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Radon transform, Buffon's needle and Integral geometry

  1. May 15, 2014 #1


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    In all the literature that I have seen it is mentioned that these two are "branches" of integral geometry, but no where I can see the exact connection since one is connected with probability and the other is an integral.

    I have seen this, but it is not clear.


    Can somebody explain the connection in a clear way. Thanks
  2. jcsd
  3. May 20, 2014 #2

    Stephen Tashi

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    While we're waiting for someone who really knows the answer, I'll make some comments.

    From the abstract point of vew, a probability distribution is a "measure" on a space of things. When you do integration from the abstract point of view, you integrate functions on a space of things with respect of a "measure" on the space. Much of probability theory (such as finding the expected value of a random variable, the variance of a random variable etc.) involves doing integrals. If [itex] p(x) [/itex] is a probability density function on the real line, you can regard the expected value of [itex] E(f(x)) = \int f(x) p(x) dx [/itex] as the integral of the product [itex] f(x)p(x) [/itex] with respect to the ordinary way of measuring length on the real line (denoted by [itex] dx [/itex]) or you can regard it as an integral of [itex] f(x) [/itex] with respect to another way of assigning a "measure" to an interval on the real line given by [itex] p(x) dx [/itex].

    The high class way to think about [itex] E(f(x)) [/itex] is to think about it as an integral of [itex] f(x) [/itex] with respect to the measure [itex] p(x) dx [/itex] because this view generalizes to cases where integration with respect to the ordinary notion of length doesn't work. For example, suppose [itex] X [/itex] is a random variable realized as follows. Flip a fair coin. If the coin lands heads then [itex] X = 1/3 [/itex]. If the coin lands tails then pick the value of [itex] X [/itex] from a uniform distribution on [itex] [0,1] [/itex]. To find the expected value of [itex] X [/itex] you can't do a simple Riemann integral since it would assign zero length to the point [itex] 1/3 [/itex] and the correct calculation of the expected value of [itex] X [/itex] somehow has to justify adding the term [itex] (1/2)(1/3) [/itex] to the result. If you think about a kind of measure on the real line where the point [itex] 1/3 [/itex] has measure [itex] 1/2 [/itex] then you can justify doing that.

    So there is an intimate connection between integration and measures. A probability disribution defines a special kind of measure.

    If a transform is defined conceptually as an integration "over all possible lines" that satisfy a certain condition, you may be able to parameterize such a line by an n-tuple of real numbers and do an n-variable Riemann integral in the ordinary way, thinking of the measure as the ordinary measure of n-dimensional volume. But if parameterizing the integral of [itex] f(x,y,z...) [/itex] that way introduces other functions as factors in the integral, the high class way of thinking about it may be to think of those factors as defining a new sort of measure on the space of lines. Someone who really knows integral geometry will have to comment on whether that's the way to look at it.
    Last edited: May 20, 2014
  4. May 22, 2014 #3


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    Thanks for the reply. I am still searching for answers, it is getting a bit complicated.
  5. Jun 20, 2014 #4


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    I have found an answer for it in here. The connection is in the second solution.
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