NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Brad Morris
Date: 2013 Mar 17, 22:59 -0400
Hi Geoffrey
Now that the Monty Hall Problem is dispensed with, we can return to the original point of your post, Bayesian Decision Theory. I fondly remember that class. You wrote:
This statement of P(x,y) assumes that all three LOPs are independent and the that probability function for each LOP is symmetrical about each LOP. But consider: Suppose you make your first two sightings and you plot the resulting LOPs out on a chart. You will get the two LOPs crossing at some point, which will become your most probable position. Now, suppose you then take your third sighting and on plotting that LOP on the chart, you see that the most probable position from the first two sightings is to the left of the this third LOP. Is it still correct that the probability of the true position being to the left of the third LOP is exactly the same of it being to the right of that third LOP, as assumed in John Karl's probability function? Or should the probability distribution around the third LOP be informed by the previous data? This is the basis of what is called Bayesian statistics, which is the fasted growing area of statistical theory today
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Using the knowledge of the first two LOPs, in deciding which side of the third LOP the probability distribution should lie on, leads to an interesting conclusion. Essentially you imply that for each LOP, the probability distribution should lie towards the intersection of the other two. When all three LOPs are thus treated, we find that the most probable position is inside the cocked hat. Why? Because the bias for each LOP is now towards that intersection and the only place where all three biases agree is inside.
And yet I have heard many state, with utter certainty, that the most probable location is outside the cocked hat.
That does place you at odds with navigational dogma! Surely you wish to include consideration of the size of the cocked hat relative to 1, 2 & 3 standard deviations of the distribution about the LOP.
Brad
John Karl wrote in /f1-Cocked-Hat-V2.pdf:
================================
what is the probability that the ship is located outside of the cocked hat?
The probably per area is P(x,y) = (1/A) exp( –(d12 + d22 + d32 )/2 ó 2 )
where d1, d2, and d3 are the perpendicular distances to the three LOPs, and A is a normalizing factor (the
integral of P(x,y) over the entire 2D surface – over all infinity). So we see that the point of maximum
probability minimizes the sum squared distances from the three LOPs. And as I later learned from Herbert
Prinz, this is called the symmedian point.
===============================
This is the usually quoted starting point to obtain the probability density distribution for the cocked hat problem, but I am going to throw a spanner in the works here.
This statement of P(x,y) assumes that all three LOPs are independent and the that probability function for each LOP is symmetrical about each LOP. But consider: Suppose you make your first two sightings and you plot the resulting LOPs out on a chart. You will get the two LOPs crossing at some point, which will become your most probable position. Now, suppose you then take your third sighting and on plotting that LOP on the chart, you see that the most probable position from the first two sightings is to the left of the this third LOP. Is it still correct that the probability of the true position being to the left of the third LOP is exactly the same of it being to the right of that third LOP, as assumed in John Karl's probability function? Or should the probability distribution around the third LOP be informed by the previous data? This is the basis of what is called Bayesian statistics, which is the fasted growing area of statistical theory today.
To point up what I am suggesting, I will pose a little puzzle. There used to be a show on American TV back in the 1970's hosted by Monty Hall called, "Let's make a deal". The idea was that the contestant was confronted by three doors. Behind one door was a big prize, like an automobile, and the contestant was asked to choose a door and so win the prize. Having chosen a particular door, Monty Hall would then open one of the two doors not chosen by the contestant to show that it was empty. He would then ask the contestant if he (or she) would like to change his (or her) choice of door to the other of the two doors left.
The question then: Would you change your choice of door if you were a contestant? If so, why?
Geoffrey Kolbe
