2
$\begingroup$

Is there any sequence of probability distributions such that their characteristic functions converge pointwise, but the sequence of prob. distributions itself does not converge weakly? :S

$\endgroup$
4
$\begingroup$

Yes, for example when $\mu_n$ is normally distributed with mean $0$ and standard deviation $n^2$: the sequence of characteristic functions converges pointwise to the map $\varphi$ such that $\varphi(t)=0$ if $t\neq 0$ and $\varphi(0)=1$. But such a sequence of measure cannot converge weakly (it's not tight).

However, if the sequence of characteristic functions converges pointwise to a function which is continuous at $1$, then one of the numerous theorems of Lévy gives weak convergence.

$\endgroup$
  • $\begingroup$ So, the explicit sequence of probability distributions is...? $\endgroup$ – JohnD May 4 '13 at 23:29
  • $\begingroup$ Sorry, I meant $\mu_n$ instead of $\mu$. Is it clearer? $\endgroup$ – Davide Giraudo May 8 '13 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.