# Is there a difference between marginal likelihood and likelihood of a marginal distribution?

Assume that we have $$n$$ i.i.d. samples $$x_i \sim f(.|\theta )$$, where $$\theta$$ is a parameter. Furthermore, the parameter is also distributed as $$\theta \sim \mathcal{g}(.|\alpha)$$.

Now let's say that we can derive a marginal distribution: $$p(x|\alpha) = \int f(x|\theta)\mathcal{g}(\theta) d\theta$$

The likelihood function is then $$L_1(\alpha| x) = \prod_{i=1}^n p(x_i|\alpha)$$

Alternatively, we can use a marginal likelihood:

$$L_2(\alpha| x) = \int \left(\prod_{i=1}^n f(x_i|\theta)\right) \mathcal{g}(\theta)d\theta$$

What's the difference between the two approaches?

• You seem to be saying that your $x_i$ are conditionally independent given $\theta$. But they need not be conditionally independent given $\alpha$, so I am not sure your $L_1$ works Oct 18, 2021 at 21:26

The problem here is that although the observations $$x_1,...,x_n$$ are independent conditional on $$\theta$$, they are not independent conditional on $$\alpha$$ instead, so as a general rule:
$$f(\mathbf{x}_n|\alpha) \neq \prod_{i=1}^n f(x_i|\alpha).$$
Indeed, marginalising over a distribution on $$\theta$$ will tend to induce positive correlation between $$x_1,...,x_n$$ (see e.g., O'Neill 2009, Theorem 2, p. 244). Consequently, your $$L_1$$ is not a valid likelihood function in this case --- the function $$L_2$$ is the correct likelihood function.