A horizontal spindle rotates freely in a fixed bearing. Three light rods are each attached by one end to the spindle so that they rotate in a vertical plane. A particle of mass \(m\) is fixed to the other end of each of the three rods. The rods have lengths \(a\), \(b\) and \(c\), with \(a > b > c\,\) and the angle between any pair of rods is \(\frac23 \pi\). The angle between the rod of length \(a\) and the vertical is \(\theta\), as shown in the diagram. \vspace*{-0.1in}
Five independent timers time a runner as she runs four laps of a track. Four of the timers measure the individual lap times, the results of the measurements being the random variables \(T_1\) to \(T_4\), each of which has variance \(\sigma^2\) and expectation equal to the true time for the lap. The fifth timer measures the total time for the race, the result of the measurement being the random variable \(T\) which has variance \(\sigma^2\) and expectation equal to the true race time (which is equal to the sum of the four true lap times). Find a random variable \(X\) of the form \(aT+b(T_1+T_2+T_3+T_4)\), where \(a\) and \(b\) are constants independent of the true lap times, with the two properties: (1) \ whatever the true lap times, the expectation of \(X\) is equal to the true race time; (2) \ the variance of \(X\) is as small as possible. Find also a random variable \(Y\) of the form \(cT+d(T_1+T_2+T_3+T_4)\), where \(c\) and \(d\) are constants independent of the true lap times, with the property that, whatever the true lap times, the expectation of \(Y^2\) is equal to \(\sigma^2\). In one particular race, \(T\) takes the value 220 seconds and \((T_1 + T_2 + T_3 + T_4)\) takes the value \(220.5\) seconds. Use the random variables \(X\) and \(Y\) to estimate an interval in which the true race time lies.
A pack of cards consists of \(n+1\) cards, which are printed with the integers from \(0\) to \(n\). A~game consists of drawing cards repeatedly at random from the pack until the card printed with 0 is drawn, at which point the game ends. After each draw, the player receives \(\pounds 1\) if the card drawn shows any of the integers from \(1\) to \(w\) inclusive but receives nothing if the card drawn shows any of the integers from \(w+1\) to \(n\) inclusive.
In this question, you may use the result \[ \displaystyle \int_0^\infty \frac{t^m}{(t+k)^{n+2}} \; \mathrm{d}t =\frac{m!\, (n-m)!}{(n+1)! \, k^{n-m+1}}\;, \] where \(m\) and \(n\) are positive integers with \(n\ge m\,\), and where \(k>0\,\). The random variable \(V\) has density function \[ \f(x) = \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \quad \quad (0 \le x < \infty) \;, \] where \(a\) is a positive integer. Show that \(\displaystyle C = \frac{(2a+1)!}{a! \, a!}\;\). Show, by means of a suitable substitution, that \[ \int_0^v \frac{x^a}{(x+k)^{2a+2}} \; \mathrm{d}x = \int_{\frac{k^2}{v}}^\infty \frac{u^a}{(u+k)^{2a+2}} \; \mathrm{d}u \] and deduce that the median value of \(V\) is \(k\). Find the expected value of \(V\). The random variable \(V\) represents the speed of a randomly chosen gas molecule. The time taken for such a particle to travel a fixed distance \(s\) is given by the random variable \(\ds T=\frac{s}{V}\). Show that \begin{equation} \mathbb{P}( T < t) = \ds \int_{\frac{s}{t}}^\infty \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}}\; \mathrm{d}x \tag{\( *\)} \end{equation} and hence find the density function of \(T\). You may find it helpful to make the substitution \(\ds u = \frac{s}{x}\) in the integral \((*)\). Hence show that the product of the median time and the median speed is equal to the distance \(s\), but that the product of the expected time and the expected speed is greater than \(s\).
Solution: \begin{align*} && f(x) &= \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \\ \Rightarrow && 1 &= \int_0^{\infty} f(x) \d x \\ &&&= \int_0^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ &&&= Ck^{a+1} \int_0^{\infty} \frac{x^a}{(x+k)^{2a+2} }\d x \\ &&&= Ck^{a+1} \frac{a!(2a-a)!}{(2a+1)!k^{2a-a+1}} \\ &&&= C \frac{a!a!}{(2a+1)!} \\ \Rightarrow && C &= \frac{(2a+1)!}{a!a!} \end{align*} \begin{align*} && I &= \int_0^v \frac{x^a}{(x+k)^{2a+2}} \d x\\ u = k^2/x, \d x = -k^2u^{-2} \d u: &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a}u^{-a}}{(k^2u^{-1} +k)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a-2a-2}u^{2a+2-a}}{(k +u)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{ k^2/v}^{\infty} \frac{u^{a}}{(k +u)^{2a+2}} \d u \\ \end{align*} At the median we want a value \(M\) such that \(M = k^2/M\) ie \(M = k\) \begin{align*} && \mathbb{E}(V) &= \int_0^{\infty} x f(x) \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \int_0^{\infty} \frac{x^{a+1}}{(x+k)^{2a+2}} \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \frac{(a+1)!(2a-(a+1))!}{(2a+1)!k^{2a-(a+1)+1}}\\ &&&= \frac{k^{a+1}}{a!} \frac{(a+1)(a-1)!}{k^{a}} \\ &&&= \frac{k(a+1)}{a} = \frac{a+1}a k \end{align*} \begin{align*} && \mathbb{P}(T < t) &= \mathbb{P}(\frac{s}{V} < t) \\ &&&= \mathbb{P}(V > \frac{s}{t}) \\ &&&= \int_{s/t}^{\infty} f(x) \d x \\ &&&= \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ \\ \Rightarrow && f_T(t) &= \frac{\d}{\d t} \left ( \mathbb{P}(T < t)\right) \\ &&&= \frac{\d}{\d t} \left ( \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \right) \\ &&&= - \frac{C \, k^{a+1} \, \left ( \frac{s}{t} \right)^a}{(\frac{s}{t}+k)^{2a+2}} \cdot \left (-\frac{s}{t^2} \right) \\ &&&= \frac{Ck^{a+1}s^{a+1}t^{2a+2}}{t^{a+2}(s+kt)^{2a+2}} \\ &&&= \frac{C(ks)^{a+1}t^a}{(s+kt)^{2a+2}} \\ &&&= \frac{C(\frac{s}{k})^{a+1}t^a}{(\frac{s}{k}+t)^{2a+2}} \end{align*} Therefore \(T\) follows the same distribution, but with parameter \(s/k\) rather than \(k\). In particular it has median \(s/k\) (and the product of the medians is \(s\)). However, the product of the expected time and expected speed is \(\frac{a+1}{a} k \frac{a+1}{a} \frac{s}{k} = \left ( \frac{a+1}{a} \right)^2s > s\)