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Suppose I have a nice distribution given by its pdf $p(x)$ on $\mathbb{R}^n$. It is usually problematic to condition on sets with zero measure (eg the Borel-Kolmogorov paradox). Nonetheless, given a point $x^*\in\mathbb{R}^n$ and $r>0$, I would like to be able to condition on $\|x-x^*\|_2=r$ with the purpose of then integrating. Namely something along the lines of writing $$\mathbb{E}_{R\sim p(r)}[\mathbb{E}_{X\sim p(x\mid \|x-x^*\|_2=R)}[f(x)]]$$ for some function $f$, where $p(r)=\tfrac{\partial}{\partial s}P_{p(x)}[B_s(x^*)]$ is the pdf of the radius.

Is this valid? If not in general, what do I need in order to make it a valid operation?

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  • $\begingroup$ I don't fully understand your notation, but there is a notion of conditional expectation where instead of considering on an event like $\{R \in A \}$ you consider on $R$ directly. Here, $E[f(X) \vert R]$ will still be random but in a sense a function of $R$. If then $X$ has a conditional density with respect to the conditional expectation you write it as an integral. See for example chapter 4.1 of Durrett's book, services.math.duke.edu/~rtd/PTE/PTE5_011119.pdf $\endgroup$
    – David
    Commented Jun 26 at 7:40

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