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mindy
10-21-2008, 10:03 PM
Suppose that the number of light absorbing molecules or particles along the path of a ray of light entering a body of water is poisson. The concentration of particles in the water is .13 particles/m

the expected number of particles along the path of 10m = 1.3 particles

What fraction of surface light rays will make it past the 10m mark? in order to make it past 10m the light needs to hit no particles in the 10m

z=depth
I(z)= light intensity at depth z
I(o)= light intensity at surface= 100%

how do I start this?

zmogggggg
10-22-2008, 02:59 AM
Consider a poisson random variable X which represents the number of light absorbing particles along this ray of light (assumed infinitely long). You are given E[X] = .13 (p/m) = 1.3 (p/10m). One quality of the Poisson distribution is the intensity parameter is equivalent to the expected value of a random variable following this distribution.

In general, P(X = x; L) = e^(-L)*L^x/x! (here L is the intensity parameter). For this experiment, P(X = x) = e^(-1.3)*1.3^x/x!, for x >=0 and counting particles.

You are asked to find what fraction of surface rays make it past 10 m and informed that in order for this to occur, all light must reach 10 m. Let z be the depth of your water, I(z) the intensity of light at depth z (non-negative almost surely). If all light reaches 10 meters, I(10) = 100%, i.e. {X = 0} (no particles have absorbed light). We can look at all future absorption of light from this point equivalently to the absorption of light upon entry. That is for the next 10m, X describes the number of particles that will absorb light (since nothing in your post mentioned varying absorption rates, I am assuming they're fixed).

Here is where I get confused. Since all light reaches the 10m mark (we are assuming this), all light technically goes beyond the 10m mark, with absorption of light following again a Poisson distribution. With this interpretation, 100% of light passes the 10m mark, given that 100% of light reached the 10m mark. I doubt this is the answer, so you're left considering future absorption in terms of expectations. That is for 10 < z < 20, 1.3 particles absorb light (on average); for 20 < z < 30, 1.3 of the particles absorb the remaining light (on average), ... Since you aren't given a max number of particles that must be absorbed before the light ceases, this process continues onward. We can modify the problem somewhat to our benefit however: suppose it takes k particles to force I(z) = 0%. We know I(10) = 100%. We also know now that for z > 10, P(I(z) < i%) = P(X > i*k) = 1 - P(X <= i*k) = 1 - sum(P(x = i), for i = 0, ...,k), for i = 0, 1/k, 2/k, ..., k (certain percentages). This at least has a closed form as what is called the incomplete gamma function. You could use this for finite k, and also notice that as P(X > k) approaches 0 as k approaches infinity.

Sorry I can't really provide a clear solution, maybe someone else can fill us both in!