In the above book (first edition), the author proves the following theorem:

If $(X_n)_{n=1}^{\infty}$ and $(Y_n)_{n=1}^{\infty}$ are increasing sequences of positive integrable random variables on the probability space $(\Omega, \mathcal{F},P)$ and if $$\lim_{n \to \infty} X_n = \lim_{n \to \infty} Y_n$$ almost surely, then $$\lim_{n\to\infty}\mathbb{E}\lbrack X_n\rbrack = \lim_{n\to\infty}\mathbb{E}\lbrack Y_n\rbrack.$$

In course of the proof, the author needs to prove, among other things, that $\lim_{n\to\infty} X_n(\omega) = \lim_{n\to\infty} X_{n}^{\lbrack j_n \rbrack}(\omega)$ for every $\omega \in \Omega,$ where $X^{\lbrack m \rbrack}$ denotes the $m-$th truncation of $X$ and $j_n$ is a certain previously defined strictly increasing ($j_{n+1} > j_n$) sequence of natural numbers. Here, he distinguishes two cases: 1) in case $\lim_{n\to\infty} X_{n}(\omega) < \infty,$ we have $X_{n}(\omega) < j_n$ for every $n$ bigger than certain $n_0$ so that we plainly have (due to the definition of $X^{m}(\omega)$) that $X_{n}^{\lbrack j_n \rbrack}(\omega) = X_n(\omega)$ for all $n \geq n_0.$ 2) In case $\lim_{n\to\infty} X_{n}(\omega) = \infty,$ he says:

for all $n,$ we have either $X_{n}(\omega) - X_{n}^{\lbrack j_n \rbrack}(\omega) \leq 1/2^{j_n}$ or $X_{n}^{\lbrack j_n \rbrack} = j_n.$ Since $j_n \geq n,$ we have, in either case, $\lim_{n \to \infty} X_n^{\lbrack j_n \rbrack} = \infty.$

This would clearly establish the claim that $\lim_{n\to\infty} X_n(\omega) = \lim_{n\to\infty} X_{n}^{\lbrack j_n \rbrack}(\omega)$ but I have a couple of questions here:

  • First, from where does it follow that $$X_{n}(\omega) - X_{n}^{\lbrack j_n \rbrack}(\omega) \leq 1/2^{j_n}$$ and for what $n'$s is it true? My guess is that it has something to do with pointwise convergence of the truncations $X^{\lbrack m \rbrack}$ to $X$ for $m \to \infty$ (this seems obvious).
  • Second, why is it not enough to say: if $\lim_{n \to \infty} X_n(\omega) = \infty,$ then $X_{n}^{\lbrack j_n \rbrack} = j_n$ for $n$ big enough and since $j_n \geq n,$ we have $\lim_{n\to \infty}X_{n}^{\lbrack j_n \rbrack} = \infty$ and we ar done (does it have sth in common with the fact that $j_n$ depends on $n$?)
  • $\begingroup$ @Math1000: I am not saying the theorem is not rigorous, I just do not understand one tiny detail of its proof. This detail is somehow captured in my questions above. And yes, it uses truncations, if only implicitly: they are defined on p. 123 in the first edition. $\endgroup$ – Jorge.Squared Jul 25 '18 at 11:05
  • $\begingroup$ @Math1000: maybe you are referring to a similar theorem dealing with positive bounded random variables only, whereas the theorem I am referring to deals with positive integrable RVs? $\endgroup$ – Jorge.Squared Jul 25 '18 at 11:26
  • $\begingroup$ Ah, indeed I had the wrong theorem; the one I was looking at states "increasing sequences of simple positive random variables." I will delete my previous comments. $\endgroup$ – Math1000 Jul 25 '18 at 11:37
  • $\begingroup$ In the proof of the (correct) theorem, the author still is not using the aforementioned truncations, but instead the "canonical sequence" of simple random variables $\{X_n\}$ defined explicitly such that $|X_n(\omega)-X(\omega)|\leqslant 2^{-n}$ for all $n$ and $\omega$. I suggest you study the construction of the $X_n$ carefully. $\endgroup$ – Math1000 Jul 25 '18 at 11:44
  • $\begingroup$ @Math1000: Thank you, your response suggests that the second edition has a different proof. This raises the question whether the present "proof" is correct or not. I still wonder. $\endgroup$ – Jorge.Squared Jul 25 '18 at 11:52

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.