Hypothesis testing: Defining H0, HA hypotheses so that ( H_A)_A' makes sense

[WWGD]

TL;DR

The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive,

The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense.
Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with $(M^{T})^{T}=M$, and just wanted to see if it made sense to talk about the " Alternative of the Alternative " returned the initial hypothesis.

 

[Dale]

TL;DR: The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive,

Since H0, HA aren't exhaustive

Why wouldn’t they be exhaustive?

 

[WWGD]

Why wouldn’t they be exhaustive?

Well, at least in the Frequentist set up , we don't get to actually accept, just conclude we don't have enough evidence to reject or not reject. I don't understand the Bayesian approach to tell.

 

[Dale]

Ah, I see your question and my confusion.

Let’s say that you are doing a placebo controlled randomized test for a medicine to reduce the duration of a cold with $\mu_p$ being the mean duration of a cold with the placebo and $\mu_d$ being the mean duration of a cold with the drug.

Usually the hypotheses would be $H_0: \ \mu_p=\mu_d$ and $H_A: \ \mu_p\ne \mu_d$. So those are exhaustive. No matter the actual values of $\mu_d$ and $\mu_p$ exactly one of those two hypotheses will be true. So the hypotheses are mutually exclusive and collectively exhaustive.

But you are looking not at the hypotheses but the decisions. And I agree with you there.

So you could make a decision to reject a hypothesis, fail to reject a hypothesis, accept a hypothesis, or fail to accept a hypothesis. So that is four possibilities for each of two hypotheses for a total of sixteen possible outcomes. Maybe accept both is not a possibility so maybe it is fifteen. But certainly reject and fail to reject the null hypothesis is not exhaustive.

Is that more or less what you meant?

 

[Dale]

With frequentist statistics you cannot speak of the probability of hypotheses, so you can only calculate $P(D|H_0)$. With Bayesian statistics the big difference is that you can calculate $P(H_0|D)$