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This is a simple question from measure theory.

Fix a measurable space $(E,\mathcal{E})$ and a family $(P_i)_{i\in I}$ of probability measures on $(E,\mathcal E)$ ($I$ is any non-empty set). Let $n\geq 2$ be a positive integer.

Is it true that the following statements are equivalent?

  1. There exists a $\sigma$-finite measure $\mu$ on $(E,\mathcal E)$ such that $P_i$ is absolutely continuous with respect to $\mu$, for all $i\in I$;
  2. There exists a $\sigma$-finite measure $\nu$ on $(E^n,\mathcal E^{\otimes n})$ such that $P_i^{\otimes n}$ is absolutely continuous with respect to $\nu$, for all $i\in I$.

In other (statistical) words, is it true that a statistical model corresponding to one observation is dominated if and only if the statistical model corresponding to $n$ iid copies of that one observation is dominated?

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Yes, this is true.

For 1. $\Longrightarrow$ 2., simply take $\nu = \mu^{\otimes n}$ to get the dominating measure for the product. This is a $\sigma$-finite measure which dominates the product of the statistical model.

For 2. $\Longrightarrow$ 1., assume w.l.o.g. that $\nu$ is a finite measure; given a $\sigma$-finite measure, there always exist a finite measure equivalent to it. Take $\mu = \pi_{1*}\nu$, where $\pi_1\colon E^n\to E$ is the projection onto the first coordinate, and $\pi_{1*}\nu$ is the pushforward measure ($\pi_{1*}\nu(A) := \nu(\pi_1^{-1}(A))$). You can check that this choice of $\mu$ dominates the statistical model, and it is a finite measure since $\nu$ is. The reason we took $\nu$ to be finite is because otherwise $\nu$ being $\sigma$-finite is not sufficient to guarantee that its pushforward $\pi_{1*}\nu$ is $\sigma$-finite.

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  • $\begingroup$ Hello, thank you! For 1. $\rightarrow$ 2., I agree, the proof is straightforward. For 2. $\rightarrow$ 1. though, it is not as simple. When $\nu$ is finite (hence, wlog, it is a probability measure), simply take its marginals. However, if $\nu$ is not finite, the marginals (push-forwards by the projections) are not $\sigma$-finite, so the proof does not carry on. So, I do not agree with your « w.l.o.g. », it is still unclear whether 2. implies 1. when $\nu$ is not finite. $\endgroup$ Commented Nov 25 at 8:24
  • $\begingroup$ I agree that this case requires care, let me clarify what I meant by "w.l.o.g. that 𝜈 is a finite measure; given a 𝜎-finite measure, there always exist a finite measure equivalent to it". If $\nu$ is not finite, we can just take a finite $\tilde \nu$ which is equivalent to $\nu$ (see en.wikipedia.org/wiki/Σ-finite_measure "Equivalence to a probability measure"). Then we can choose $\mu = \pi_{1*}\tilde \nu$. $\endgroup$ Commented Nov 25 at 8:30
  • $\begingroup$ Ooh that's a super nice fact to know, thank you! $\endgroup$ Commented Nov 25 at 8:43

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