A bullet of mass \(m\) is fired horizontally with speed \(u\) into a wooden block of mass \(M\) at rest on a horizontal surface. The coefficient of friction between the block and the surface is \(\mu\). While the bullet is moving through the block, it experiences a constant force of resistance to its motion of magnitude \(R\), where \(R>(M+m)\mu g\). The bullet moves horizontally in the block and does not emerge from the other side of the block.
The infinite series \(S\) is given by \[ S = 1 + (1 + d)r + (1 + 2d)r^2 + \cdots + (1+nd)r^n +\cdots\; ,\] for \(\vert r \vert <1\,\). By considering \(S - rS\), or otherwise, prove that \[ S = \frac 1{1-r} + \frac {rd}{(1-r)^2} \,.\] Arthur and Boadicea shoot arrows at a target. The probability that an arrow shot by Arthur hits the target is \(a\); the probability that an arrow shot by Boadicea hits the target is \(b\). Each shot is independent of all others. Prove that the expected number of shots it takes Arthur to hit the target is \(1/a\). Arthur and Boadicea now have a contest. They take alternate shots, with Arthur going first. The winner is the one who hits the target first. The probability that Arthur wins the contest is \(\alpha\) and the probability that Boadicea wins is \(\beta\). Show that \[ \alpha = \frac a {1-a'b'}\,, \] where \(a' = 1-a\) and \(b'=1-b\), and find \(\beta\). Show that the expected number of shots in the contest is \(\displaystyle \frac \alpha a + \frac \beta b\,.\)
Solution: Notice that \begin{align*} && S - rS &= 1 + dr + dr^2 + \cdots \\ &&&= 1 + dr(1 + r+r^2+ \cdots) \\ &&&= 1 + \frac{rd}{1-r} \\ \Rightarrow && S &= \frac{1}{1-r} + \frac{rd}{(1-r)^2} \end{align*} The number of shots Arthur takes is \(\textrm{Geo}(a)\), so it's expectation is \(1/a\). The probability Arthur wins is: \begin{align*} \alpha &= a + a'b'a + (a'b')^2a + \cdots \\ &= a(1+a'b' + \cdots) \\ &= \frac{a}{1-a'b'} \\ \\ \beta &= a'b + a'b'a'b + \cdots \\ &= a'b(1+b'a' + (b'a')^2 + \cdots ) \\ &= \frac{a'b}{1-a'b'} \end{align*} The expected number of shots in the contest is: \begin{align*} E &= a + 2a'b + 3a'b'a + 4a'b'a'b + \cdots \\ &= a(1 + 3a'b' + 5(a'b')^2 + \cdots) + 2a'b(1 + 2(a'b') + 3(a'b')^2 + \cdots) \\ &= a \left ( \frac{1}{1-a'b'} + \frac{2a'b'}{(1-a'b')^2} \right) + 2a'b \left ( \frac{1}{1-a'b'} + \frac{a'b'}{(1-a'b')^2}\right) \\ &= \frac{a}{1-a'b'} \left (1 + \frac{2a'b'}{(1-a'b')} \right) + 2\frac{a'b}{1-a'b'} \left ( 1 + \frac{a'b'}{(1-a'b')}\right) \\ &= \alpha \frac{1+a'b'}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ &= \alpha \frac{1+1-a-b+ab}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ \end{align*}
In this question, \({\rm Corr}(U,V)\) denotes the product moment correlation coefficient between the random variables \(U\) and \(V\), defined by \[ \mathrm{Corr}(U,V) \equiv \frac{\mathrm{Cov}(U,V)}{\sqrt{\var(U)\var(V)}}\,. \] The independent random variables \(Z_1\), \(Z_2\) and \(Z_3\) each have expectation 0 and variance 1. What is the value of \(\mathrm{Corr} (Z_1,Z_2)\)? Let \(Y_1 = Z_1\) and let \[ Y_2 = \rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2\,, \] where \(\rho_{12}\) is a given constant with $-1<\rho _{12}<1$. Find \(\E(Y_2)\), \(\var(Y_2)\) and \(\mathrm{Corr}(Y_1, Y_2)\). Now let \(Y_3 = aZ_1 + bZ_2 + cZ_3\), where \(a\), \(b\) and \(c\) are real constants and \(c\ge0\). Given that \(\E(Y_3) = 0\), \(\var(Y_3) = 1\), \( \mathrm{Corr}(Y_1, Y_3) =\rho^{{2}}_{13} \) and \( \mathrm{Corr}(Y_2, Y_3)= \rho^{{2}} _{23}\), express \(a\), \(b\) and \(c\) in terms of \(\rho^{2} _{23}\), \(\rho^{2}_{13}\) and \(\rho^{2} _{12}\). Given constants \(\mu_i\) and \(\sigma_i\), for \(i=1\), \(2\) and \(3\), give expressions in terms of the \(Y_i\) for random variables \(X_i\) such that \(\E(X_i) = \mu_i\), \(\var(X_i) = \sigma_ i^2\) and \(\mathrm{Corr}(X_i,X_j) = \rho_{ij}\).
Solution: \begin{align*} \mathrm{Corr} (Z_1,Z_2) &= \frac{\mathrm{Cov}(Z_1,Z_2)}{\sqrt{\var(Z_1)\var(Z_2)}} \\ &= \frac{\mathbb{E}(Z_1 Z_2)}{\sqrt{1 \cdot 1}} \\ &= \frac{\mathbb{E}(Z_1)\mathbb{E}(Z_2)}{\sqrt{1 \cdot 1}} \\ &= \frac{0}{1} \\ &= 0 \end{align*} \begin{align*} && \mathbb{E}(Y_2) &= \mathbb{E}(\rho_{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \mathbb{E}(\rho_{12} Z_1) + \mathbb{E}( (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \rho_{12}\mathbb{E}( Z_1) + (1 - {\rho_{12}^2})^{ \frac12}\mathbb{E}( Z_ 2) \\ &&&= 0\\ \\ && \textrm{Var}(Y_2) &= \textrm{Var}(\rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \textrm{Var}(\rho_{12} Z_1)+\textrm{Cov}(\rho_{12} Z_1,(1 - {\rho_{12}^2})^{ \frac12} Z_ 2 ) + \textrm{Var}((1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \rho_{12}^2\textrm{Var}( Z_1)+\rho_{12} (1 - {\rho_{12}^2})^{ \frac12} \textrm{Cov}(Z_1, Z_ 2 ) + (1 - {\rho_{12}^2})\textrm{Var}(Z_ 2) \\ &&&= \rho_{12}^2 + (1-\rho_{12}^2) = 1 \\ \\ && \textrm{Cov}(Y_1, Y_2) &= \mathbb{E}((Y_1-0)(Y_2-0)) \\ &&&= \mathbb{E}(Z_1 \cdot (\rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2)) \\ &&&= \rho_{12} \mathbb{E}(Z_1^2) + (1-\rho_{12}^2)^{\frac12}\mathbb{E}(Z_1, Z_2) \\ &&&= \rho_{12} \\ \Rightarrow && \textrm{Corr}(Y_1, Y_2) &= \frac{\textrm{Cov}(Y_1, Y_2)}{\sqrt{\textrm{Var}(Y_1)\textrm{Var}(Y_2)}} \\ &&&= \frac{\rho_{12}}{1 \cdot 1} = \rho_{12} \end{align*} Suppose \(Y_3 =aZ_1 +bZ_2+cZ_3\) with \(\mathbb{E}(Y_3) = 0\) (must be true), \(\textrm{Var}(Y_3) = 1 = a^2+b^2+c^2\) and \(\textrm{Corr}(Y_1, Y_3) = \rho_{13}, \textrm{Corr}(Y_2, Y_3) = \rho_{23}\). \begin{align*} && \textrm{Corr}(Y_1,Y_3) &= \textrm{Cov}(Y_1, Y_3) \\ &&&= \textrm{Cov}(Z_1, aZ_1 +bZ_2+cZ_3) \\ &&&= a \\ \Rightarrow && a &= \rho_{13} \\ \\ && \textrm{Corr}(Y_2,Y_3) &= \textrm{Cov}(Y_2, Y_3) \\ &&&= \textrm{Cov}(\rho_{12}Z_1+(1-\rho_{12}^2)^\frac12Z_2, \rho_{13}Z_1 +bZ_2+cZ_3) \\ &&&= \rho_{12}\rho_{13}+(1-\rho_{12}^2)^\frac12b \\ \Rightarrow && \rho_{23} &= \rho_{12}\rho_{13}+(1-\rho_{12}^2)^\frac12b \\ \Rightarrow && b &= \frac{\rho_{23}-\rho_{12}\rho_{13}}{(1-\rho_{12}^2)^\frac12} \\ && c &= \sqrt{1-\rho_{13}^2-\frac{(\rho_{23}-\rho_{12}\rho_{13})^2}{(1-\rho_{12}^2)}} \end{align*} Finally, let \(X_i = \mu_i + \sigma_i Y_i\)