# Radon transform, Buffon's needle and Integral geometry

1. May 15, 2014

### ftr

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.

http://www.encyclopediaofmath.org/index.php/X-ray_transform

Can somebody explain the connection in a clear way. Thanks

2. May 20, 2014

### Stephen Tashi

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 $p(x)$ is a probability density function on the real line, you can regard the expected value of $E(f(x)) = \int f(x) p(x) dx$ as the integral of the product $f(x)p(x)$ with respect to the ordinary way of measuring length on the real line (denoted by $dx$) or you can regard it as an integral of $f(x)$ with respect to another way of assigning a "measure" to an interval on the real line given by $p(x) dx$.

The high class way to think about $E(f(x))$ is to think about it as an integral of $f(x)$ with respect to the measure $p(x) dx$ because this view generalizes to cases where integration with respect to the ordinary notion of length doesn't work. For example, suppose $X$ is a random variable realized as follows. Flip a fair coin. If the coin lands heads then $X = 1/3$. If the coin lands tails then pick the value of $X$ from a uniform distribution on $[0,1]$. To find the expected value of $X$ you can't do a simple Riemann integral since it would assign zero length to the point $1/3$ and the correct calculation of the expected value of $X$ somehow has to justify adding the term $(1/2)(1/3)$ to the result. If you think about a kind of measure on the real line where the point $1/3$ has measure $1/2$ 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 $f(x,y,z...)$ 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
3. May 22, 2014

### ftr

Thanks for the reply. I am still searching for answers, it is getting a bit complicated.

4. Jun 20, 2014

### ftr

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