Suppose that I have three dice. The first die is an ordinary six-sided die, the second die is eight-sided, with each of the numbers 1 through 8 equally likely. The third die is six-sided, but it only has the numbers 1 through 5, the space where the six should be is blank (counts as zero). Let the parameter \(\theta\) denote which of the three dice I roll. I will choose one of the dice and roll it, and I'll tell you the number it lands on, but not which of the three dice it was. The next three questions are all about this situation.
Suppose that I pick a die and roll it. If I get a 3, what is the likelihood function for \(\theta\)?
Compute the posterior distribution based on the uniform prior \(\pi(\theta) = \frac{1}{3}\) assuming that the die I picked landed on a 3.
What would the posterior distribution be if the die I picked landed on a 6?
Suppose you toss a slightly bent coin and put a Uniform\(([0.4,0.6])\) prior on \(p\), the probability of getting a head on a single toss.
If you toss the coin \(n\) times and obtain \(n\) heads, then determine the posterior density of \(p\). You'll have to calculate an integral to find the right constant for the posterior distribution.
Suppose the true value of \(p\) for our coin is actually \(0.9\). Notice that no matter what happens, the posterior distribution will never have any weight around 0.9. What could you do when you choose your prior to avoid this problem?
Suppose that 0.1% of the population has a disease called brittle bones disorder. Everyone else breaks about 0.5 bones per decade on average, and the number of broken bones per decade for these healthy people follows a \(\operatorname{Pois}(0.5)\) distribution. For the people with brittle bones disorder, the average number of broken bones is 4 per decade, so you can model distribution of bone breaks with a \(\operatorname{Pois}(4)\) distribution. If you randomly select one person, and then find out that they have broken 3 bones in the last decade, then what is your posterior probability that they have brittle bone disorder?
The mean of a \(\operatorname{Gamma}(\alpha, \lambda)\) distribution is \(\alpha/\lambda\) and the variance is \(\alpha/\lambda^2\). How would you choose \(\alpha\) and \(\lambda\) to get a Gamma distribution with mean 900 and variance 400?
German Tank Problem During World War II, the allies were able to estimate how many tanks the Germans had manufactured by looking at the serial numbers of captured tanks. You can model this situation by assuming that the serial number of each captured tank is uniformly distributed between 1 and \(N\), where \(N\) is the total number of tanks the Germans have made. Although this is a discrete problem, it is actually a little easier to solve if you pretend that \(N\) doesn't have to be an integer. Then you can use a \(\operatorname{Gamma}(\alpha, \lambda)\) prior for \(N\). A good choice for the values of \(\alpha\) and \(\lambda\) might be \(\alpha = 9\) and \(\lambda = 1/100\), so that our prior estimate is that the Germans have around 900 tanks with a variance of 90,000 (which is a standard deviation of 300). If the allies have captured 20 tanks and the largest serial number is 700, then what is the posterior probability that the Germans have less than 800 tanks total? You'll need to set up two integrals to solve this, and I recommend using WolframAlpha to calculate the integrals.