Units for likelihoods and probabilities In this discussion by comments Is the exact value of any likelihood meaningless?, it was suggested (firmly!) that likelihoods and probabilities calculated from continuous data not only have units, but those units are the reciprocal of units of the data. In contrast, likelihoods and probabilities from discrete data have no units. Is that correct?
I thought that probabilities and likelihoods were dimensionless and had not thought about such a stark contrast coming from the continuity of observed values.
 A: Probabilities (also called "probability masses") are unitless, but probability densities have units of 1/(units of the variable).
Let's say we have a probability density $p\left(x\right)$ of some variable $x$. If we integrate this density over some range in $x$, we obtain the probability that $x$ falls in this range:
$$
P\left(a < x < b\right) = \int_a^b p\left(x\right) dx .
$$
The quantity $P\left(a < x < b\right)$ is, of course, a probability mass, and must therefore be a unitless real number between 0 and 1. But in order for $P\left(a < x < b\right)$ to be unitless, we can see from the above integral that $p\left(x\right)$ must have units of 1/(units of $x$). The quantity $p\left(x\right)$ is the amount of probability per unit of $x$, which is why the term "probability density" is applied to it in the first place.

likelihoods and probabilities calculated from continuous data not only have units, but those units are the reciprocal of units of the data

"Probabilities calculated from continuous data" only have units of 1/(units of the data) if they are probability densities of the data. The likelihood is the probability density of the data (assuming the data to be made up of continuous variables - for discrete data, the likelihood is a unitless probability mass), given a choice of model. The likelihood is usually given as a function of the model parameters, $\theta$:
$$
L\left(\theta\right) := p\left(D|\theta\right) ,
$$
where $D$ is the data. As such, the likelihood has units of 1/(units of $D$), and its integral over all possible values of $D$ must come to 1. However, very importantly, $L\left(\theta\right)$ is not a probability density of $\theta$, and the integral of $L\left(\theta\right)$ over all values of $\theta$ does not have to come to 1. In fact, that integral might not even yield a unitless quantity at all: it will have units of (units of $\theta$)/(units of $D$).
This often leads to considerable confusion, because students are rightly taught that $L\left(\theta\right)$ is not a probability density in $\theta$, but then incorrectly conclude that it's not a probability density at all. In fact, it's the probability density of the data (given the model).
