow up to this question.

Improve this question
$\endgroup$
2
  • $\begingroup$ @Zhanxiong Thanks for your comment, but these questions are not the same. This one has a much stronger premise (a.s. convergence to a constant), and I did not say $E[X_1]$ is defined and finite (indeed, the question is whether we can conclude this)---I have edited it now to further emphasize this. $\endgroup$
    – Golden_Ratio
    Commented Mar 20, 2023 at 1:55
  • $\begingroup$ Is the "constant" a finite real number? $\endgroup$
    – Zhanxiong
    Commented Mar 20, 2023 at 2:13

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.