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options("scipen" = 9999, "digits" = 4)Q1
Every time you and your friend get together to watch movies, you flip a coin to determine who gets to choose the movie. Your friend always picks heads, and every Friday for 10 weeks, the coin lands on heads. You develop a hypothesis that the coin has two heads sides, rather than both a heads side and a tails side. Set up a Bayes factor for the hypothesis that the coin is a trick coin over the hypothesis that the coin is fair. What does this ratio alone suggest about whether or not your friend is cheating you?
Let \(H_1\) be the hypothesis that our friend’s coin has heads on both sides and let \(H_2\) be the hypothesis that our friend’s coin is fair. If \(H_1\) were true, then the probability of every flip being heads would be \(1\) no matter how many flips we do. However, under \(H_2\), the probability of every coin being heads would be \[P(D \mid H_2) = 0.5^{10} = \frac{1}{1024}.\]
So our bayes factor is then \[B = \frac{1}{1 / 1024} = 1024.\]
If we believed that both scenarios were equally likely, we would have incredibly strong evidence that our friend were cheating.
Q2
Now imagine three cases: that your friend is a bit of a prankster, that your friend is honest most of the time but can occasionally be sneaky, and that your friend is very trustworthy. In each case, estimate some prior odds ratios for your hypothesis and compute the posterior odds.
For the first case, that our friend is a bit of a prankster, we’ll say our posterior odds of being tricked are somewhat high, maybe \(0.8\) – that is, out of every 5 times we interact with our friend, we expect one prank. The posterior odds would then be \[0.8 \times 1024 = 819.2,\] and we would still be quite sure that our friend was tricking us.
If our friend is honest most of the time, say we believe that our prior odds of being tricked are \(1/100\). Then our posterior odds would be \[0.01 \times 1024 = 10.24,\] so we should at least be amenable to the idea that our friend is tricking us, although the evidence is not quite conclusive.
Finally, if our friend is really trustworthy, let’s say our prior odds are \(1/10000\). Then we would get a posterior odds of \[0.0001 \times 1024 = 0.1024,\] and we would actually be somewhat confident that our friend is not cheating (note that this corresponds to a Bayes factor of about \(10\) for the relative odds of a fair coin).
Q3
Suppose you trust this friend deeply. Make the prior odds of them cheating \(1/10000\). How many times would the coin have to land on heads before you feel unsure about their innocence–say a posterior odds of 1?
We need to solve the equation \[1 \leq \frac{1}{10000}\frac{1}{0.5^x}\] for \(x\), which yields \[x \approx 14.\]
So we would need to see 14 heads in a row before we would entertain the notion that our friend is cheating.
Q4
Another friend of yours also hangs out with this same friend and, after only four weeks of the coin landing on heads, feels certain you’re both being cheated. This confidence implies a posterior odds of about 100. What value would you assign to this other friend’s prior belief that the first friend is a cheater?
We can solve for the prior odds in the equation \[100 = O(H_1)\times \frac{1}{0.5^4},\] which gives \[O(H_1) = 6.25.\]