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options("scipen" = 9999, "digits" = 4)Recall that from Bayes’ theorem, \[P(H \mid D) \propto P(H) \times P(D\mid H).\] If we want to compare the relative likelihood of two events, we do not need the normalizing factor \(P(D)\), because it will cancel out, so we can use the ratio \[\frac{P(H_1) \times P(D\mid H_1)}{P(H_2) \times P(D\mid H_2)}.\] The second part, \[B = \frac{P(D\mid H_1)}{P(D\mid H_2)},\] is called the Bayes factor and measures the relative likelihood of two hypothesis if our prior belief in both is equal. The first part, \[O(H_1) = \frac{P(H_1)}{P(H_2)},\] is called the prior odds even though it’s a risk ratio (it is only really the “odds” if \(H_1\) and \(H_2\) are mutually exclusive which is rarely true in real life). When we combine the two, we get the posterior odds: \[\tilde{O}(H_1) = O(H_1) \times B.\]
Q1
Returning to the dice problem, assume that your friend made a mistake and suddenly realized that there were, in fact, two loaded dice and only one fair die. How does this change the prior and therefore the posterior odds for our problem? Are you more willing to believe that they die being rolled is the loaded die?
The likelihoods of the die rolls are still the same, so we don’t need to edit that. However, we need to update our prior possibilities to \[P(H_1) = 2/3; \text{ and } P(H_2) = 1/3,\] where \(H_1\) is the hypothesis that the die is loaded and \(H_2\) is the hypothesis that the die is fair.
When we recompute teh posterior odds, we get \[\frac{2/3}{1/3} \times 3.77 = 2 \times 3.77 = 7.54,\] so given the data and our prior knowledge, we now believe that the probability that the rolled die was loaded is \(7.5 \times\) higher than the probability that the rolled die was fair.
Q2
Returning to the rare diseases example, suppose you go to the doctor and after having your ears cleaned you notice that your symptoms persist. Even worse, you have a new symptom: vertigo. The doctor proposes another possible explanation, labyrinthitis, which is a viral infection of the inner ear in which 98 percent of cases involve vertigo. However, hearing loss and tinnitus are less common in this disease; hearing loss occurs only 30 percent of the time and tinnitus occurs only 28 percent of the time. Vertigo is also a possible symptom of vestibular schwannoma, but occurs in only 49 percent of cases. In the general population, 35 people per million contract labyrinthitis annually. What is the posterior odds when you compare the hypothesis that you have labyrinthitis against the hypothesis that you have vestibular schwannoma?
If you had vestibular schwannoma (\(H_\text{VS}\)), the probability of experiencing hearing loss, tinnitus, and vertigo would be \[P(\text{symptoms} \mid \text{VS}) = 0.94 \times 0.83 \times 0.49 = 0.38.\]
If you had labyrinthitis (\(H_\text{L}\)), the probability of experiencing all three symptoms simultaneously would be \[P(\text{symptoms} \mid \text{L}) = 0.30 \times 0.28 \times 0.98 = 0.08.\]
The bayes factor for the ratio of labyrinthitis to vestibular schwannoma is then \[B_{\text{L} / \text{VS}} = \frac{0.08}{0.38} = 0.21,\] indicating that vestibular schwannoma is about 5 times as likely given your currently symptoms, if both diseases were equally likely. Not good!
However, if we compute the prior odds of labyrinthitis relative to vestibular schwannoma, we get \[O(\text{L}) = \frac{35/1000000}{11/1000000} = 3.18.\]
Thus, the posterior odds of labyrinthitis relative to vestibular schwannoma is \[3.18 \times 0.21 = 0.67,\] so given our data and the prevalence of both conditions, so unfortunately it seems that vestibular schwannoma is a slightly better explanation than labyrinthitis, but the evidence is not extremely strong either way.