3

A stick of length $2$ m is made of uniformly dense material. A point is chosen randomly on the stick and the stick is broken at that point. The left portion of the stick is discarded and now again another point is chosen randomly on the remaining right portion of the stick and the stick is broken again at that point and the left part is again discarded.The process is continued indefinitely.What is the probability that one of the discarded left parts has length $>1$ m?

Formulating this problem we basically have a sequence of random variables {$X_n$} where $X_1 \sim U(0,2)$ , $X_2|X_1 \sim U(0,2-X_1)$,$X_3|X_1,X_2 \sim U(0,2-X_1-X_2)$ and so on. The probability that any one of the discarded parts is more than $1$ m is equivalent to say that it is $1-P(\cap${$ X_i<1$}) But I cannot find the probability explicitly as it is dependent on $X_1$. Help!

  • Nice question. I tried to work it out. For $n=1$, the answer is obvious ($1/2$). For $n=2$, the answer is $1-\ln(2)/2$. For $n=3$, however, I end up needing to integrate $\ln(x)/x$, for which I do not know the solution. – EdG Aug 08 '19 at 05:57
  • I was too quick with my previous comment. For $n=3$, the probability of one part being larger than $1$ is $3/2-\ln(2)(1+\ln(2)/4)$. I guess there is a certain pattern, maybe this is of help for you. – EdG Aug 08 '19 at 06:03
  • So the first related question here is exactly the same up to multiplicative constant. – dEmigOd Aug 08 '19 at 06:16

1 Answers1

0

The answer linked to in the comments is extremely elegant and gives the exact right answer of $\ln(2)\approx .693$. For those of us that aren't that smart, I figured I'd give a way you could get provable and decent upper and lower bounds fairly quickly. Notice that in your notation, the events $\{X_i >1\}$ are mutually disjoint, hence \begin{equation} \Pr(\sup_{i} X_i>1)=\sum_{i=1}^{\infty} \Pr(X_i>1). \end{equation}

Obviously $\Pr(X_1>1)=1/2$. It's not too hard to see that \begin{align*} \Pr(X_2>1)&=\frac{1}{2}\int_0^1 dx_1\bigg(\int_1^{2-x_1}\frac{1}{2-x}dx_2\bigg)\\ &=\frac{1}{2}\int_0^1 \frac{1-x_1}{2-x_1}dx\\ &=\frac{1}{2}(1-\ln(2))\\ &\approx .1530. \end{align*}

Similarly, \begin{align*} \Pr(X_3>1)&=\frac{1}{2}\int_0^1 dx_1\bigg(\int_0^{1-x_1-x_2}\frac{1}{2-x}dx_2\bigg(\int_1^{2-x_1-x_2}\frac{1}{2-x_1-x_2}dx_3\bigg)\\ &=\frac{1}{2}-\frac{1}{4}\ln(2)(2+\ln(2))\\ &\approx .0333. \end{align*} (I used Wolfram Alpha...) In general, one has \begin{equation} \Pr(X_n>1)=\frac{1}{2}\int_0^1 dx_1\bigg(\int_0^{1-x_1} \frac{1}{2-x_1}dx_2\bigg(\ldots \bigg(\int_1^{2-x_1-\ldots-x_{x_{n-1}}} \frac{1}{2-x_1-\ldots-x_{n-1}}dx_n\bigg)\ldots\bigg). \end{equation}

Using Wolfram again, you can find that \begin{equation} \Pr(X_4>1)=\frac{1}{12}(6-\ln(2))(6+\ln^2(2)+\ln(8))\approx .0056. \end{equation}

This tells us already that \begin{equation} 0.6919=.5+.1530+.0333+.0056\leq \Pr(\sup_i X_i>1). \end{equation}

Now, to get an upper bound, simply note that for any $n$, \begin{equation} \Pr(X_i>1, i>n)\leq \Pr(\sum_{i=1}^n X_i<1), \end{equation} using the fact that the first left hand side event implies the right hand side event. Notice then that \begin{align} \Pr(\sum_{i=1}^n X_i<1)&=\frac{1}{2}\int_0^1 dx_1\bigg(\int_0^{1-x_1} \frac{1}{2-x_1}dx_2\bigg(\ldots \bigg(\int_0^{1-x_1-\ldots-x_{x_{n-1}}} \frac{1}{2-x_1-\ldots-x_{n-1}}dx_n\bigg)\ldots\bigg)\\ &=\frac{1}{2}\int_0^1 dx_1\bigg(\int_0^{1-x_1} \frac{1}{2-x_1}dx_2\bigg(\ldots \bigg(\int_1^{2-x_1-\ldots-x_{x_{n-1}}} \frac{1}{2-x_1-\ldots-x_{n-1}}dx_n\bigg)\ldots\bigg)\\ &=\Pr(X_n>1), \end{align} where we use the fact that the only difference is the last integrand is shifted, but the function does not depend on $x_n$. In particular, for all $n$, we have \begin{equation} \sum_{i=1}^n \Pr(X_i>1)\leq \Pr(\sup_i X_i>1)\leq \sum_{i=1}^{n} \Pr(X_i>1) +\Pr(X_n>1). \end{equation} Applying this with $n=4$, you get \begin{equation} .6919\leq \Pr(\sup_i X_i>1) \leq .6975. \end{equation}

Jason Gaitonde
  • 5,284