The original definition for Dominated Convergence Theorem is that $X_n\xrightarrow { a.s.}X$ and $|X_n|\le Y \ \ \ \ \forall n\implies E(X_n)\to E(X)$.

However if we replace almost sure conevergence with convergence in probability, then also the theorem holds.

So I want to know what is the most general version of DCT? In particular, in what other types of convergences does the theorem work? What are the other conditions that we can relax and still have DCT holding?

  • 1
    $\begingroup$ You have $L^1$ convergence if and only if you have uniform integrability and convergence in probability. Since this is an equivalency, it doesn't get more general than that. But of course you may want to find different characterizations of convergence in probability and uniform integrability. $\endgroup$ – Calculon Oct 1 '16 at 13:21
  • 1
    $\begingroup$ DCT or uniform integrability + convergence in probability give indeed a stronger conclusion, i.e. $\mathbb{E}[|X_n-X|]\to 0$. However, if one assumes DCT or uniform integrability + convergence in law only, then still $\mathbb{E}[X_n]\to \mathbb E[X]$, but, of course the convergence $\mathbb{E}[|X_n-X|]\to 0$ cannot be true in general. $\endgroup$ – zhoraster Oct 1 '16 at 13:46
  • 1
    $\begingroup$ @ClementC. The statement I gave is different than the one you gave a link to. I have convergence in probability in lieu of convergence in distribution and $L^1$ convergence in lieu of convergence of expectations. The question seems to be about convergence of expectations but as you said the "classical" DCT is about $L^1$ convergence, which is stronger. So there is some confusion here I think. $\endgroup$ – Calculon Oct 1 '16 at 13:54
  • 1
    $\begingroup$ @ClementC. Yes. $\endgroup$ – Calculon Oct 1 '16 at 13:58
  • 2
    $\begingroup$ @Qwerty If you are talking about convergence of expectations, then there is another equivalency. Namely, the expectations converge if and only if there is convergence in distribution and asymptotic uniform integrability. $\endgroup$ – Calculon Oct 1 '16 at 14:06

Your Answer

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

Browse other questions tagged or ask your own question.