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Paraphrasing Casella and Berger (2002): A hypothesis test is defined by a null hypothesis $H_0: \theta \in \Theta_0 $ and an alternative hypothesis $H_1: \theta \in \Theta_0^c = \bar{H_0}$, where $\Theta$ is a set of possible values of the population parameter $\theta$, and $\Theta_0^c = \Theta - \Theta_0 $.

In the Frequentist setting, $\theta$ is an unknown but fixed quantity (meaning, it is non-random). As a result, for a given set $\Theta_0$, the null hypothesis $H_0$ is strictly True or False (but we do not know which). The alternative hypothesis, being the negation, is correspondingly False or True.

Let our test statistic be $\hat{\theta} = f(X_1, ..., X_n)$. We define the rejection region $RR \subset Range(\hat{\theta})$ as the set of values of the test statistic $\hat{\theta}$ for which we choose to reject $H_0$. As our sample $X_1, ..., X_n$ is random, $\hat{\theta}$ is a random variable, and so the quantity $p(\hat{\theta} \in RR)$ is a valid probability, similar to simpler probabilities like $p(X \le 5)$

So far so good. However, my confusion begins when we start to define Type I errors. A Type I error is when we reject $H_0$ even though it is True. This is defined mathematically as:

$$ \alpha = p(\hat{\theta} \in RR | H_0 \text{ is True}) = p(\hat{\theta} \in RR | \theta \in \Theta_0) $$

This probability makes no sense to me. $\theta$ is not a random variable, so $\theta \in \Theta_0$ is not a probabilistic event, it is a True-or-False value. The framework of probability (as far as I understand it) does not allow us to meaningfully condition on real values or booleans, only on events.

What is going on here? The only meaningful interpretation I can make is if $\theta$ is a random variable, which violates the core idea of Frequentist statistics.

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    $\begingroup$ I am not sure I understand your concern? When we use the notation $P(A|B)$ to express conditional probabilities, the condition B is never a random variable; by definition it is assumed to be true, to have occured, and hence its probability is always 1. So $P(\hat \theta \in RR|\theta = {\theta}_0)$ is perfectly valid. As an answer noted (+1), it is a bit more complex for composite nulls, but usually that gets back to the above case, because it is sufficient to compute this in the "worse case", which is a single $\theta_0$ value. $\endgroup$
    – jginestet
    Commented Nov 26 at 18:56
  • $\begingroup$ @jginestet That seems to be at odds with Bayes' theorem. $\endgroup$
    – Dave
    Commented Nov 26 at 19:00
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    $\begingroup$ @Dave, "seem" being the proper term. Yes, if you apply Bayes' theorem to P(A|B), you end up with a $P(B)$ term; do you set it to 1? No. That term is a prior: prior probability of B being true. So what is the prior for $\theta=\theta_0$? See e.g. Colquhoun's paper (in Appendix A) royalsocietypublishing.org/doi/10.1098/rsos.140216 where he uses Bayes' theorem for significance tests (whic is basically the context of the OP's question). $\endgroup$
    – jginestet
    Commented Nov 26 at 20:02
  • $\begingroup$ @jginestet In the Frequentist setting, $\theta$ is fixed but unknown, so we do not know whether “$\theta \in \Theta_0$” is True. This term is also not an event in the probabilistic sense. How, then, do we justify conditioning on it within the framework of conditional probability? That is the core of my question. $\endgroup$ Commented Nov 27 at 1:59

2 Answers 2

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You're right that the notation of the form $P(X|H_0)$ is misleading if this is supposed to be a frequency distribution for $X$ under a particular model assumption. I have seen people who know what they're talking about use it as shorthand, but I've also seen people sleepwalk into using it without fully appreciating what's going on.

If we're being careful, we will use different notation. In grad school, I think we used the notation $P(X;H_0)$. This is not written as a conditional probability, and is not meant to be interpreted as one. Rather, it should be read as something like "the probability (density) of $X$ under the probability model parameterized according to $H_0$." You'll notice that Casella and Berger, who you cite, are also careful and use notation such as $P_{\theta_0}(X)$ (where $H_0$ is such that $\theta=\theta_0$; page 388).

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  • $\begingroup$ Thanks @JohnMadded, this makes sense. Do you mind expanding your answer to flesh out the exact statistical model we assume in hypothesis testing? I have not found a text which mentions it in detail (I have been searching in Casella and Wackerly) $\endgroup$ Commented Nov 26 at 12:34
  • $\begingroup$ (Sorry, I meant @JohnMadden) $\endgroup$ Commented Nov 26 at 13:15
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    $\begingroup$ @AbhishekDivekar a statistical model (for us) is nothing more than a specification of a distribution for observable quantities. So for example, "this coin is fair e.g. P(X=Heads)=$\theta$=0.5" is an example of a statistical model. The difference is that for the Frequentist, this is the entire model. Whereas for the Bayesian, the statistical model consists of (whether implicitly or explicitly) a joint distribution placed on $\theta,X$ together. Subsequently, the Bayesian achieves a conditional distribution on $X|\theta$. (In practice, this joint distribution is often specified implicitly by... $\endgroup$ Commented Nov 27 at 16:04
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    $\begingroup$ ... way of first defining $P(X|\theta)$ and $P(\theta)$). $\endgroup$ Commented Nov 27 at 16:04
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    $\begingroup$ @Alexis "best" refers to $H_0$, in this sence the "best" knowledge, the current knowledge. But you're right, it can just be an approximation of anything, so yes, it's a chosen model $\endgroup$
    – Mayou36
    Commented Nov 27 at 23:34
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The issue here seems to be more about simplified notation than a conceptual error. Let’s break it down in more detail:

You wrote:
$$ P(\hat{\theta} \in RR \mid \theta \in \Theta_0) = \alpha $$

This expression doesn't make sense when $\Theta_0$ contains more than one element, i.e., when you have a composite null hypothesis. In this case, you are not specifying a single probability distribution, as $\theta$ can take multiple values within $\Theta_0$.

Case 1: Simple Null Hypothesis

When the null hypothesis $H_0$ is simple, for example, $\Theta_0 = \{\theta_0\}$, the computation of $\alpha$ is straightforward and can be expressed as:

$$ P_{\theta_0}(\hat{\theta} \in RR) = \alpha, $$

Here, $\alpha$ represents the probability that the statistic $\hat{\theta}$ falls within the rejection region $RR$, given that the true value of $\theta$ is $\theta_0$. The subscript in $P_{\theta_0}$ indicates that the probability is calculated under the assumption that $\theta = \theta_0$, and the distribution is fully specified.

Case 2: Composite Null Hypothesis

When $\Theta_0$ contains multiple possible values, as in a composite null hypothesis, there isn’t a single $\theta$ value tied to $H_0$. This makes the interpretation of $\alpha$ more complex. In this case, the significance level is typically defined as the supremum of the rejection probability over all values of $\theta$ in $\Theta_0$:

$$ \alpha = \sup_{\theta \in \Theta_0} P_{\theta}(\hat{\theta} \in RR). $$

This ensures that $\alpha$ represents the maximum Type I error rate, regardless of the specific value of $\theta$ within $\Theta_0$. In other words, this definition guarantees that the significance level does not exceed $\alpha$ for any parameter value allowed under the null hypothesis.

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    $\begingroup$ +1 as this answer corrects my definition of Type I error to be more precise. However it does not answer the core question, which is about the interpretation of the conditional term $\theta \in \Theta_0$. $\endgroup$ Commented Nov 27 at 1:41

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