Charles Link said:
The Poisson process assigns a mean ## \lambda=Np ## to the process. In this case, we can select ## \lambda =15 ## if we want to work with an area of 1 km^2 and write the probability we find ## k ## diseased trees in that 1 km^2 area is ## P(k ) =e^{-\lambda} \lambda^k/k! ##.(The mean ## \bar{k} ## for this distribution is ## \bar{k}=\lambda ##). If we want to work with A=2 km^2, then ## \lambda ## is equal to 30. (## N ## is proportional to the selected area).## \\ ## If we want to work with an area such that we have a mean equal to 1 diseased tree, that means ## A_{single}=1/15 \, km^2 ## and ## \lambda =1 ## for this case. I believe a correct answer for the mean radius ## r_{single} ## to have one diseased tree would be ## A_{single}=\pi r_{single}^2 ##. There are probably more complicated ways of computing this that might be equally correct, but this is how I would solve it.
I think this is basically right. I had a half typed response when you posted, with a much more complicated response asking for a first moment and second moment, which leads to variance, then square root that to get standard deviation which should be an upper bound on the distance (which I presume is standard Euclidean) and then...
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But as you point out, it's a lot simpler. The goal is find expected distance till next tree -- call it ##r##. Area of the associated circle is ##\pi r^2##.
##k * AreaOfThatCircle =## Expected Trees In Our Big Square.
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Poisson processes
are memoryless, which makes them a lot easier to think about... memorylessness allows us to 'chop up' the circles, use fractional portions, and re-arrange the pieces up to any level of precision, such there is ultimately no 'dead space' left when we tile our square with them (or I suppose I should say that the 'dead space' gets arbitrarily close to zero).
This is quite different than when, say, you put golf balls in a box (in ##\mathbb R^3##) or look at the maximal number of balls on a billiards table (##\mathbb R^2##). Hence ##k## is the amount of circles that fit in that square, inclusive of our ability to chop them up however we want, and use less than whole portions. So it becomes a geometry problem of making sure the "surface areas" match, and if we know the first moment, i.e. expected number of trees in our square, we can solve for ##r##.
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For what it's worth, I think we can justify the above entirely with linearity of expectations and memorylessness of Poisson processes. However, I can't quite shake the feeling that using Wald's Equality makes it cleaner.
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edit:
one technical nit: the OP asks of distance from a point to a diseased tree. This could be interpreted as meaning the minimal distance, on average, from a point to a diseased tree (i.e. a nearest neighbor). It could also be interpreted in terms of averaging the distance to all diseased trees, from that point. I think the question is how far is the nearest neighbor on average, but strictly speaking the language is too loose to tell. (In a manner similar to the Inspection Paradox, the answer depends quite a bit on what you
exactly want to calculate.) There are even more subtleties than usual here since we're dealing in higher dimensions than 1-d.
