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Oscar_Cunningham comments on Open thread, 24-30 March 2014 - Less Wrong Discussion
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Am I confused about frequentism?
I'm currently learning about hypothesis testing in my statistics class. The idea is that you perform some test and you use the results of that test to calculate:
P(data at least as extreme as your data | Null hypothesis)
This is the p-value. If the p-value is below a certain threshold then you can reject the null hypothesis (which is the complement of the hypothesis that you are trying to test).
Put another way:
P(data | hypothesis) = 1 - p-value
and if 1 - p-value is high enough then you accept the hypothesis. (My use of "data" is handwaving and not quite correct but it doesn't matter.)
But it seems more useful to me to calculate P(hypothesis | data). And that's not quite the same thing.
So what I'm wondering is whether under frequentism P(hypothesis | data) is actually meaningless. The hypothesis is either true or false and depending on whether its true or not the data has a certain propensity of turning out one way or the other. Its meaningless to ask what the probability of the hypothesis is, you can only ask what the probability of obtaining your data is under certain assumptions.
Your conclusion
is correct. Frequentists do indeed claim that P(hypothesis | data) is meaningless for exactly the reasons you gave. However there are some little details in the rest of your post that are incorrect.
The hypothesis you are trying to test is typically not the complement of the null hypothesis. For example we could have:
where theta is some variable that we care about. Note that the region theta<0 isn't in either hypothesis. If we were instead testing
then frequentists would suggest a different test. They would use a one-tailed test to test H1 and a two-tailed test to test H1'. See here.
No. This is just mathematically wrong. P(A|B) is not necessarily equal to 1-P(A|¬B). Just think about it for a bit and you'll see why. If that doesn't work, take A="sky is blue" and B="my car is red" and note that P(A|B)=P(A|¬B)~1.