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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
distribution,
 >

As George has said, I (and not he) am responsible for invoking that
simplification.

 >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
outside.
 >

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
being
 >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
close
 >to that X on the chart, because that is where the Probability Density is
 >highest.
 >
 >[snip]
 >
 >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.
 >
 >[snip]
 >
 >
 >For the LLL case or the RRR case, it seems that it is only possible to
have
 >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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