Poisson distribution regarding expected distance

[Nick Jarvis]

Hi

The question is about diseased trees in an area (Poisson process), and states that λ = 15 diseased trees in a km square. I need to calculate expected distance from a point in the square to a diseased tree.

Now I thought that this means that P(diseased tree = 0) ~ Po(15) = 3.059 x 10^-7

Or do I calculate this as 1/15 which is P(distance = 0) ~ Po(1/15) = 0.936, then (1 - 0.936) gives me the expected distance??

This is a question that I know should be so straightforward.

Cheers

 

[Charles Link]

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.

 

[StoneTemplePython]

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.

 

[Nick Jarvis]

Apologies for the lateness. Many thanks for your replies. May need to speak to tutor as we have not covered area at all!

Cheers

 

[WWGD]

Apologies for the lateness. Many thanks for your replies. May need to speak to tutor as we have not covered area at all!

Cheers

No problemif you missed less than 1/15th around a tree, not likely any trees in there $Km^2$ ;).