3 problems found
A box contains \(n\) pieces of string, each of which has two ends. I select two string ends at random and tie them together. This creates either a ring (if the two ends are from the same string) or a longer piece of string. I repeat the process of tying together string ends chosen at random until there are none left. Find the expected number of rings created at the first step and hence obtain an expression for the expected number of rings created by the end of the process. Find also an expression for the variance of the number of rings created. Given that \(\ln 20 \approx 3\) and that \(1+ \frac12 + \cdots + \frac 1n \approx \ln n\) for large \(n\), determine approximately the expected number of rings created in the case \(n=40\,000\).
Solution: Let \(X_i\) be the indicator variable a loop is formed when there are \(i\) strings in the bag, so \(\mathbb{P}(X_i = 1) = \frac{1}{2i-1}\). Therefore \begin{align*} && Y_n &= X_n + Y_{n-1} \\ && Y_n &= X_n + \cdots + X_1 \\ \Rightarrow && \E[Y_n] &= \frac{1}{2n-1} + \frac{1}{2n-3} + \cdots + \frac{1}{1} \\ && \var[Y_n] &= \sum_{i=1}^n \frac{1}{2i-1} \frac{2i-2}{2i-1} \\ &&&= 2\sum_{i=1}^n \frac{i-1}{(2i-1)^2} \end{align*} \begin{align*} && \E[Y_{n}] &= 1 + \frac13 + \cdots + \frac{1}{2n-1} \\ &&&= 1 + \frac12 + \cdots + \frac1{2n} - \frac12\left (1 + \frac12 + \cdots + \frac1n \right) \\ &&&\approx \ln 2n -\frac12 \ln n \\ &&&= \ln 2 \sqrt{n} \\ \\ && \E[Y_{40\,000}] &= \ln 2 \sqrt{40\,000} \\ &&&= \ln 400 \\ &&&= 2 \ln 20 \approx 6 \end{align*}
Bar magnets are placed randomly end-to-end in a straight line. If adjacent magnets have ends of opposite polarities facing each other, they join together to form a single unit. If they have ends of the same polarity facing each other, they stand apart. Find the expectation and variance of the number of separate units in terms of the total number \(N\) of magnets.
Solution: There are \(N-1\) gaps between the magnets which are independently gaps or not gaps. Therefore the total number of gaps is \(X \sim Binomial(N-1, \frac12)\) and \begin{align*} \mathbb{E}(X) &= \frac{N-1}{2} \\ \textrm{Var}(X) &= \frac{N-1}{4} \end{align*}
Integers \(n_{1},n_{2},\ldots,n_{r}\) (possibly the same) are chosen independently at random from the integers \(1,2,3,\ldots,m\). Show that the probability that \(\left|n_{1}-n_{2}\right|=k\), where \(1\leqslant k\leqslant m-1\), is \(2(m-k)/m^{2}\) and show that the expectation of \(\left|n_{1}-n_{2}\right|\) is \((m^{2}-1)/(3m)\). Verify, for the case \(m=2\), the result that the expection of \(\left|n_{1}-n_{2}\right|+\left|n_{2}-n_{3}\right|\) is \(2(m^{2}-1)/(3m).\) Write down the expectation, for general \(m\), of \[ \left|n_{1}-n_{2}\right|+\left|n_{2}-n_{3}\right|+\cdots+\left|n_{r-1}-n_{r}\right|. \] Desks in an examination hall are placed a distance \(d\) apart in straight lines. Each invigilator looks after one line of \(m\) desks. When called by a candidate, the invigilator walks to that candidate's desk, and stays there until called again. He or she is equally likely to be called by any of the \(m\) candidates in the line but candidates never call simultaneously or while the invigilator is attending to another call. At the beginning of the examination the invigilator stands by the first desk. Show that the expected distance walked by the invigilator in dealing with \(N+1\) calls is \[ \frac{d(m-1)}{6m}[2N(m+1)+3m]. \]