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I've seen the following claim in the context of likelihood ratio tests but I'm having a hard time verifying the result.

Suppose I have an iid sample $x_1, \ldots, x_n$ drawn from a distribution with parameter $\theta \in \Theta$. The likelihood ratio for testing $H_0: \theta = \theta_0$ versus some alternative $\theta$ is $$ \Lambda_n(\theta) = \frac{L(\theta \,|\, x_1, \ldots, x_n)}{L(\theta_0 \,|\, x_1, \ldots, x_n)} $$ is a martingale under $H_0$ with respect to $\{\mathcal{F}_n: n \geq 1\}$, the filtration generation by $x_1, \ldots, x_n$.

The claim is that the mixture of this likelihood ratio with respect to a Normal prior $H$ on $\theta$ is also a martingale, i.e. letting $$ M_n = \int_{\Theta} \Lambda_n(\theta) \, dH(\theta)$$ then $\{M_n: n \geq 1\}$ is also a martingale with respect to $\{\mathcal{F}_n: n \geq 1\}$.

I've tried writing out the likelihood explicitly to show $E(M_n \,|\, \mathcal{F}_{n-1}) = M_{n-1}$, but I'm not sure how to proceed further past: $$ E \left[ \int_{\Theta} \prod_{i=1}^{n-1} \frac{L(\theta \,|\, x_1, \ldots, x_{n-1})}{L(\theta_0 \,|\, x_1, \ldots, x_{n-1})} \cdot \frac{L(\theta \,|\, x_n)}{L(\theta \,|\, x_n)} dH(\theta) \; \Big| \; \mathcal{F}_{n-1} \right] $$

gogurt
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    You ought to be able to interchange the integral over $\Theta$ with the conditional expectation, via an appropriate version of Fubini's theorem. An axiom I learned in graduate school: any time you see two integrals, they will need to be interchanged. – Nate Eldredge Dec 08 '17 at 06:23
  • @NateEldredge Hah. That's what I was thinking. But I'm not familiar with the right version of Fubini. Guess I'll do some Googling. Thanks. – gogurt Dec 08 '17 at 06:45
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    To show $E[X \mid \mathcal{F}] = Y$, one has to show that for every $A \in \mathcal{F}$ we have $E[X 1_A] = E[Y 1_A]$. Now you are working with unconditional expectation and you can try to interchange that with another integral. – Nate Eldredge Dec 08 '17 at 06:59
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    Have a look at this: https://math.stackexchange.com/q/1165066/36150 (You can replace the Lebesgue measure $du$ by some other (finite) measure; doesn't change anything about the proof.) – saz Dec 08 '17 at 07:29

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