1
$\begingroup$

It is obvious that probability density does not necessarily exist for a given probability distribution. However, many statistical applications assume the existence of density and even use closed form density functions. For example, when calculating posterior distributions we usually use density functions of data and prior. In Neyman-Pearson school of hypothesis testing, likelihood ratio is used. There are many more examples. So I am wondering if the existence of probability densities is essential to statistics, in either theory or applications?

My background is mostly in probability and I have less experience with statistics. Maybe I miss a lot of commonly used statistical models, but I would really appreciate if anyone points that out.

Thanks!

$\endgroup$
3
$\begingroup$

Density against which dominating measure? The question is somewhat unclear, but I take it as asking why a statistical analysis usually makes the assumption that all distributions within the set $\mathfrak F$ of potential distributions of the observed dataset $\mathbf x$ are dominated by or absolutely continuous wrt the same measure $\text{d}\mu$, hence all admit densities $f_\theta$ against this measure $\text{d}\mu$ $$ \mathfrak F=\left\{ f_\theta(\cdot)\text{d}\mu;\ \theta\in\Theta \right\}$$

Unless there is a realistic reason to introduce a family of probability distributions with orthogonal components, this assumption is minor when considering that the assumption that the data comes from a distribution is a much stronger leap of faith. Note further that in the event that the elements of $\mathfrak F$ are not all absolutely continuous wrt the same dominating measure the central notion of likelihood cannot be defined. But in such unusual cases, since the measures all have orthogonal components, there exist realisations of the data that identify uniquely the member of $\mathfrak F$ that generated these realisations. Asymptotically, the distribution is thus uniquely identified by the data.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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