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I consulted with a noted statistician friend, Roger Pinkham, because
this is a complex problem and making guesses didn't appeal to me.

He is thinking about it, but immediately made a few points that will
shed some light.

1. You can't assume the errors are normally distributed. There will be
numerous instances where the error will be asymmetrical, including for
example some consistent instrument error that you don't catch, and
observations of phenomena where your only possible error is one-sided (I
don't think that second one is too relevant for navigation). But
assuming that you are spot on except for random, normally distributed
errors in each reading.... You further have to assume equal errors on
all readings for the moment.

2. Making a statement like p(inside the triangle) = 0.25 assumes that
each reading is independent. They are not. Going outside the triangle
can put you further out on the tail on each of your readings, so the
probability of being outside the triangle is much lower than that.


So, while he is thinking about the problem, in the meantime, he mentions
a related unsolved problem just to show how "unsimple" this is:

Consider a two-variable bivariate gaussian, (shooting at a target). For
n shots, what is the distribution of the size of the smallest circle
that can enclose all the points?
This is an unsolved problem. It isn't trivial, and he therefore doubts
that the navigation problem is any easier to come up with an analytical
answer.

My own take on this problem:

Estimating the error: You can't talk about the probability of being
outside the triangle without knowing the variance of your measurements.
I think the probability of being outside the triangle has nothing to do
with the size of the triangle, but has everything to do with the
consistency of your readings.
The reason a small triangle is an indication that you've hit the
jackpot, is that the three points will only coincide if a) they just
happen to do so (a small probability), or b) they really are correct.
So what you really want are a couple of more readings of the same bodies
to give you some indication of how consistent your first readings were.

I am short on time at the moment, but it seems to me the best way to
calculate the probabilities of being in the triangle are:

1. select three points and the observer's location
2. compute the bearing to each point
3. add normally distributed errors to each bearing (as George says, this
one is easy)
4. Determine if the position is within the resulting triangle
repeat step 3-4 as long as you want.

alternatively, just  walk through the range of possible values for each
observation, ie do the triple integration, but then you have to factor
in the probability of each reading.

This procedure, however, only gives the distribution for one set, and
the problem is that different configurations appear from those diagrams
to yield radically different answers.
Still, by performing this computation on different configurations, you
can at least see the range of probabilites. I suspect that the
probability of being within the triangle is at least 0.5.
That's just a guestimate, but I am looking forward to the first
simulation to see if that's right.
I will also guess that the probability of being within the triangle goes
up at least quadratically if the triangle is increased in size by
projecting lines parallel to the sides outside it. So if p(outside) =
0.5, with a similar triangle containing the original one that is bigger
by 100%, I will guess that the probability of being outside the
augmented triangle goes down to  0.125

Dov

   

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