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What is the meaning of an integral like $$ \int dx_1 x_1 P(x_1,\dots,x_n)$$ for a joint probability distribution $P$? Can this be interpreted as the first moment of $x_1$ held conditional on the other $x_2, \dots, x_n$?

kevinkayaks
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    Short answer: yes. – herb steinberg Nov 21 '21 at 00:47
  • @herbsteinberg what I find confusing about this is that if $\int v W(x,v) dv = \langle v \rangle P(x)$, where $W(x,v)$ is the joint distribution and $P(x)$ is a reduced distribution, then how do joint moments come out? Is $\langle xv \rangle = \int x \langle v \rangle P(x,t) dx$? – kevinkayaks Nov 21 '21 at 18:15
  • What do you mean by (v)P(x)? Last statement has a parameter t. Typo? – herb steinberg Nov 21 '21 at 18:35
  • $\langle v \rangle$ is the "expectation of $v$ conditional on $x$" as in the original question: $\langle v \rangle P(x) = \int dv W(x,v).$ Maybe this would be better written as $\langle v|x \rangle$? The last statement should be $\langle x v \rangle = \int x \langle v \rangle P(x) dx$. – kevinkayaks Nov 21 '21 at 18:41
  • Sorry for the searchy questions. Largely I'm confused by the meaning of $\bar{v}$ used here: https://journals.aps.org/pr/abstract/10.1103/PhysRev.93.1169 – kevinkayaks Nov 21 '21 at 18:43
  • Confusing notation: $(xv)=\int x(v)P(x)dx$. $x$ is dummy in RHS. What is meaning of $(xv)$? – herb steinberg Nov 21 '21 at 18:49
  • $\langle x v \rangle$ is the joint moment of $x$ and $v$ over the distribution $W(x,v)$. $P(x)$ is the reduced distribution $\int W(x,v) dv = P(x)$ – kevinkayaks Nov 21 '21 at 19:08
  • Joint moment? Product, sum, two-d? Use different letters for dummy than argument for expression. – herb steinberg Nov 21 '21 at 22:40

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