Let $X$ be a random variable with a probability density $f(\cdot;\theta)$ where $\theta \in \Theta \subset \mathbb R$ is an unknown parameter. Suppose that we have a prior density $\pi(\theta)$, with associated random variable for the parameter $\tilde\Theta$. Suppose that we have a random sample $X_1, \dots, X_n$ from $X$.

The interest is in the hypothesis testing problem $$ H_0: \theta \in \Theta_0 \quad\text{vs.}\quad H_1: \theta \in \Theta_1,$$ where $\Theta_1 = \Theta \setminus \Theta_0$.

Now, according to my course notes, we should proceed as follows. We determine the ratio of the posterior probabilities

$$ \frac{P(\tilde\Theta \in \Theta_0 \mid X_1,\dots,X_n)}{P(\tilde\Theta \in \Theta_1 \mid X_1,\dots,X_n)} = \frac{P(\tilde\Theta \in \Theta_0) \int_{\Theta_0} \prod_{i=1}^n f(X_i;\theta) \pi(\theta) \,d\theta}{P(\tilde\Theta \in \Theta_1) \int_{\Theta_1} \prod_{i=1}^n f(X_i;\theta) \pi(\theta) \,d\theta}.$$

So we have $$ \frac{P(\tilde\Theta \in \Theta_0 \mid X_1,\dots,X_n)}{P(\tilde\Theta \in \Theta_1 \mid X_1,\dots,X_n)} = \frac{ \int_{\Theta_0} \prod_{i=1}^n f(X_i;\theta) \pi(\theta) \,d\theta}{ \int_{\Theta_1} \prod_{i=1}^n f(X_i;\theta) \pi(\theta) \,d\theta} \times \frac{\int_{\Theta_0} \pi(\theta)}{1 - \int_{\Theta_0} \pi(\theta)}.$$

However, we also have that the posterior density is given by $$f_{\tilde \Theta \mid X_1,\dots,X_n}(\theta\mid x_1,\dots,x_n) = \frac{\prod_{i=1}^n f(X_i;\theta) \pi(\theta)}{\int_{\Theta}\prod_{i=1}^n f(X_i;\theta) \pi(\theta)\,d\theta}.$$

So it seems to me that the factor $\frac{ \int_{\Theta_0} \prod_{i=1}^n f(X_i;\theta) \pi(\theta) \,d\theta}{ \int_{\Theta_1} \prod_{i=1}^n f(X_i;\theta) \pi(\theta) \,d\theta}$ is already the ratio of the posterior probabilities (since it is just the posterior distribution integrated over $\Theta_0$ and $\Theta_1$). Where is the mistake in my reasoning?


1 Answer 1


You’re right, and the course notes are wrong.

You can perhaps see this most clearly if you take both $f$ and $\pi$ constant. Then they both drop out of these equations, and the course notes are left with

$$ \frac{P(\tilde\Theta \in \Theta_0 \mid X_1,\dots,X_n)}{P(\tilde\Theta \in \Theta_1 \mid X_1,\dots,X_n)} = \frac{P(\tilde\Theta \in \Theta_0) \int_{\Theta_0} \mathrm d\theta}{P(\tilde\Theta \in \Theta_1) \int_{\Theta_1} \,\mathrm d\theta}. $$

But now the left-hand side is

$$ \frac{P(\tilde\Theta \in \Theta_0)}{P(\tilde\Theta \in \Theta_1)}\;, $$

whereas the right-hand side is

$$ \frac{P(\tilde\Theta \in \Theta_0)^2}{P(\tilde\Theta \in \Theta_1)^2}\;. $$

  • $\begingroup$ In an independent source from my course notes (onlinelibrary.wiley.com/doi/full/10.1111/spsr.12375), I also find the same formulas as in my course notes. So I actually still think they are probabily correct. $\endgroup$
    – Krup'a
    Commented Jan 16, 2020 at 9:27
  • $\begingroup$ Although that source stresses that in the integral one should have the priors of the models there, not of the prior of the parameter. Could it then be that the prior of the model is given by $p(\theta\mid \tilde\Theta \in \Theta_0) = \pi(\theta)/(\int_{\Theta_0}\pi(\theta)\,d\theta)$? $\endgroup$
    – Krup'a
    Commented Jan 16, 2020 at 9:53

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