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Setup. Let $p\in[0,1]$ be a continuous random variable with density $f(\cdot)$. Assume that $f$ is bounded, continuously differentiable, and has full support $[0,1]$.

Let $a_1$ and $a_2$ be distinct, known constants from $(0,1)$. Conditional on $p$, iid data $\{Z_t\}_{t=1}^{\infty}$ are generated by one of two models: model 1 has $Z_t\overset{\text{iid}}{\sim}\text{Bernoulli}(a_1 p)$; model 2 has $Z_t\overset{\text{iid}}{\sim}\text{Bernoulli}(a_2 p)$.

Assume that model 1 is the true model and fix realization $p^*\in [0,\min\left\{a_2/a_1,1\right\}]$. Finally, for each $t\in\{1,2,3,...,\}$ define partial sums $S_t:=\sum_{\tau=1}^t Z_{\tau}$ and conditional likelihood ratio $ L_t:=\frac{\int_{0}^1f(p)(a_1p)^{S_t}(1-a_1p)^{t-S_t} dp}{\int_{0}^1f(p)(a_2 p)^{S_t}(1-a_2 p)^{t-S_t} dp}. $

Question. I want to show that $L_t\overset{a.s.}{\to}{\frac{f(p^*)}{f(\smash{\frac{a_1}{a_2}}p^*)}}$ as $t\to\infty$. This seems to be the case when $p$ is a discrete random variable with finite support. This is quite intuitive, so I conjecture that $L_t$ will almost surely converge to the same limit when $p$ is a continuous random variable. However, I am having difficulty characterizing the a.s. limit of $L_t$. $\ $

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  • $\begingroup$ Notes: First, my assumptions on $f$ are really just me saying "assume that $f$ is nice/well behaved." This is just an eyeball guess on what's "nice enough" given what I want to show. Second, regarding the "$\min\{a_2/a_1,1\}$," please feel free to assume that $a_2<a_1$ if the proof is identical in the case where $a_2\geq a_1.$ $\endgroup$ Commented Jun 28 at 2:50
  • $\begingroup$ I would have thought $p^* \in [0, \max\{a_1,a_2\}]$ though if it is between $\min\{a_1,a_2\}$ and $\max\{a_1,a_2\}$ then perhaps the limit of the likelihood ratio could be $0$ or $\infty$. But your suggestion of ${\dfrac{f(p^*)}{f\left({\frac{a_1}{a_2}}p^*\right)}}$ looks unlikely to me - why that rather than $\dfrac{f(p^*/ a_1)}{f(p^*/ a_2)}$? $\endgroup$ Commented Jun 29 at 21:55
  • $\begingroup$ Hello @Henry, many apologies for the very late response, it's been very busy on my end! So, the intuition behind my suggestion for the a.s. limit of $L_t$ is as follows. First, recall that $L_t$ is the CLR of model 1 vs. model 2; there are just two cases that "perfectly explain the data" asymptotically; first case is that model 1 is the true model and that $p$ was realized as $p^*$, i.e. $Z_t\overset{{iid}}{\sim}\text{Bern}(a_1 p^*)$; second case is model 2 is the true model and that $p$ was realized as $\hat{p}=a_1 p^*/a_2$, i.e. $Z_t\overset{{iid}}{\sim}\text{Bern}(a_2 \hat{p})$.... $\endgroup$ Commented Jul 21 at 19:14
  • $\begingroup$ ... therefore, intuition seems to suggest that the CLR of model 1 to model 2 should boil down to the ratio of the prior probability of $p=p^*$ to the prior probability of $p=\hat{p}$. This is how it works out under the setup where $p$ is a discrete random variable with finite support $\mathcal{P}\subseteq [0,1]$; I would assume that it works out similarly when p is a continuous r.v., at least under certain conditions (hopefully not very restrictive ones!) Also, having $p^*\in[0,\min\{a_2/a_1,1\}]$ was just meant to ensure that $\hat{p}:=a_1 p^*/a_2\leq 1$. $\endgroup$ Commented Jul 21 at 19:21

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