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There seems to a fair amount of confusion about statistics in general, and
in particular, George's contention that the probability of being within the
"cocked hat" is only 25%.

With some trepidation, I will give a pot a brief stir which may muddy
things, but will, I hope, make a few points clear.

First of all, I would like to introduce the concept of "Probability
Density". Probability Density (in this case) is simply the probability that
you are inside a particular unit area. That unit area can be anything you
like, but let us be more specific and talk about the Probability Density as
being the probability of being within any particular square mile on a chart.

Suppose I am in a squall off the South coast of England. Through the rain,
I catch sight of the Needles lighthouse and Hurst Castle and manage to take
bearings on both these landmarks. These two bearings cross just South East
of Christchurch. I mark my Most Probable Position on the chart and wonder
if it is wise to head North of the Isle if Wight through the Solent. How
sure am I of my position?

There will be some error in the bearings I have taken on the two landmarks,
so I know that the X where the two bearings cross on the chart is just an
approximation of where I actually am.

What I can say is that the Probability Density (probability that I am
within any given square mile on the chart)  is highest at the point where
the two bearings cross on the chart. As I go further from the point where
the two bearings cross, the Probability Density will get smaller and
smaller. As the distance from my X on the chart increases, the Probability
Density will fall off in the manner of a bell curve, (or normal, or
Gaussian distribution).

If I add up all the Probability Densities of all the square miles on the
surface of the Earth, the total sum will be (very, very close to) one. This
proves that I am (almost certainly somewhere) on the planet Earth!

So there is a chance that I am within a certain square mile of downtown Des
Moines, Iowa! But the Probability Density for that area will be very small.

The Probability Density is much higher for the area around the English
Channel, so the chances are good that I am somewhere around here rather
than in Iowa. 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.

I think it is a misnomer to call the point X on the chart my "Most Probable
Position" and I think that this is where a lot of trouble arises. This
statement can be re-written as "the probability is highest that my position
is at X".

But you cannot talk about the probability of a point! To find the
probability that you are within a certain area, you must multiply that area
by the Probability Density of that area. A point has no area, so the
probability that you are at a certain point on the chart will always be
zero! Too, lines have no width, so they have no area, to the probability
that you are on a particular LOP will always be zero!

What I can say is that the Probability Density of my position is highest at
the point X, so it is most likely that I am pretty close to here.

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.

            -----------------------------------

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. Judged by this criteria, there seems
little doubt that measurements of altitudes for LOP's or taking bearings
using a compass are independent. Perhaps George would like to share the
cause of his doubts.


             ------------------------------------

I am worried that Georges' analysis of the cocked hat problem has an
assumption or two too many.

Take George's diagram of three lines crossing at 120 degrees to each other
and meeting at a common point. These represent three perfect LOP's or
bearings (no error) taken from that point.

Suppose now we introduce error in the manner which George has done. Suppose
we take the LLR case. We shade everything to the left of one line,
everything to the left of the next line, and everything the right of the
third line. In this instance, our position will be in that region which has
been shaded three times and will lie outside the cocked hat.

For the LRR case, we once again shade those areas to the left of one line,
the right of the next line and the right of the third line. A different
area of the drawing is now shaded three times. We are somewhere is this area.

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.

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. 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. In only two cases
will the cocked hat actually enclose the position. 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! 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.

Yours aye,

Geoffrey Kolbe.














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