betanalpha
commented
4 years ago First, thank you for providing these studies. I have been looking for probability theory intro material to provide to my data science students & advisees, and I think your studies will be quite useful for their intermediate-level study of probability.
Thanks!
The first is the pushforward map definition. Since $f$ is not restricted to be one-to-one at this point, isn't $f^{-1}$ a set? Use later in 4.1 would be consistent with this understanding. If $f^{-1}$ is a set, $f^{-1} \in A$ didn't make sense to me; $f^{-1} \cap A \ne \emptyset$ seems to be the meaning here.
f^{-1} is definitely meant to be treated as a set everywhere, except in a few places where I explicitly assume a 1-1 function and consider only point pullbacks, f^{-1}(y). For example in the intro to Section 4.1, right before Section 4.1.1, I talk about how a pullback can be computed on discrete spaces by pulling back each element in the set one by one.
Looking at the current version I see f^{-1} = A, where A \in \mathcal{X}, the sigma-algebra of the input space, everywhere. Is there a particular place in the text with confusion notation, or should I try to make the role of A more clear?
The second is the definition of absolutely continuous in 4.3. The statement:
A measure $\nu$ is absolutely continuous with respect to another measure $\mu$ when $\nu$ allocates zero volume only to those sets which $\mu$ also allocates zero volume
My translation of this, consistent with the 'only if' in the following formula, would be that $\nu$ absolutely continuous w.r.t. $\mu$ means $\nu(A) = 0 \implies \mu(A) = 0$ ($\nu(A) = 0$ only if $\mu(A) = 0$). However, this is the converse of the definition I find in Athreya and Lahiri (p. 53) and the Encyclopedia of Mathematics https://www.encyclopediaofmath.org/index.php/Absolute_continuity: $\nu$ is absolutely continuous w.r.t. $\mu$ if $\mu(A) = 0 \implies \nu(A) = 0$. If I'm understanding correctly, it should say:
A measure $\nu$ is absolutely continuous with respect to another measure $\mu$ when $\nu$ allocates zero volume to all sets to which $\mu$ also allocates zero volume
It's quite possible I'm missing something, as I am pretty new to measure theory, but I don't see how these definitions don't contradict.
Thanks! The “only” should indeed by “every”.
I’ll include a fix when I update the case study next.
mdekstrand
First, thank you for providing these studies. I have been looking for probability theory intro material to provide to my data science students & advisees, and I think your studies will be quite useful for their intermediate-level study of probability.
There are two things that tripped me up, though.
The first is the pushforward map definition. Since $f$ is not restricted to be one-to-one at this point, isn't $f^{-1}$ a set? Use later in 4.1 would be consistent with this understanding. If $f^{-1}$ is a set, $f^{-1} \in A$ didn't make sense to me; $f^{-1} \cap A \ne \emptyset$ seems to be the meaning here.
The second is the definition of absolutely continuous in 4.3. The statement:
My translation of this, consistent with the 'only if' in the following formula, would be that $\nu$ absolutely continuous w.r.t. $\mu$ means $\nu(A) = 0 \implies \mu(A) = 0$ ($\nu(A) = 0$ only if $\mu(A) = 0$). However, this is the converse of the definition I find in Athreya and Lahiri (p. 53) and the Encyclopedia of Mathematics: $\nu$ is absolutely continuous w.r.t. $\mu$ if $\mu(A) = 0 \implies \nu(A) = 0$. If I'm understanding correctly, it should say:
It's quite possible I'm missing something, as I am pretty new to measure theory, but I don't see how these definitions don't contradict.