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 \(\frac{a+1}{a} k \frac{a+1}{a} \frac{s}{k} = \left ( \frac{a+1}{a} \right)^2s > s\)