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Problem: Recall the probability distribution of a Poisson random variable X:

$P(X=x)=\frac{\lambda^x{e^{-\lambda}}}{x!}$

where $\lambda$ is the rate parameter that equals the expected value of $X$.

You are a professor and assign your TAs to type up a very important homework assignment. You have three TAs: David makes an average of one typo per page, Amy makes an average of two typos per page, and Joe makes an average of three typos per page. A one-page typed homework assignment is turned into your box that has ten typos! Assuming that typos follow a Poisson distribution and you have no prior knowledge about which TA typed the assignment, what is the posterior probability that the TA who typed the homework assignment was Joe?

Hints:

  • The likelihood of the data given the grader is the probability of getting ten typos, given that the data follow a Poisson distribution with the typo averages of the graders as the rate parameter.

  • Use Bayes’ rule to compare multiple hypotheses about a discrete random variable

The answer is one of the following:

  • 0.334
  • 0.547
  • 0.866
  • 0.954

How I tried to solve the problem:

Likelihood = P(10 typos|Joe) = $\frac{3^{10}e^{-3}}{10!} = 0.00081$

Prior (each of the TAs can be considered to be equally likely) = P(Joe) = $0.5$

Data = $P(10 Typos)$ = 0.5($\frac{3^{10}e^{-3}}{10!} + \frac{2^{10}e^{-2}}{10!} + \frac{1^{10}e^{-1}}{10!})$ = 0.000848

Posterior = P(Joe|10 Typos) = $\frac{0.00081*0.5}{0.000848}$ = 0.4775

This does not equal any of the options. Insight on where my reasoning is incorrect would be much appreciated.

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    There are three TAs so P(Joe) = 1/3, although I don't think fixing this will change the final answer – 2017-02-17
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    @spaceisdarkgreen: I tried it with P(each TA) = 1/3, and I get P(Joe|10 Typos) = 0.5. This is incorrect too, as you've pointed out. Do you have any other suggestions? I'll edit my question. – 2017-02-17
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    You forgot the factor of $.5$ (which as I said above should actually be 1/3) when you calculated P(10 typos). As a result you got an answer too small by a factor of 2. – 2017-02-17
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    Do you agree the answer should be $$ \frac{\frac{3^{10}e^{-3}}{10!}(1/3)}{(1/3)(\frac{1^{10}e^{-1}}{10!}+\frac{2^{10}e^{-2}}{10!}+\frac{3^{10}e^{-3}}{10!})}$$? That's something around 95% – 2017-02-17
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    Thank you so much @spaceisdarkgreen. I was making some silly calculation errors. I had the formula right after you pointed out that the prior is 1/3 but I wasn't calculating properly. I feel a bit silly now. – 2017-02-17

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