Among competing hypotheses that predict equally well, the one with the fewest assumptions should be selected.

In ML, you have some set of data $D$ and a model for generating this data. This model has parameters $\theta$. The probability of observing data is $\mathrm{Pr}(D \, | \, \theta)$. The best parameter point estimate $\hat{\theta}$ is simply the value that maximizes $\mathrm{Pr}(D \, | \, \theta)$.

For example, if we have data $D$ from a Bernoulli observation model representing $k$ successes in $n$ trials, then the probability of observing $k$ and $n$ given coin flip probability parameter $p$ is simply $$\mathrm{Pr}(k,n \, | \, p) = p^k \, (1-p)^{n-k}.$$

For the Bernoulli model $\mathrm{Pr}(k,n \, | \, p) = p^k \, (1-p)^{n-k}$, we have $\hat{p} = \cfrac{k}{n}$. For example, with $k=8$ and $n=10$, $\hat{p}=0.8$ and the likelihood curve follows

In phylogenetics, $D$ are the observed tip sequences and $\theta$ is the phylogenetic tree including topology and branch lengths.

Inference becomes a search for the tree that maximizes the likelihood of observing tip sequences. Lots of computation goes into this.

Generally, it's difficult to make probability statements using frequentist statistics. You cannot
directly say that model 1 is twice as likely as model 2.
People misuse *p* values in this sort of fashion all
the time.

Bayes' rule forms the basis of Bayesian inference, it states: $$ \mathrm{Pr}(A \, | \, B) = \cfrac{ \mathrm{Pr}(B \, | \, A) \, \mathrm{Pr}(A) }{ \mathrm{Pr}(B) } $$

For example, let's say we have an Ebola test that is 99% sensitive and 99% specific (meaning if someone has Ebola it will report true 99% of the time and if someone doesn't have Ebola it will report false 99% of the time). Let's further say that 0.1% of the population has Ebola. If we select a random individual and observe a positive test result, what is the probability that they actually have Ebola?

If we select a random individual and observe a positive test result, what is the probability that they
actually have Ebola? I.e.
$\mathrm{Pr}(\mathrm{P} \, | \, \mathrm{E}) = 0.99$,
$\mathrm{Pr}(\mathrm{N} \, | \, \mathrm{E}) = 0.01$,

$\mathrm{Pr}(\mathrm{P} \, | \, \mathrm{NE}) = 0.01$,
$\mathrm{Pr}(\mathrm{N} \, | \, \mathrm{NE}) = 0.99$

and $\mathrm{Pr}(\mathrm{E}) = 0.001$.

In this case, what is $\mathrm{Pr}(\mathrm{E} \, | \, \mathrm{P})$?

Bayesian inference applies Bayes' rule in a likelihood context, so that $$ \mathrm{Pr}(\theta \, | \, D) = \cfrac{ \mathrm{Pr}(D \, | \, \theta) \, \mathrm{Pr}(\theta) }{ \mathrm{Pr}(D) }, $$ where $D$ is data and $\theta$ are parameters. $\mathrm{Pr}(D)$ is constant with respect to $\theta$, so that $ \mathrm{Pr}(\theta \, | \, D) \propto \mathrm{Pr}(D \, | \, \theta) \, \mathrm{Pr}(\theta)$. This relationship is often referred to as $ \mathrm{posterior} \propto \mathrm{likelihood} \times \mathrm{prior}$.

Following our previous Bernoulli example, we've observed $k$ successes in $n$ trials, and so the likelihood $\mathrm{Pr}(k,n \, | \, p) = p^k \, (1-p)^{n-k}$. We'll assume a flat prior $\mathrm{Pr}(p) = 1$. In this case, the marginal likelihood follows $$\mathrm{Pr}(k,n) = \int_0^1 \mathrm{Pr}(k,n \, | \, p) \, \mathrm{Pr}(p) \, dp = \cfrac{k! \, (n-k)!}{(n+1)!}.$$ And the full posterior follows $$\mathrm{Pr}(p \, | \, k,n) = \cfrac{(n+1)! \, p^k \, (1-p)^{n-k}}{k! \, (n-k)!}.$$

If $k=8$ and $n=10$, the mean posterior $\mathrm{E}[p] = 0.75$, while the 95% credible interval extends from $0.482$ to $0.896$, and the posterior distribution follows

Here, we are interested in the posterior distribution $\mathrm{Pr}(\tau, \mu \, | \, D)$, where $D$ represents sequence data, $\tau$ represents the tree topology and $\mu$ represents mutational parameters (like transition vs tranversion rate). In this case, $$ \mathrm{Pr}(\tau, \mu \, | \, D) \propto \mathrm{Pr}(D \, | \, \tau, \mu) \, \mathrm{Pr}(\tau) \, \mathrm{Pr}(\mu). $$

In the case of the coalescent model, we are interested in coalescent rate parameter like $\lambda$. Here, we use $\lambda$ to give the likelihood of observing a particular tree topology $\mathrm{Pr}(\tau \, | \, \lambda)$. This probability is the likelihood of observing the coalescent intervals seen in the tree.

Thus, the full model becomes $$ \mathrm{Pr}(\tau, \mu, \lambda \, | \, D) \propto \mathrm{Pr}(D \, | \, \tau, \mu) \, \mathrm{Pr}(\tau \, | \, \lambda) \, \mathrm{Pr}(\lambda) \, \mathrm{Pr}(\mu). $$ Bayesian approaches work well to build these sorts of nested models.