Convergence of likelihood implying convergence of marginal likelihood?

Question

I will ask my question through a toy motivating example.

It is well known that a Poisson process is the continuous time analog to a Bernoulli process (for example: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-262-discrete-stochastic-processes-spring-2011/course-notes/MIT6_262S11_chap02.pdf). There is also a similar well known relationship between the Gamma distribution and the Beta distribution (for example: https://math.stackexchange.com/questions/190670/how-exactly-are-the-beta-and-gamma-distributions-related). I would like to make one related point that seem less obvious in the literature:

If we place a Beta prior upon that Bernoulli process and then compute the marginal likelihood, can we prove that this is the discrete time analog to the corresponding marginal likelihood when we place a Gamma prior upon the Poisson process... only using the underlying correspondences between Beta and Gamma, and between Bernoulli process and Poisson process? Is there a theorem we may invoke and under what conditions may we invoke it, or do we have to show correspondence by working with the marginals directly?

This example in particular is a toy -- it is relatively straightforward to work with the marginal likelihoods in the two cases, starting with the discrete case and shrinking the time increments down to zero. However for a research project I am working on, it is much easier to show correspondence between likelihoods, and between priors, than between marginal likelihoods. In my research, I am considering a Bernoulli($p_1$) process which is `stopped' at some time $\tau$ (i.e., all observations after time point $\tau$ equal 0 with probability 1), where $\tau$ is geometric($p_2$) distributed. Both $p_1$ and $p_2$ have independent Beta priors.

To put this in formulae using generic notation, if the limiting form of the pmf is equal in distribution to another pmf: $$ \lim_{n \rightarrow \infty} p(x | \theta_n) = p(x | \theta) $$

and the limiting form of the prior pdf is equal in distribution to another pdf:

$$ \lim_{n \rightarrow \infty} p(\theta_n | \psi_n) = p(\theta_n | \psi) $$

then under what conditions (any?) can we `take the limit within the integral' and state that marginal likelihoods are equal in distribution as well:

$$ \lim_{n \rightarrow \infty} \int p(x | \theta_n) p(\theta_n | \psi_n) d\theta_n = \int \lim_{n \rightarrow \infty} p(x | \theta_n) p(\theta_n | \psi_n) d\theta_n = \int p(x | \theta) p(\theta | \psi) d\theta \ \ \text{?} $$

Any help would be most appreciated. I've tried to make this as clear as possible.

Answer 1

I think what you want is the dominated convergence theorem. From Wikipedia:

Let $f_n$ be a sequence of real-valued measurable functions on a measure space $(S, Σ, \mu)$. Suppose that the sequence converges pointwise to a function $f$ and is dominated by some integrable function $g$ in the sense that $$|f_n(x)| \le g(x)$$ for all numbers $n$ in the index set of the sequence and all points $x \in S$. Then $f$ is integrable and $$\lim_{n\to\infty} \int_S |f_n-f|\,d\mu = 0$$ which also implies $$\lim_{n\to\infty} \int_S f_n\,d\mu = \int_S f\,d\mu.$$

Here $f_n = p(x |\theta_n) p(\theta_n | \psi_n)$. Since it is a product of probabilities, it seems hopeful that you can bound the $f_n$ by some integrable $g$.