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    Re: That darned old cocked hat
    From: George Huxtable
    Date: 2010 Dec 11, 00:05 -0000

    I think John Karl and I may be at cross-purposes, obtaining different 
    answers because we are calculating different things, as George Brandenburg 
    has suggested.
    
    Let me explain to John Karl exactly what is the event for which I claim the 
    probability to be 75%.
    
    A vessel is at some position. It does not matter whether the exact position 
    is known to those aboard the vessel, or whether it isn't known. 
    Nevertheless, the vessel has a certain position.
    
    Three position lines are measured to three stars. There is no systematic 
    error, only a scatter which makes a towards error, with respect to the 
    vessel's position, just as likely as an  away error. It does not matter 
    whether the distribution is Gaussian or otherwise. Nor does it matter how 
    wide that distribution is. From arguments that are by now familiar, if we 
    construct a triangle from those three position lines, the probability that 
    the triangle will not embrace the position will be exactly 75%..
    
    And if we repeat the exercise, from that same position or from any other 
    position, the probability will be 75% each time.
    
    Does John agree?
    
    Next, could he please define, in similar terms, what exactly is the 
    situation for which he obtains a probability of 86%?
    
    George.
    
    contact George Huxtable, at  george{at}hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ----- Original Message ----- 
    From: "John Karl" 
    To: 
    Sent: Friday, December 10, 2010 9:55 PM
    Subject: [NavList] Re: That darned old cocked hat
    
    
    George writes:
    
    Take an anchored vessel at a precisely known location, say
    by GPS.
    
    Measuring the altitude of a star does not alter that location. The 
    probability of the observation being towards or away depends only on the 
    random scatter of the observation itself, being precisely 50:50. Instead of 
    actually measuring it, me might just as well simulate the observation by 
    tossing a coin. And we could simulate the observations of the other two 
    stars in exactly the same way.
    
    In which case the probability of the three observations being in the same 
    direction, TTT or AAA, is exactly the same as the probability of getting 
    three coins the same (all-heads or all-tails), which is 1 in 4.
    --------------------------------------
    
    Ah, Ha, that’s what I thought and suggested in my [14746] post.
    George is knowingly at the exact fix location acquiring many LOPs, and 
    plotting many cocked hats from them.  Thus he’s computing the in/out 
    statistics of the exact known position relative to the plotted cocked-hat 
    locations.  (This is very similar to simply measuring the random error 
    distribution of his LOP measuring gadget.)  In short, he’s doing the 
    opposite of what navigators do:
    
    He’s using his known position to measure the cocked-hat probability 
    distribution.  Navigators measure one cocked hat and estimate the MPP from 
    a priori knowledge of the gadget’s error distribution.
    
    These are very, very, different things.  And this highlights  the 
    significance of the Bayesian approach to estimation theory.  It may seem 
    subtle, but the difference is rational thinking -- which isn’t always easy 
    for many of us (including me).
    
    JK
    
    
    
    
    

       
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