The function \(\log^+ (x)\) is defined by \[\log^+ (x) = \begin{cases} \log_e (x) & (x \geq 1) \\ 0 & (x < 1) \end{cases}\] Positive numbers \(\lambda_1 > \lambda_2 > ... > \lambda_n\) and \(\mu_1 > \mu_2 > ... > \mu_n\) satisfy \[\lambda_1 \lambda_2 ... \lambda_j \geq \mu_1 \mu_2 ... \mu_j \quad \text{for} \quad 1 \leq j \leq n.\] Show that \[g(x) = \sum_{j=1}^{n} \log^+ (\lambda_j x) \geq h(x) = \sum_{j=1}^{n} \log^+ (\mu_j x),\] for all \(x\). By considering \[\int_0^{\infty} \frac{g(x)}{x^{s+1}} dx \quad \text{and} \quad \int_0^{\infty} \frac{h(x)}{x^{s+1}} dx,\] show that \[\lambda_1^s + ... + \lambda_n^s \geq \mu_1^s + ... + \mu_n^s, \quad \text{for} \quad s > 0.\]
\(X\) is an integer-valued random variable, with distribution given by \[\text{Pr}[X = k] = \frac{c}{k \cdot 2^k}, \quad k \geq 1.\] Find the probability generating function of \(X\), and hence deduce the value of \(c\). A car insurance company observes that the number \(N\) of claims in any year is distributed as a Poisson random variable with mean \(\mu\), and that the sums of money paid out on the different claims are distributed, independently of \(N\) and of each other, in the same way as \(X\). By considering probability generating functions, or otherwise, find the mean and variance of the total sum \(S\) paid out per year. [Hint. If the probability generating function of \(S\), given that \(N = n\), is denoted by \(G_n(z)\), then the probability generating function of \(S\) is given by \(\sum_{n \geq 0} G_n(z) \cdot \text{Pr}[N = n]\).]
Solution: \begin{align*} G_X(z) &= \sum_{k=1}^{\infty} \text{Pr}[X = k]z^k \\ &= \sum_{k=1}^{\infty} \frac{c}{k \cdot 2^k}z^k \\ &= c\sum_{k=1}^{\infty} \frac1k \left ( \frac{z}{2} \right)^k \\ &= -c \ln \left (1 - \frac{z}{2} \right) \end{align*} \begin{align*} S = \end{align*} Since \(G_X(1) = 1\), we must have \(-c \ln \tfrac12 = 1 \Rightarrow c = \frac1{\ln 2}\)
Show that, if \(X_1\), \(X_2\) and \(X_3\) are independent, and have a common continuous distribution, \(\text{Pr}[X_1 > \max(X_2, X_3)] = \frac{1}{3}\). Independent random samples \(X_1, X_2, ..., X_n\) and \(Y_1, Y_2, ..., Y_n\) are drawn from two unknown continuous distributions. It is suspected that, in fact, the distribution of the \(X\)'s is the same as that of the \(Y\)'s, and, to test this, it is proposed to consider the statistic \[W = \sum_{i,j=1}^{n} Z_{ij},\] where \[Z_{ij} = \begin{cases} 1 & \text{if} \quad X_i < Y_j \\ 0 & \text{if} \quad X_i \geq Y_j. \end{cases}\] Show that, if the two distributions are indeed identical, \(W\) has expectation \(n^2/2\) and variance \(n^2(2n + 1)/12\).* Show also that if, instead, \(P[X_i > Y_j] = p \neq \frac{1}{2}\), the expectation of \(W\) is \((1-p)n^2\). Assuming that \(W\) is approximately normally distributed, investigate whether the following data support the view that values from the \(Y\) distribution are typically larger than values from the \(X\) distribution: \(X\) \(3.41\) \(3.63\) \(3.77\) \(4.00\) \(4.54\) \(4.82\) \(4.91\) \(5.08\) \(Y\) \(3.91\) \(4.70\) \(4.71\) \(4.93\) \(4.95\) \(5.12\) \(5.37\) \(5.90\) * [Hint. Find the variance of \(W\) by first evaluating the expectation of \(W^2\).]
A particle of mass \(m\) moves in a horizontal straight line under a force equal to \(mn^2\) times the displacement from a point \(O\) in the line and directed towards \(O\). In addition, the motion of the particle is resisted by a force equal to \(mk\) times the square of its speed. The particle is projected with speed \(V\) from \(O\) along the line; by considering the motion in dimensionless form, show that the displacement \(x\) is of the form \[x = \frac{V}{n}f\left(\frac{v}{V}, \frac{kV}{n}\right),\] where \(v\) is the non-zero speed of the particle when at \(x\). Show that the particle first comes to rest at distance \(X\), of the form \[X = \frac{V}{n}g\left(\frac{kV}{n}\right).\] If \(\frac{kV}{n}\) is small, show that \(g\left(\frac{kV}{n}\right)\) is approximately \[1-\frac{2kV}{3n}.\]
A particle moves in a horizontal circle on the inner surface of a smooth spherical shell of radius \(a\) and is slightly disturbed. Show that the period of the small oscillation about the steady motion is \[2\pi \left(\frac{a \cos \alpha}{g(1+3\cos^2 \alpha)}\right)^{\frac{1}{2}},\] where \(a \sin \alpha\) is the radius of the circle.
A particle of unit mass is projected from level ground with speed \(u\sqrt{2}\) at an elevation of \(\frac{1}{4}\pi\) above the horizontal. It experiences a resisting force directly opposing its motion whose magnitude is \(k\) times the square of the particle's speed, where \(ku^2/g\) is a small number. Show that, after a lapse of time \(t\) during the flight, the vertical component of the acceleration is approximately equal to \[-g-k(u-gt) \{(u-gt)^2 + u^2t^2\}^{\frac{1}{2}}.\] Deduce that the highest point of the trajectory is attained when \(t\) is approximately equal to \[\frac{u}{g} - \frac{ku^3}{3g^2}(2^{\frac{1}{2}}-1).\]
The moment of relative momentum of a particle \(P\), of mass \(m\), about an arbitrary point \(O'\) is defined as \(m\mathbf{r'} \wedge \dot{\mathbf{r'}}\), where \(\mathbf{r'} = \overrightarrow{O'P}\) and the dot denotes differentiation with respect to time. A collection of particles has centre of gravity \(G\), total mass \(M\), and moment of momentum \(\mathbf{h}\) about the origin \(O\) of a fixed coordinate system. Show that the moment of relative momentum about an arbitrary point \(O'\) is \(\mathbf{h'}\), given by \[\mathbf{h'} = \mathbf{h} - M\mathbf{s} \wedge \dot{\mathbf{f}} - M(\mathbf{f}-\mathbf{s}) \wedge \dot{\mathbf{s}},\] where \(\mathbf{f} = \overrightarrow{OG}\) and \(\mathbf{s} = \overrightarrow{OO'}\). Show also that, if \(\mathbf{L'}\) is the moment of the external forces about \(O'\), then \[\dot{\mathbf{h'}} = \mathbf{L'} - M(\mathbf{f}-\mathbf{s}) \wedge \ddot{\mathbf{s}}.\]
Show that for three vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) \[(\mathbf{a} \wedge \mathbf{b})\cdot(\mathbf{a} \wedge \mathbf{c}) = (\mathbf{a}\cdot\mathbf{a})(\mathbf{b}\cdot\mathbf{c})-(\mathbf{a}\cdot\mathbf{b})(\mathbf{a}\cdot\mathbf{c})\] and \[(\mathbf{a} \wedge \mathbf{b}) \wedge (\mathbf{a} \wedge \mathbf{c}) = \mathbf{a}(\mathbf{a}\cdot\mathbf{b} \wedge \mathbf{c}).\] [You may assume \[\mathbf{x} \wedge \mathbf{y}\cdot\mathbf{z} = \mathbf{x} \cdot \mathbf{y} \wedge \mathbf{z}\] and \[(\mathbf{x} \wedge \mathbf{y}) \wedge \mathbf{z} = \mathbf{y}(\mathbf{x}\cdot\mathbf{z}) - \mathbf{x}(\mathbf{y}\cdot\mathbf{z}).\] Three points \(A\), \(B\) and \(C\) lie on a sphere with centre \(O\). Let \(\hat{A}\), \(\hat{B}\) and \(\hat{C}\) be the angles \(BOC\), \(COA\) and \(AOB\), and let \(\alpha\), \(\beta\) and \(\gamma\) be the angles between the pairs of planes \(AOB\) \& \(AOC\), \(BOC\) \& \(BOA\) and \(COA\) \& \(COB\). Deduce the spherical triangle cosine and sine formulae \[\cos \hat{A} = \cos \hat{B} \cos \hat{C} + \sin \hat{B} \sin \hat{C} \cos \alpha\] and \[\frac{\sin \alpha}{\sin \hat{A}} = \frac{\sin \beta}{\sin \hat{B}} = \frac{\sin \gamma}{\sin \hat{C}}.\]