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    Re: Still on LOP's
    From: Trevor Kenchington
    Date: 2002 Apr 20, 19:49 -0300

    Collection of responses to bits and pieces culled from the recent spate
    of messages:
    Rodney Myrvaagnes wrote:
     >Taking George's simplifying assumption of a rectangular probability
    As George has said, I (and not he) am responsible for invoking that
     >One possibility is that they are just wide enough to make all three
    observations possible. That results in the traditronal center point of
     >the triangle being the only non-zero probability.
     >Another possibility is to make the rectangles just cover all the other
    observations. If the cocked hat happens to be equilateral, the
     >non-zero area expands to fill the triangle, and is zero everywhere
    Not as I plot it out. I have a non-zero area covering a large
    equilateral triangle, comprising the cocked hat and three identical
    triangles, each abutting one side of the 'hat. In short, that model fits
    George's 25% probability that the true position is inside the 'hat.
     > As the triangle changes away from equilateral, a band of
     >non-zero appears outside the two long sides, but the inside of the
    triangle remains as strong.
    I plot that one as having the non-zero area in the form of an irregular
    hexagon, each side of the "probable band" around each LOP forming part
    of the boundary. I suspect that in all cases the non-zero area outside
    the 'hat is much more than 3 times the area inside the 'hat, hence there
    would be a much less than 25% probability of the true position falling
    inside the 'hat -- IF this particular model were realistic, which I
    don't think it is. The reason that the areas shift, relative to the
    proportions with an equilateral 'hat is that the 'hat itself becomes
    narrower relative to the width of the "probability band" around each
    LOP. It must get narrower, rather than longer, as the 'hat moves away
    from the equilateral shape because the apex of the 'hat has to remain
    within the "probability band" around the LOP which forms the base of the
    'hat. (That limitation does not, of course, apply with a more realistic
    probability distribution, such as the Gaussian.)
    If that is hard to understand, plot it out and I think it will make sense.
    Rodney also wrote, in response to a question of mine:
     >A probability function P{x} must be integrated over an interval to
    determine the probability of the instance falling within that interval.
     >The correct way to say it is "The probability of a position lying
    within (some distance) (delta x) of a value for x is (whatever). In the
     >cases we are talking about, measurements of LOPs, the probability of
    it lying within [delta x] of the MPP is higher than the probability
     >of it falling within that same (delta x) of any less-probable position.
     >The confusion in this discussion seems to come from comparing a
    probablity over a zero interval with probabilities over large areas.
     >It doesn't work that way.
    I don't think that George or I have been confused in that way at all. We
    have simply been defining the LOP as an infinitely-thin line dividing
    two the areas on either side of it. We have then been expressing the
    probabilities of the true position lying within an infinite distance on
    one or the other side of such a line (or in some sector bounded by two
    or more LOPs, as the case may be). Of course, a full quantitative answer
    needs to replace the summation of probabilities over distances of zero
    to infinity with summations over various distances. Unfortunately,
    nobody who has yet joined the discussion seems both willing and able to
    supply such a quantitative solution.
    And still with Rodney:
     >Williams says-
     >>"Fig 14.6 is the cocked hat formed by three position lines. If no
     >>information is available on the distribution of error the probability of
     >>being in each of the seven zones defined by the cocked hat is as shown in
     >>the figure. The proof is so simple that an outline of it is given in note
     >If a mariner takes a series of LOPs, and they come somewhere near to
     >defining a position, he has a lot of information about the error
     >distributions. The assumption of no information at all implies equal
     >probability anywhere in the world.
    I really doubt that that was what Williams meant. I suspect he was
    taking it for granted that there was some sort of probability density
    distribution around each LOP that, in very rough terms, approximated to
    Gaussian but that he could prove his point without assuming that the
    distribution was Gaussian, let alone having to specify the standard
    deviation. (He did, however, make an implicit assumption that the
    density distribution was symmetrical around each LOP.)
    Martin Gardner wrote:
     >The scariest suggestion I have heard, and I can't recall who said it, was
     >that perhaps there is no single number that defines our likelihood of
     >inside the hat in all cases.
    That was me and I am as certain of it as of anything in this debate.
    Consider the results of Navigator X who, after careful study of his
    observations including making many replicate observations in careful
    tests, has concluded that, under reasonable observing conditions, his
    LOPs are usually close to being right but are subject to random errors
    with a probability distribution that approximates a Gaussian
    distribution with standard deviation S. [This guy is a paragon amongst
    navigators but many another will have the same sort of consistency. They
    just won't be able to tell you what their value of S is.] So one day, X
    takes three observations which he is reasonably happy with. He is
    confident that they are not unusually precise nor unusually erroneous.
    Yet, when he plots them, the three LOPs pass through the same pencil dot.
    Should X believe, as many do, that his three observations each chanced
    to be so accurate that he has a 25% (or whatever) chance of being in the
    area of that dot? Or should be conclude, I think correctly, that his
    three observations chanced to produce LOPs which all passed through the
    same point -- within his high-probability ellipse but not necessarily at
    his true position? If the latter, then it is impossible for the tiny
    area of intersection of the three LOPs to encompass the same probability
    of occurrence of the true position as the cocked hat from less lucky
    observations would have done.
    Or consider Navigator Y, who is a lot less competent than X, and ends up
    with a huge cocked hat covering half of whatever sea he is sailing in.
    Are we to suppose that there is still a 75% (or whatever) chance that he
    is outside that area?
    Finally, Geoffrey Kolbe wrote:
     >There seems to a fair amount of confusion about statistics in general
    Our problem exactly!
     >The point where the Probability Density is highest of all, is
     >my X on the chart. As a betting man, I am prepared to put my money where
     >the odds are shortest and say that it is most likely that I am pretty
     >to that X on the chart, because that is where the Probability Density is
     >If we now think about the centre of the cocked hat being the point of
     >highest Probability Density, rather than a Most Probable Position, we can
     >escape from the intuitive problems to which this approach gives rise.
    That is clearly true and was a point I tried to make at the end of my
    last contribution to this thread -- with the caveat that prudent seamen
    (not trusting their vessels to the long odds that appeal to a good
    betting man) will also bear in mind the not-insignificant probability
    that they are some way away from the X that they mark on the chart. Sods
    Law being what it is, they will particularly not forget that there is
    some probability that they are in a much more dangerous position than
    the one marked by the X.
     >George seemed worried about the independence of the measurements of three
     >LOP's. One measurement is independent of another if one measurement does
     >not influence the other in any way.
    I am sorry but that is not the definition of independence. If I take
    measurements of two variables, each of which is correlated to a third,
    then those two variable are NOT independent, even if neither of the
    things that I am measuring has any influence over the other.
    The case of the multiple LOPs that we are considering is not as simple a
    matter as correlation. However, the fundamental principle remains. The
    suggestion that I and others have made of non-independence is not a
    matter of one observation "knowing" the error in another but of the
    error in each LOP being a matter of the position of _that_ LOP relative
    to the _same_ true position as we use to consider the error in each
    other LOP.
    I return to a limiting case that I have invoked before and which at
    least one contributor has, in my view, misinterpreted: Our incompetent
    Navigator Y takes bearings from two distant objects that are very close
    to one another. Because of the inevitable random errors, he could find
    that the LOPs either run parallel or diverge, in which case there would
    be no intersection and no fix. But let us concentrate on those cases
    (about half of them) in which the LOPs cut, though at a very narrow
    angle. I hope that we can all agree that where they cut gives Y little
    or no information on how far off the two sighted objects he is but also
    that the two LOPs (which approximate to two sighting on the same object)
    do give him a fair idea of his bearing from either object. When he plots
    the first of those LOPs, I trust we also agree that there is a 50%
    chance that his true position is to the left of the line (as seen from
    seaward) or rather, in deference to Rodney's point, that there is a 50%
    chance that his true position lies to the left of the infinitely-thin
    centreline of the pencil mark.
    Let us further limit the discussion and consider only those 50% of cases
    in which the true position does lie to the left of that line. If the
    errors in Y's two lines are independent, then we must conclude that,
    within these cases (25% of all observations, 50% of those in which the
    two LOPs cross) there is a 50/50 chance that the true position lies to
    the right of the second LOP, despite the two LOPs lying very, very close
    together and the true position lying to the left of the first one. That
    can only be so if we are prepared to accept that there is a 50% chance
    that Y is actually in one of the thin sectors delimited by the two LOPs
    -- indeed in only one, not either, of them (since one lies left of the
    first LOP and right of the second, whereas the other is vice versa). To
    extend that logic a bit: We would then have to conclude that if a
    navigator took a single bearing with a compass and concluded that he
    bore 090 from some lighthouse with a 50% chance that his actual bearing
    from it was between 085 and 095 (on a bad day for sights!), he could
    take a second (equally precise) bearing and if it indicated that he bore
    091 from the lighthouse, he could now conclude that there was a 50%
    chance that he actually bore somewhere between 090 and 091. That
    conclusion would be simply WRONG. Combining the two observations would
    certainly narrow the 50% confidence interval but not by that much.
    Besides, his second observation could have been 092 or 090.1. If the
    previous conclusion were right, we would have to accept that the width
    of the 50% confidence interval was directly related to the difference
    between the first and second bearings, which makes no sense at all.
    Thus, I find it very difficult not to reject the idea that the LOPs are
    independent. And if they are not, all of the arguments yet advanced that
    lead to conclusions about true positions falling within cocked hats only
    in 25% or 14.3% or whatever of cases are simply invalid. However, I
    would stress that I am still putting an "if" into that conclusion,
    pending comment from a statistician or counter-argument from anyone who
    can find fault with my logic.
    To return to Geoffrey's posting:
     >I am worried that Georges' analysis of the cocked hat problem has an
     >assumption or two too many.
     >For the LLL case or the RRR case, it seems that it is only possible to
     >any area of the drawing which is shaded three times, when the three
     >bearings or azimuths of the three positions are occupy less than 180
     >degrees. Thus your three objects could be at bearings of 25 degrees 78
     >degrees and 198 degrees and you can have a common shaded area. But if the
     >third object was at 208 degrees, there would be no common shaded area.
    Try it.
    I have and I get an RRR area, with "right" and "left" as seen from
    seaward. It is the cocked hat. with a (roughly) northwest corner at the
    intersection of the 025 and 078 bearings, northeast corner at the
    intersection of the 078 and 208 bearings, and a southwest corner where
    the 025 and 208 bearings cut. If the 208 bearing chanced to lie
    northwest of the intersection of the other two, there would be no RRR
    area but their would be an LLL, which would again be the cocked hat.
    As a couple of us have noted, there cannot be both an LLL and an RRR
    area with the same cocked hat. More to the point, there would be a lot
    less confusion from resorting to the "+" and "-" designations, in which
    case any cocked hat with three LOPs is always "+++" and no area can have
    three minus rankings.
     >But here is another way of looking at the problem.
     >Start out with three lines crossing at a common point. This is the zero
     >error scenario and the crossing point represents your actual position.
    That is a false deduction: Toss a coin three times -- just because it
    comes up heads every time does not not mean that each individual toss
    did not have a 50% chance of coming up tails. If a thousand tosses all
    produced heads, you could conclude that you probably (though not quite
    certainly) had a two-headed coin. In the same way, if every one of your
    cocked hats fits within a pencil point, you can conclude that your LOPs
    have very high precision. But if a single one does, all it indicates is
    that each LOP chanced to pass through the point of intersection of the
    other two. It does not indicate that that is also your true position.
     > Now
     >draw out the 8 cases that George outlines, where position lines are offset
     >to one side or the other of the lines by a certain amount, or the angle is
     >changed by a certain amount clockwise or anticlockwise.
    Only 7 of which 8 cases can exist with any particular set of three LOPs.
     > In only two cases
     >will the cocked hat actually enclose the position.
    Only one case with any particular set of three LOPs.
     > If the bearings or
     >azimuth of the three positions occupies less than 180 degrees, the two
     >cases will not be the LLL and RRR cases!
    That is only a matter of labelling, which amounts to deciding, in the
    case of a compass bearing, whether to look from seaward or landward when
    saying which side is "left". In the case of a celestial LOP, the two
    directions are symmetrical and so you can call "left" "right" and "right"
    "left" as you please. Hence, it is better to use the +/- (or in you
    prefer "in"/"out") designations, in which case the cocked hat will
    always be +++ or "in in in".
     > Thus the cocked hat only encloses
     >the actual position 25% of the time. QED, the probability of being within
     >the cocked hat is 0.25.
    That is only a re-statement of George's proof. I don't find it any more
    satisfying than it was the first time around.
    Trevor Kenchington

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