Exercise 3.10 - Taxicab problem

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Some similar entertaining problems are guessing the number of piano tuners from the average time for a tuner to arrive in one guest’s house, etc.

For question (a), we begin with hyperparameters $K = 0$ , $b = 0$ , which is improper since the Pareto distribution cannot normalize. With $𝒟 = {100}$ , we have the posterior distribution another Pareto distribution with $K = 1$ and $b = 100$ , i.e.,

p (𝜃 | 𝒟) = \frac{100}{𝜃^{2}} \cdot 𝕀 [𝜃 \geq 100] .

For question (b), we firstly derive the distribution of the taxi index:

\begin{aligned} p (x | 𝒟, K, b) & = \int_{0}^{\infty} p (x, 𝜃) d 𝜃 \\ = \int_{0}^{\infty} p (x | 𝜃) \cdot p (𝜃 | 𝒟, K, b) d 𝜃 \\ = \int_{100}^{\infty} 𝕀 [x \leq 𝜃] \cdot \frac{1}{𝜃} \cdot \frac{100}{𝜃^{2}} d 𝜃 \\ = \int_{\max (x, 100)}^{\infty} \frac{100}{𝜃^{3}} d 𝜃 \\ = 50 \cdot \max {(x, 100)}^{- 2}, \end{aligned}

whose plots look very much similar to that of electrical potential along an axis that penetrates the center of a conductor sphere with radius 100, through declines exponentially faster.

The posterior mode of $x$ is any number in $[0, 100]$ .

The posterior mean of $x$ is:

𝔼 (x) = \sum_{x = 0}^{100} \frac{x}{200} + \sum_{x = 100}^{\infty} \frac{50}{x},

whose second term diverges, so the posterior mean does not exist.

The posterior median is 99.5, since:

\sum_{x = 0}^{99} \frac{1}{200} < 0.5 < \sum_{x = 0}^{100} \frac{1}{200} .

Question (c) is identical to (b), as we have adopted a Bayesian treatment for (b).

For question (d), we have:

p (x = 100 | 𝒟, K, b) = \frac{1}{200},

p (x = 50 | 𝒟, K, b) = \frac{1}{200},

p (x = 150 | 𝒟, K, b) = \frac{1}{450} .

For question (e), we might adopt better $K$ and $b$ with expert knowledge and collect more samples.

solour_lfq

2021-03-24 13:42

Exercise 3.10 - Taxicab problem

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