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Olbers' paradox - Poisson model

  1. Jul 8, 2011 #1
    Olbers' paradox states that if the universe is infinite, static and homogeneous then why is the night sky dark. Of course it's been resolved but it brings up an interesting probability question:

    If we model the universe with a spatial Poisson model (probability that a small element is occupied is proportional to the volume) and ignoring decay, variations in star brightness and relativistic effects etc we get

    [tex]L \propto \int_r^R \tfrac{1}{s^2}dN(s)[/tex]

    as the amount of light reaching your eye originating from stars between r and R units of distance away, where N(s) is a Poisson process with rate at time s proportional to s^2. Does L go to infinity as R increases?
  2. jcsd
  3. Jul 9, 2011 #2
    If [itex]N(s)[/itex] is a Poisson process with intensity [itex]\rho s^2[/itex], the compensated process whose differential is [itex]d M(s)=d N(s)-(\rho s^2)d s[/itex] is a martingale, [itex]\mathbb{E}\left[d M(s)\right]=0[/itex].

    0 & = & \mathbb{E}\left[{\int_0^R \frac{d M(s)}{s^2}}\right] \\
    & = & \mathbb{E}\left[L\right] - \rho R \\
    \mathbb{E}\left[L\right] & = & \rho R
    The mean of [itex]L(R)[/itex] increases without bound. To show that L itself almost surely goes to infinity, consider the variance. The spherical shell [itex]s \leq r < s+ds[/itex] contains on average [itex]\rho s^2 ds[/itex] stars, with variance [itex]\rho s^2 ds[/itex]. This shell contributes [itex]\frac{\rho s^2 ds}{s^4}=\frac{\rho ds}{s^2}[/itex] to the variance of L. All the shells are independent, so their variances add.

    Var(L) = \int_0^R \frac{\rho ds}{s^2}

    which is bounded. Since [itex]\mathbb{E}\left[L\right][/itex] increases without bound while its variance is bounded, [itex]L[/itex] is almost surely unbounded. (I.e., the probability that [itex]L \leq b[/itex] for any finite b goes to 0 as R goes to infinity.)
    Last edited: Jul 9, 2011
  4. Jul 9, 2011 #3
    That's neat! I didn't think to check that the variance is finite or at least slowly growing. For the variance to be bounded the lower integration limit would have to be r>0, but that's fine if the observer is not inside a star.

    It looks like the argument works in >3 dimensions as well. To work in 1 or 2 dimensions apply Chebyshev's inequality:

    Pr[L<=C] <= Pr[|X-M|>=|M-C|] <= V/|M-C|^2

    where M and V are the mean and variance and C < M is arbitrary. We have M=k.(R-r) for all dimensions d and V=k.(R^(2-d)-r^(2-d))/(2-d) for d<>2 or V=k.log(R/r) so the RHS is O(1/R) or less in all cases.
  5. Jul 9, 2011 #4
    Oops! You're right. What saves you, of course, is that stars have finite size, so it's not Poisson at low r.
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