Consider the probability, $\theta$, that a randomly selected person in NYC is from Manhattan. We collect data as we walk down Times Square. We ask three people whether they are from Manhattan, and all three respond with yes.

Assuming a flat prior for $\theta$,

a) Find the posterior distribution of $\theta$, conditional on observed data. Also find the mean and mode of $\theta$, while also deriving an expression for median of $\theta$ as well.

b) Find the predictive distribution that a randomly selected fourth person is also from Manhattan.

By flat prior does it mean that in this case, since it's a yes or no question, I am to assume that the probability of a yes on any given answer is 0.5? And if this is the case, how would the normal situation, where we have posterior $\propto$ likelihood $\times$ prior apply in this case?

Would really appreciate detailed help for the other parts as well. Thanks in advance!


I doubt that sampling in Times Square will give unbiased information about New York City as a whole, but ignoring that:

  • A flat prior is one with a constant density, here on the interval $[0,1]$, so your prior is $\pi_0(\theta)=1$ for $0 \le \theta \le 1$

  • The likelihood here is proportional to the conditional probability that all three would say yes, which is $\theta^3$

  • So the posterior distribution is proportional to the product of these, i.e. also proportional to $\theta^3$ in $[0,1]$

I will leave it to you to find the constant of proportionality, mean, mode, median etc.

  • $\begingroup$ I will find the mean, median and mode, but would it be possible give me some direction for the constant of proportionality? $\endgroup$ – user603218 Oct 24 '18 at 7:35
  • $\begingroup$ @MatthijsBjornlund You want the posterior density to give probabilities, so if it is $k \theta^3$ on $[0,1]$ then you want $\int\limits_0^1 k \theta^3\, d\theta = 1$ $\endgroup$ – Henry Oct 24 '18 at 7:40
  • $\begingroup$ Thus, $k=4$. That really simplifies things. However, while the mean and median are easy enough, what would the median look like for this case? Since it is the "mid-point" of the distribution, would it just be $$\int_0^{0.5) 4\theta^3 = 0.5^4 = 0.0625?$$ That seems smaller than it should be. $\endgroup$ – user603218 Oct 24 '18 at 8:10
  • $\begingroup$ @MatthijsBjornlund The median would be the value of $m$ for which $\int\limits_0^m 4 \theta^3\, d\theta = \frac12$ and also $\int\limits_m^1 4 \theta^3\, d\theta = \frac12$ $\endgroup$ – Henry Oct 24 '18 at 9:07

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