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I learned that the posterior odds is the ratio of the two posterior probabilities of hypothesis:

\begin{align} PO[H_1:H_2] &= \frac{P(H_1|\text{data})}{P(H_2|\text{data})} \\ &= \frac{P(\text{data}|H_1)}{P(\text{data}|H_2)} * \frac{P(H_1)}{P(H_2)} \\ &= \text{Bayes factor} * \text{prior odds} \end{align}

And $$\text{BF}[H_1:H_2]=\frac{P(\text{data}|H_1)}{P(\text{data}|H_2)}$$

The answer reads that because of the above equation it does not depend on any prior probability of $H_1$ or $H_2$.

But in my understanding, it seems that if we set $H_1$ and $H_2$ differently(we can set the two priors as any distribution pairs) the Bayes factor would also vary, hence making the Bayes factor sensitive to the choice of prior. I cannot get my head around why Bayes factor is insensitive to prior?

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2 Answers 2

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It is easier to check the impact of the prior when using Harold Jeffreys' decomposition of the prior measure: $$\pi(\theta) = \varpi_1 \pi_1(\theta)\mathbb I_{H_1}(\theta) + \varpi_2 \pi_2(\theta)\mathbb I_{H_2}(\theta)$$ (assimilating the hypotheses to subsets of the parameter space). In that case, the Bayes factor writes $$\mathfrak B_{12}(x) = \dfrac{\int_{H_1} \pi_1(\theta) f_1(x|\theta)\,\text d\theta}{\int_{H_2} \pi_2(\theta) f_2(x|\theta)\,\text d\theta}$$ which shows that

  1. the prior weights $\varpi_1$ and $\varpi_2$ have no influence on $\mathfrak B_{12}(x)$
  2. the submodel priors $\pi_1$ and $\pi_2$ influence the value of $\mathfrak B_{12}(x)$
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In your query we see that:

$$\text{Bayes factor} = \frac{P(\text{data}|H_1)}{P(\text{data}|H_2)}$$

Examining this expression we see that it does not contain the prior distribution and hence can't be sensitive to it. The posterior odds are, however, a different matter.

To the extent that value of the Bayes factor is conditional on $H_1$ and $H_2$, yes, it varies. However in no sense do $H_1$ and $H_2$ have a distribution there, it's just a deterministic function of two arguments. It's the priors which state at what rate particular values of $H_1$ and $H_2$ get fed into that function.

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    $\begingroup$ You're effectively asking why a function e.g. $f(x)=3x$ not depend on a second distinct function $g(x)$. The answer is not deep. $\endgroup$ Commented Jan 28, 2021 at 15:10

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