FWIW, 'jsprogrammer is smuggling here the basic definition of probability - that P(event) = (number of observations in which an event happens) / (total number of observations).

That only is the case if you work on a discrete uniform distribution, which you usually don't.

The axioms are only defined on discrete uniform distributions. All other distributions must be constructed from the discrete uniform.

This is not true. If you have a discrete uniform distribution, you can only have finitely many elements in your probability space, so it wouldn't make sense to have an axiom which refers to countably-many disjoint sets. The axioms are most useful for continuous distributions, and are closely related to measure theory.

The discrete, uniform nature is obvious in the rule of computation:

Σ P(E)

Each "unit" of the computation is weighted equally (though, the value of the units may differ).

I'll give you the benefit of the doubt and assume that you aren't deliberately trolling me, but you do seem to be deeply confused. I'll just give you a link to some lecture notes that go all the way from the axioms of probability (lecture 1) to p-values (lecture 12) and leave it at that: http://www.win.tue.nl/~rmcastro/2DI90/index.php?page=lecture....

Thanks for the massive set of pages. I already went through a similar course a long time ago.

You'll note that the notes don't derive the calculation of p-value. It merely gives a trivial example (curiously, the sum of two probabilities) with a fiat interpretation.

It's curious that my last post, merely quoting the third axiom and showing its properties plain, makes you think that I am deeply confused, when you cannot even refute trivial points and must resort to arguments (poor, in that they do not address our issue here) from authority.

I assume you refer to page 28 in the first slide deck? If you would have read closely you would have noticed that it says:

> We will restrict ourselves to discrete sample spaces for now, to avoid some technical difficulties…

In the later lectures they take a look at sample spaces that are uncountably infinite. Good luck working with a sum there.

No. I am referring to lecture 12 (ie. Ch9).

What reformulation of the probability axioms are you using to eliminate the sum in the third axiom?

The axioms may be trivially extended to bounded intervals (a la calculus).

>It doesn't make sense to say...

That depends on your hypothesis.

>The p-value is the probability of seeing a set of values

That is a circular definition. What is, "probability"?

P-values are not prescriptive statements about future observations, but are descriptions of actual observations.

As TeMPOraL points out, you seem to be defining probability in general. The p-value is a very specific concept in statistics, not an abbreviation for "probability value" (https://en.wikipedia.org/wiki/P-value).

Your link defines p-value as a probability.

>More specifically, the p-value is defined as the probability of...

Edit: A good exercise would be to start with the axioms[0] of probability theory and then derive the p-value for a simple experiment, using only those axioms (and your measured values).

[0] https://en.wikipedia.org/wiki/Probability_axioms

Example: The null hypothesis is that X has a standard normal distribution, and you have one observation, with value 0.7. Since you only have one observation, your definition of p-value can only give the values 0 or 1, depending on whether you consider 0.7 to agree or disagree with the null hypothesis. The actual answer is 0.2419637 (or double that, depending on whether you do a one-tailed or two-tailed test), the probability of drawing a value >= 0.7 from a standard normal distribution (you can get this value by computing an integral on the density function of the standard normal).

You are only taking a single observation in that example, making analysis through probability theory effectively useless.

Note that, probabilities of 0 or 1 are completely allowed under the theory and the computed value in your example is still consistent with my statement and probability theory.

An issue is that probability distributions (such as standard normal) don't exist in the axioms and so must be constructed from the axioms. So, a p-value test using a standard normal distribution is taking a shortcut that hides some of the calculations, making it appear that p-value is not actually counting a ratio, however, if you look at the full calculation from the axioms, p-value represents the ratio of agreeing observations to total observations. Basically, a single observation at the level of testing a standard normal distribution may actually represent multiple observations at the level of the math required to construct the full calculation (this entirely depends on the hypothesis being tested).

Let Ω = R (real line), F = the set of Lebesque-measurable subsets of R, and P(E) = the integral over E of the density function of the standard normal. This is a probability space in the Kolmogorov sense, and my previous example carries through perfectly well in this framework. You observe x = 0.7, and compute the probability of the event X >= 0.7 as an integral. There are no ratios, and you don't need to label the observation as agreeing or disagreeing with the null hypothesis.

See my response here: https://news.ycombinator.com/item?id=10630389

You are doing it wrong. A p-value based on a single observation is meaningless. P-values are much more useful when they are continuously computed on a running experiment with sequential observations.

See the work at the LHC for an example.