I came across the following Bayesian equation in textbook of evolutionary biology:

$f(t, r, \theta|X) \propto f(X|t, r, \theta)f(t|\theta)f(r|t,\theta)f(\theta)$

$f(X|t, r, \theta)$ is the likelihood and $f(t|\theta)$ is the prior distribution on times $t$ $f(r|t,\theta)$ is the prior distribution on rates $r$, and $f(\theta)$ is the prior of the substitution model.

I'm having trouble understanding what it means for a prior distribution to be conditioned on other values, as we have in $f(t|\theta)$ and $f(r|t,\theta)$. Does this mean that our prior beliefs about $t$ depend on $\theta$ and those about $r$ depend on both $t$ and $theta$? I.e., that without knowing $t$ and $theta$, we have no prior beliefs about $r$?

Does the specification of the priors in this way make this an example of a hierarchical model? If so, what exactly is the hierarchy?

If anyone can recommend any literature that might help me understand this, I'd be grateful for it.


2 Answers 2


It is often easier to reason about one random quantity at a time than work with all random quantities simultaneously. In Bayesian statistics, where everything is a random quantity, this is especially true. You often have to fix one random quantity to work with another. In more technical terms, it is often easier to work with conditional distributions than joint distributions.

You can think of conditional probability as tool to set a random quantity to a particular, fixed value. So rather than thinking about $f(r | t, \theta)$ as prior knowledge of $r$ depending on $t$ and $\theta$, imagine this conditional distribution as a way to reason about $r$ without the interference of random fluctuations in $t$ and $\theta$. You set $t$ and $\theta$ to specific values $t^*$ and $\theta^*$ while allowing $r$ to vary: $f(r | t = t^*, \theta = \theta^*)$

As @Peter Pang notes, you can factor the joint prior distribution differently than in your original post: $$ \begin{aligned} p(t, r, \theta) &= p(r \vert \theta, t)p(\theta \vert t)p(t)\\ &= p(\theta \vert r, t)p(r \vert t)p(t)\\ &= p(\theta \vert r, t)p(t \vert r)p(r)\\ &=\vdots \end{aligned} $$ Depending on the specific problem you're working on, it may be simpler (conceptually, mathematically, or numerically) to work with the distribution of, say, $f(t | r, \theta)$ instead of $f(r | t, \theta)$. Since the joint prior can be factor differently, it is always your option choose what quantities, if any, are fixed in place (i.e. conditioned on) at each step in the factorization.


Let's take a step back and revisit the Bayes theorem; $$ p(t, r, \theta \vert X) \propto p(X\vert t, r, \theta)p(t, r, \theta). $$ The expression you shown has done is to factorize the prior distribution $p(t, r, \theta)$. Of course, the factorization is not unique, for instance; $$ \begin{aligned} p(t, r, \theta) &= p(r \vert \theta, t)p(\theta \vert t)p(t)\\ &= p(\theta \vert r, t)p(r \vert t)p(t)\\ &= p(\theta \vert r, t)p(t \vert r)p(r)\\ &=\vdots \end{aligned} $$ All these factorizations are valid, but not all of them are useful. The particular factorization is chosen based on the physical picture of the system of interest. And if it is easy to model the behavior of the parameters $r$ given $t$ and $\theta$ and the behavior of $t$ given $\theta$, the factorization you have shown would be a sensible choice.

You can also model the joint distribution altogether without any factorization, this is more of a modelling problem instead of a statistical one.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.