# Probability distribution of fragment lengths

I would like to compute the probability distribution for the length of the fragments which I would obtain by fragmenting a linear rod of length $L$ in the following way:

1. I choose at random (uniformly) $n$ breakpoints
2. I cut the rod at those breakpoints, creating $(n+1)$ fragments.

Now, while it is easy to see that the probability that a stretch of length $x$ does not contain any breakpoint goes like a negative exponential, I don't know how to throw in the information about the length of the rod.

• You can choose your units of measurement--meters, yards, parsecs, whatever--without materially changing the problem. So, choose a unit in which the rod has length 1. Done! :-) – whuber May 17 '11 at 16:25
• @whuber, thanks but I am not yet there. Basically I get a negative exp by computing the product of many small steps, each of them with probability (1-p), in the limit where the length of the step is very small. Now, how do I introduce in this computation the fact that I can't overstep the "right end" of the rod? – XenophiliusLovegood May 17 '11 at 17:02
• @X I'm a little lost, because I don't see how you are making a connection between the "many small steps" and the situation you have presented. – whuber May 17 '11 at 17:25
• @X I also wonder about your "negative exponential" assertion. Fixing a segment of length $x$ in a unit rod (without any loss of generality), the chance that any single breakpoint misses it is $1-x$. Because the breakpoints are independent, the chance that all of them miss it is $(1-x)^n$. That's not a negative exponential: it's a polynomial in $x$. Perhaps you're thinking of an asymptotic characterization for small $x$ and large $n$ (and bounded $nx$)? With those asymptotics the rod's length is arbitrarily large compared to $x$ and therefore should not play any role. – whuber May 18 '11 at 4:35
• @whuber I was obtaining the probability that no breakpoint falls in a segment of length $x$ by $(1-x/M)^{nM}$, from which an exponential when $M$ goes to infinity. So, I divide $x$ in $M$ parts, and multiply the probabilities that each of these parts is not hit by a breakpoint. – XenophiliusLovegood May 18 '11 at 10:20

Let the rod have length $L$ and fix a segment of length $x$. The chance that any single breakpoint misses the segment equals the proportion of the rod not occupied by the segment, $1−x/L$. Because the breakpoints are independent, the chance that all of them miss it is the product of $n$ such chances, $(1 - x/L)^n$.

From comments following the question, it appears that $x$ is intended to be small compared to the rod's length: $x/L \ll 1$. Let $\xi = L/x$ (assumed to be large) and rewrite $n = \xi(n/\xi)$, leading (purely via substitutions) to

$$\Pr(\text{all miss}) = (1 - x/L)^n = (1 - 1/\xi)^{\xi(n/\xi)} = \left((1-1/\xi)^\xi\right)^{n/\xi}\text{.}$$

Asymptotically $\xi \to \infty$. If we assume that $n$ varies in a way that makes $n/\xi$ converge to a constant, this probability approaches a computable limit. Let this constant be some value $\lambda$ times $x$. It is the limiting value of $n/\xi/x = n/L$: notice how the length of the rod is involved here and effectively is incorporated in $\lambda$. Because $\exp(-1) = 1/e$ is the limiting value of $(1-1/\xi)^\xi$ and raising to (positive) powers is a continuous function, it follows readily that the limit is

$$\Pr(\text{all miss}) \to e^{-\lambda x}.$$

One application is when $n$ is a constant, entailing $\lambda = n/L$, and $x \ll L$. We obtain $$e^{-nx/L}$$ as a good approximation for the probability that all breaks miss the segment. This analysis shows that the approximation fails as $x$ grows large: the approximation is only as good as the approximation $1/e \sim (1-1/\xi)^\xi$. Finally, if you set $x = L$, the approximation is clearly wrong because it gives $e^{-n}$ instead of the correct answer, $0$.

• thanks for your answer. I have one residual doubt. My original quest was to find a pdf for the lengths of the fragments. now, if I use $(1-\frac{x}{L})^n$, I run into troubles. First, (let's assume $L=1,n=1$) it is not normalized $\int_0^1 (1-x)\, dx=\frac{1}{2}$. Secondly, imagine I want to compute the average length of a fragment. That'd be $\int_0^1 x(1-x)\, dx=\frac{1}{6}$, which I don't understand either. What's your take? – XenophiliusLovegood May 21 '11 at 11:42
• @X We're not computing a probability distribution here: $x$ is a fixed value, not a random quantity! (See the first line of my reply.) Thus, neither the normalization nor the expectation make any sense at all. The question I answered is the one you asked in comments: "I was obtaining the probability that no breakpoint falls in a segment of length $x$". Your original question asks for a "probability distribution for the length of the fragments...." Because there will be $n+1$ fragments, you are looking for an $n+1$-variate distribution for all the lengths. – whuber May 21 '11 at 19:33
• I don't understand well your last comment. On the other hand, I think I possibly got to a satisfactory conclusion: $(1-x)^n$ is the cdf of the pdf I am looking for (in the original question). The corresponding pdf is $n(1-x)^{n-1}$ which is normalized and has nice expectation values. – XenophiliusLovegood May 23 '11 at 10:44

Let $\{X_i\}$ be the locations of the cuts.

I'd approach this problem by finding the order statistics $\{Y_i\}$ so that $Y_1$ would be the location of the leftmost cut. Then I'd calculate the probability distributions of the differences between the variables $Y_i-Y_{i-1}$. Don't forget to also calculate $Y_1-0$ and $L-Y_n$.

Can anyone think of a better way?

• Yes: First close the rod into a circle. Then break it (uniformly, randomly, independently) at $n+1$ breakpoints. This introduces a helpful symmetry :-). – whuber May 17 '11 at 16:24
• Or, draw a connection to the Poisson process. – cardinal May 17 '11 at 16:46