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I was walking in a forest one day and saw trees all around me. I begun wondering about how far do I see in the forest on average. I was also reminded to the "proof" that the age of the universe is finite, ie. that if universe was infinitely old, the light from every star everywhere would have reached us and no matter where we look in the sky, we would see a point of light, this is not true, thus universe is of finite age.

I know that the distribution of the number of tree "cores" or centers on a given area $N(A)$ is the Poisson distribution, ie. $P(N(A)=n) = (\lambda A)^n e^{-(\lambda A)}/n!$.

So, I asked myself, if I had one of those laser distance measuring things, pointed it to a random direction parallel to the ground and waited till the distance ($X$) was shown, what would be the distribution of $X$? Assume the trees are all of the same diameter $d$, the ground is flat and all trees are perfectly smooth. How would this be generalized to higher dimensions?

I then asked the follow up question: if there was a red rectangle placed perpendicular to my line of sight, how far would it have to be placed so that I would no longer be able to see it, ie. the fraction visible behind the trees would be small, ie. smaller than some small $\varepsilon$?

edit: The answer to the first question, attempt:

Lets take that I stand in an empty spot and fire the laser to a random direction. Assume that $x$ is the distance travelled, $\frac{d}{2} $ is the radius of the trees. (I would like to add a pic, but don't really know how).

Then we need to calculate the area where there can be no tree centers. Thus around me there is a circle with radius $d/2$ where there are no trees, the laser is surrounded by area the width of which is $d$ and finally at the end there is a semicircle of radius $\frac{d}{2} $. Thus we sum all these together (negative cause I stood in a clearing of radius $\frac{d}{2} $) $$A = -\frac{\pi}{2} \left( \frac{d}{2} \right)^2 + x d + \frac{\pi}{2}\left( \frac{d}{2} \right)^2 = x d$$

Then, using the Poisson distribution I gave above, we get the cumulative distribution function of $X$: $$1-P(X\ge x) = 1-e^{- \lambda d x}.$$ This is Weibull distribution with parameters $1$ and $1/(\lambda d)$. The expected value is thus $E(X)=\frac{1}{\lambda d}.$ This answers my first question. I think it would be pretty easy to extend to other dimensions. The last question is a bit too intimidating.

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By the way, the "proof" of a finite universe is known as Olbers' paradox. – Rahul Nov 18 '12 at 21:25
Well, thanks for the WP link, appreciate it :) Also, this might give some idea what I think when I walk in a forest :D – Valtteri Nov 18 '12 at 22:11
You might be interested in this question. The forest there has the trees on lattice points instead of random – Ross Millikan Nov 18 '12 at 23:25
@RossMillikan Yeah, that's a pretty interesting page you linked, thanks. – Valtteri Nov 18 '12 at 23:38
up vote 0 down vote accepted

OK, lets try to solve this myself.

The first question is already answered.

It is simple to extend it to general dimension $n$, it is, using the approach used in the first answer: let $V$ be the size of an $n$-dimensional cylinder whose bottom is an $(n-1)$-dimensional sphere of radius $R$, $W = \frac{\pi^{n/2}}{\Gamma(n/2+1)}R^n$, ie: $V=\frac{\pi^{n/2}}{\Gamma(n/2+1)}R^n x$. Then the cumulative density function of X, ie. $$1-P(X\ge x)=1-e^{\frac{\pi^{n/2}}{\Gamma(n/2+1)} \lambda R^n x}.$$

For the last question I will simplify it by assuming that the rectangle is curved so that the distance to every point of it is same: $D$. Then we add a new random variable $\Theta$ that represents the angle of the ray. Then the joint-probability distribution $$P(X\le x \ , \ \Theta=\theta) = \frac{1-e^{-\lambda d x}}{2 \pi}.$$ Now, if the angle in which the rectangle is seen is $\varphi$, the distance $x$ where the fraction that we see less than $\varepsilon$ of it is $$x > \frac{\ln(2 \pi)-\ln(\varphi \varepsilon)}{\lambda d}.$$ I was unable to figure how to do this is the rectangle is not curved.

Edit: might as well add a couple interesting applications:

If we assume that the average density of stars in the universe is $10^{-9}$ stars per cubic light year (estimate from And we also assume that all stars have the same radius as our Sun, about $0.7*10^{-7}$ ly. Then, in order to see with $0.999$ probability a star at a random direction, the radius of the observable universe would have to be about $5*10^{29}$ ly. This is over $10^{18}$ times greater than the estimates of the current size of the universe.

Another one:

Assume that we have small silt particles floating in water, the diameter of these particles being about $30\ \ \mu m$ or $3*10^{-5}\ \ m $ (from ). Assume that the visibility is so that we see no further than $3$ meters with probability $0.999$. Then the density of the particles in water would be about $2.5*10^{12}$ per cubic meter of water.

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