Calculating the Probabilities Subjectively - Bayesian I ran across this problem when I was trying to review my probability. It seemed easy enough, but I'm having a difficult time understanding the logic behind solving it.


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*The country of Chile is divided administratively into 15 regions. The size of the country is 756,096 square kilometers. How big do you think the region of Atacama is? Let A1 be the event that Atacama is less than 10,000 square kilometers. Let A2 be the event that Atacama is between 10,000 and 50,000 square kilometers. Let A3 be the event that Atacama is between 50,000 and 100,000 square kilometers. Let A4 be the event that Atacama is more than 100,000 square kilometers. Assign probabilities to A1,...,A4.

*Atacama is the fourth largest of 15 regions. Using this information, revise your probabilities.

*The smallest region is the capital region, Santiago Metropolitan, which has an area of 15,403 square kilometers. Using this information, revise your probabilities.

*The third largest region is Aysen del General Carlos Ibanez del Campo, which has an area of 108,494 square kilometers. Using this information, revise your probabilities.


Any help would be great, thanks!
edit: Sorry I didn't mention this, the problem said that I just needed to determine the probabilities subjectively.
 A: Here's my take on it. Might be overly naive, but anyway:


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*An "unbiased" guess would be to take the entire area of Chile and divide by 15. Any other choice would require us to justify why we think it should be bigger or smaller than the other regions. As for assigning probabilities for the intervals: Without any extra information and any extra guesses, I'd naively assume a linear cumulative probability function: I.e. the probability that the area of Atacama is less than A will be $F(A)$ where $F(0) = 0$ and $F(size-of-Chile) = 1$. Or in other words, assume a uniform probability density function. Interestingly, that leads to an answer that is independent of the number of regions. Because it's overly naive. 

*Next up, with my naive assumption, I don't actually think that adds new information but I could be wrong. 

*In that case, first remove that smallest region from Chile's total area, then use a cdf that is 0 for any area below that small region's area (because it's the smallest).

*Finally, in that case, take out the region's area from the remaining area (Chile minus smallest). But this time we don't gather additional information from the placement, because the smaller and larger regions can be (taken together) anything. 


So there goes. I feel like with Bayesian magic, with which I'm not overly familiar, one could do much more interesting things.
A: Here's how I might think about it, assuming a standard distribution of areas:


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*Divide the total area by 15 to get about 50,000 square kilometers. Since we have no reason to believe that Atacama is large or small, it might be close to this number. This next part is subjective, as the problem states: maybe A2 and A3 are more likely than A1 and A4, so I arbitrarily assign them 0.70 probability together. Then I have A1 = A4 = 0.15, A2 = A3 = 0.35. It's just a guess, right? But since the mean of the 15 regions is 50,000, it stands to reason that being close to 50,000 is more likely than being further away. Likely, we could tidy this up more with standard deviations and z-scores if we wanted to, but the problem doesn't require it.

*Since Atacama is 4th largest, I intuitively think that it is unlikely to be in A4. If it was, then more than 400,000 square kilometers would be taken up by the four largest regions, which only leaves 300,000 for the 11 smaller regions. This isn't impossible, but assuming a standard distribution, it's not likely. Likewise, since it's fourth largest, it probably isn't in A1, since that would mean that the three largest took up 650,000 square kilometers between them. I'm more inclined to believe that Atacama is in A2 or A3, so I adjust my probabilities to A1 = A4 = 0.05, A2 = A3 = 0.45.

*I'm assuming that this compounds with #2. In that case, the probability of A1 = 0. The probability I previously gave to A1 I might instead give to A3, since I think A2 is less likely, as the smallest region is well within A2. Thus A1 = 0, A2 = 0.40, A3 = 0.55, A4 = 0.05.

*Again, I'm assuming compounding with #2 and #3. Since my A4 probability is already low, I'm not going to revise my estimates. It might be argued that the probability for A2 should go down further than it already is, but it's just as possible that the two largest regions account for a very large percentage of Chili, so I will leave them as before.
These are just best guesses, as stated. Hopefully my thinking behind them is clear, and offers you what you're hoping for (the logic behind it). Have fun!
