Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here are two exercises from my Probability book ("Probabilidade: um curso em nível intermediário", by Barry James). (I have translated them from Portuguese to English and modified them a bit.)

Prove the following claims:

$1$) If $\phi$ is a characteristic function and there exists $\delta > 0$ such that $\phi (t) = 1$ for all $t$ with $|t| < \delta$, then $\phi (t) = 1 \forall t$.

$2$) Let $X_{1}, X_{2}, \ldots$ be random variables whose characteristic functions $\phi _{1}, \phi_{2}, \ldots$ converge pointwise. If there exists $\delta > 0$ such that $\phi_{n} (t) \rightarrow 1$ for all $t$ with $|t| < \delta$, then $X_{n}$ converges in probability to $0 $. Hint: use Claim $1$.

I've proven Claim $1$, and I can prove Claim $2$ if I assume that the limit of the characteristic functions $\phi _{1}, \phi_{2}, \ldots$ is also a characteristic function. My question is: is Claim $2$ still true without this additional assumption?

share|cite|improve this question
up vote 2 down vote accepted

Yes, thanks to Lévy's continuity theorem, the pointwise limit is a characteristic function as soon as it is continuous at $0$.

share|cite|improve this answer
Thanks! We've only seen in class a "weaker" form of this theorem, namely: "$X_n$ converges in distribution to $X$ if and only if $\phi_{X_{n}} (t)$ converges to $\phi_{X} (t)$ for all $t$." I guess my teacher forgot about this when he assigned us this exercise. – TuringMachine Feb 25 '13 at 22:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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