Talk:Pushforward measure

Thanks for elaborating this page. I have relinked to here as much as possible. Now I will take it off my watch list. Good luck! Geometry guy 00:17, 12 February 2007 (UTC)

The first example seems to state that the measure of an arc $A$ of the circle is equal to the measure of $f^\{-1\}(A)$ on the real line, where $f:\backslash mathbb\{R\}\backslash rightarrow\; C$ is the wrap-around function. But $f^\{-1\}(A)$ has measure $\backslash infty$.

Should the correct definition define $\backslash mu(A):=\backslash inf\_\{f(B)=A\}\backslash mu\; (B)$? Am I missing something?

69.81.71.60 (talk) 11:54, 28 June 2017 (UTC)

:No, why? It is written "Let λ also denote the restriction of Lebesgue measure to the interval [0, 2π)". Also f is defined on [0, 2π). Not infinity. Boris Tsirelson (talk) 18:47, 28 June 2017 (UTC)

::I see now. Thank you! Norbornene (talk) 13:29, 9 July 2017 (UTC)

As far as I can see, the following statement is false:

Random variables are pushforward measures

A r.v. defines a pushforward measure, but there is not one-to-one identification. For example, i.i.d r.v.'s $Y\_i:X\_1\; \backslash to\; X\_2$ define the same pushforward measure $Y\_*\; P$, although they are clearly distinct mappings from the probability space $(X\_1,\; \backslash Sigma\_1,\; P)$ to a measurable space $(X\_2,\; \backslash Sigma\_2)$.

AVM2019 (talk) 12:50, 19 May 2022 (UTC)

The pushforward measure definition provided assumes there's a single pre-image point, effectively assuming that the function is invertible, but without formally stating that requirement. Do we have a source which describes a formula showing how to explicitly sum over a pseudoinverse which can handle non-injective cases? 24.236.207.173 (talk) 15:52, 2 February 2025 (UTC)