Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following is an elementary probability question related to a generalization of the famous "Buffon's needle experiment" which allows one to estimate $\pi$ by counting how many times a randomly tossed needle crosses a line on a lined sheet of paper. If we replace the needle with a rigid wire in the shape of any piecewise smooth plane curve, I believe it is well-known that the expected number of line crossings depends only on the length of the wire and not on its specific shape.

I am seeking an elementary proof of this fact in the case where the wire consists of two line segments joined end-to-end. The only parameters here are the lengths of the two line segments and the angle at which they are joined; I would like to prove that the expected number of crossings depends only on the sum of the lengths. If it helps, I am happy to assume that both line segments are very small compared to the spacing between the lines on the paper. Any ideas?


Added: Several have argued that this follows simply from the linearity of expectation, but I am not convinced. Suppose it were the case that the expectation for a single segment of length $\ell$ was given by $\ell^2$. Then if $X$ and $Y$ are the random variables representing the two needles making up the wire we would have $E(X+Y) = E(X) + E(Y) = \ell_X^2 + \ell_Y^2$, and this is not a function of $\ell_X + \ell_Y$ (though it is a function of $\ell_X$ and $\ell_Y$). Of course we secretly know that $E(X) = C \ell_X$, but my goal in asking this question is to prove this fact without actually calculating anything.

share|improve this question
    
If the lengths are equal (S) and the angle is 0, it is as if a single segment of size 2S. If the angle is pi, it is as if a segment of size S. So your hypothesis can't stand. Similarly for the more general problem, think of a spiral. You can make the spiral as long as you want, and if coiled tight, you can make it as small as you want. It behaves like it's convex hull, so area is the important thing here, but that can be made arbitrarily small as above. –  ex0du5 Jan 25 '13 at 21:53
    
Flip 0 and pi, I guess. I was thinking external angle. –  ex0du5 Jan 25 '13 at 21:54
1  
@ex0du5: If the angle is $0$, it is as if it is a segment of size $S$, but you should count each crossing twice (since it hits both segments). –  Micah Jan 25 '13 at 21:57
    
@exodu5: The point is that the segment of length $2S$ is more likely to hit a line (because it is longer), while for the "doubled" segment of length $S$ it is less likely to hit a line but each crossing will be counted twice. So in balance the expectations are the same. Turning this into a rigorous proof is tricky. –  Paul Siegel Jan 25 '13 at 23:30
    
Linearity of expectation immediately gives you the fact that the expected number of crossings is independent of the angle between the two segments. Now, apply that to two segments joined at an angle of $\pi$, and you get a proof that it's also independent of the individual lengths of the segments. –  Micah Jan 25 '13 at 23:52

3 Answers 3

up vote 0 down vote accepted

Let $E(\ell)$ denote the expected number of crossings achieved by a segment of length $\ell$. By linearity of expectation, we know that any union of two segments of lengths $\ell_1,\ell_2$ will achieve an expected crossing number of $E(\ell_1)+E(\ell_2)$, irrespective of the angle they meet at.

In particular, this holds for the line segment of length $\ell_1+\ell_2$. So we must have $$ E(\ell_1)+E(\ell_2)=E(\ell_1+\ell_2) \, ; $$ since this last quantity depends only on the total length of the segments, it follows that the expected crossing number of any union of two segments depends only on the total length of the segments.

share|improve this answer

The result simply follows from linearity of expectation. The expected total number of crossings is the sum of the expected number of crossings for each portion of the wire (even as those portions become infinitesimally small).

In fact, this leads to an elementary proof of the original Buffon result. Let the separation between the lines be $L$, and let the length of the straight wire be $X$. Curl the wire $n$ times around, into a tight circle of radius $\frac{X}{2\pi n} \ll L$. Then the number of crossings will be $2n$ if the center of the circle lands within a distance $\frac{X}{2\pi n}$ of a line, which will happen with probability $\frac{X}{\pi n L}$. The expected number of crossings is therefore $2n\cdot\frac{X}{\pi n L}=\frac{2X}{\pi L}.$ This result is correct for the original straight wire as well.

share|improve this answer

Let's reconsider the "throwing at random" part in terms of an arbitrary shape needle of overall length S. Focus on an arbitrarily short section at one end, Section #1. Throw the wiggly needle down and count the crossings only by this Section #1.

Meanwhile another observer has decided to "throw at random" by taking the position and orientation of Section #1 (from the throw above) and applying the necessary translation and rotation (from the shape of the wiggly needle) to find the position and orientation of, say, Section #2. Then he counts the crossings of only Section #2.

Repeat for all the short sections.

The proof of the Buffon's needle theorem for "short" needles shows the probability of a crossing depends in a linear fashion on the length of the short needle. So we can just add up the results for all the Section #1, Section #2, .... by adding the lengths.

QED ?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.