I was wondering, if given a Polish space $X$, and given some probability measure $p$ on $X$, can the expectation of an $X$-valued function be taken? In particular, would the integral $\int_X x dp$ make sense? I feel like it wouldn't, $X$ simply does not have enough structure, but I would like to know if anyone has any thoughts on this.

Many thanks in advance!

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    $\begingroup$ I think this question cannot be answered. First, what is so special about Polish space? Why not taking expectation/mean of any probability measure on some measure space $(X,\cal{A})$? And second, as noted in the question, we need some additive structure on $X$. Otherwise we can call anything "expectation". $\endgroup$ Apr 12 at 11:04
  • $\begingroup$ @DieterKadelka Thank you for the comment! I am actually trying to rule out Polish spaces, and am simply looking for confirmation that indeed, it does not make much sense to work with expectations on fully general Polish spaces. $\endgroup$ Apr 12 at 11:23
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    $\begingroup$ @DieterKadelka : That Polish spaces are metrizable distinguishes them from other measurable spaces. $\endgroup$ Apr 12 at 12:32
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    $\begingroup$ A very special case of expectation is: $(a+b)/2$. This operation is not defined in a general metric space. I have heard of work on semigroup-valued integrals. $\endgroup$ Apr 12 at 12:37

1 Answer 1


If $(X,|\cdot|)$ is a Hilbert space and $Z$ is a random vector in $X$ with $E|Z|^2<\infty$, then $EZ$ is the unique minimizer of $E|Z-a|^2$ in $a\in X$.

So, more generally, for any random element $Z$ of any metric space $(X,d)$ with $Ed(Z,a)^2<\infty$ for some (or, equivalenly, for all) $a\in X$, one may define an expectation of $Z$ as any minimizer of $Ed(Z,a)^2$ in $a\in X$ (if such a minimizer exists). In general, such a minimizer does not have to be unique.

For any $p\in[1,\infty]$, one can similarly define a $(p-1)$-expectation of $Z$ as any minimizer of $\|d(Z,a)\|_p$ in $a\in X$, where for any real-valued random variable $\xi$ we let $\|\xi\|_p:=(E|\xi|^p)^{1/p}$ if $p\in[1,\infty)$ and $\|\xi\|_\infty:=\lim_{p\to\infty}\|\xi\|_p$.

Then a $1$-expectation is an expectation defined above, and a $0$-expectation is a geometric median, and an $\infty$-expectation is a generalized midrange.

  • $\begingroup$ Being a Polish space is a topological property, not a metric property. The answer above regards a specific metric. In this setting, a natural axiom to put on a metric space is the CAT(0), under which the $L^2$ minimzer alluded to above does exist and it is indeed unique. $\endgroup$
    – Uri Bader
    Apr 12 at 13:58
  • $\begingroup$ @UriBader : Thank you for your comment. Can you explain how to prove what you stated about the CAT(0) spaces? $\endgroup$ Apr 12 at 15:37
  • $\begingroup$ @IosifPinelis this is Theorem 3.2.1 in "Nonpositive Curvature: Geometric and Analytic Aspects" (by Jost), for example. The idea is that for CAT(0) spaces, the function you want to minimize is strictly convex, continuous, and coercive. From that, it is not hard to prove that any minimizing sequence is Cauchy. $\endgroup$
    – FMB
    Apr 12 at 16:24
  • $\begingroup$ @FMB : Thank you for the reference. So, I guess this will work as well for $Ed(Z,a)^p$ (in place of $Ed(Z,a)^2$) with any $p\in(1,\infty)$, right? $\endgroup$ Apr 12 at 16:43

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