The diagnostic test AL has a probability 0.9 of giving a positive result when applied to a person suffering from the rare disease mathematitis. It also has a probability 1/11 of giving a false positive result when applied to a non-sufferer. It is known that only \(1\%\) of the population suffer from the disease. Given that the test AL is positive when applied to Frankie, who is chosen at random from the population, what is the probability that Frankie is a sufferer? In an attempt to identify sufferers more accurately, a second diagnostic test STEP is given to those for whom the test AL gave a positive result. The probablility of STEP giving a positive result on a sufferer is 0.9, and the probability that it gives a false positive result on a non-sufferer is \(p\). Half of those for whom AL was positive and on whom STEP then also gives a positive result are sufferers. Find \(p\).
Solution: \begin{align*} \mathbb{P}(M | P_{AL}) &= \frac{\mathbb{P}(M \cap P_{AL})}{\mathbb{P}(P_{AL})} \\ &= \frac{\frac{1}{100} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} + \frac{99}{100} \frac{1}{11}} \\ &= \frac{\frac{9}{10}}{\frac{9}{10} + \frac{9}{1}} \\ &= \frac{9}{99} = \frac{1}{11} \\ \end{align*} \begin{align*} && \frac12 &= \mathbb{P}(M | P_{STEP}, P_{AL}) \\ &&&= \frac{\frac{1}{100} \frac{9}{10} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} \frac{9}{10} + \frac{99}{100} \frac{1}{11}p} \\ &&&= \frac{81}{81+900p} \\ \Rightarrow && p &= \frac{81}{900} = \frac{9}{100} \end{align*} Therefore \(p = 9\%\)
A random variable \(X\) has the probability density function \[ \mathrm{f}(x)=\begin{cases} \lambda\mathrm{e}^{-\lambda x} & x\geqslant0,\\ 0 & x<0. \end{cases} \] Show that $${\rm P}(X>s+t\,\vert X>t) = {\rm P}(X>s).$$ The time it takes an assistant to serve a customer in a certain shop is a random variable with the above distribution and the times for different customers are independent. If, when I enter the shop, the only two assistants are serving one customer each, what is the probability that these customers are both still being served at time \(t\) after I arrive? One of the assistants finishes serving his customer and immediately starts serving me. What is the probability that I am still being served when the other customer has finished being served?
Solution: \begin{align*} && \mathbb{P}(X > t) &= \int_t^{\infty} \lambda e^{-\lambda x} \d x\\ &&&= \left[ -e^{-\lambda x} \right]_t^\infty \\ &&&= e^{-\lambda t}\\ \\ && \mathbb{P}(X > s + t | X > t) &= \frac{\mathbb{P}(X > s + t)}{\mathbb{P}(X > t)} \\ &&&= \frac{e^{-(s+t)\lambda}}{e^{-t\lambda}} \\ &&&= e^{-s\lambda} = \mathbb{P}(X > s) \end{align*} The probability both are still being served (independently) is \(\mathbb{P}(X > t)^2 = e^{-2\lambda t}\). The probability is exactly \(\frac12\). The property we proved in the first part of the questions shows the distribution is memoryless, ie we are both experiencing samples from the same distribution. Therefore we are equally likely to finish first.
The staff of Catastrophe College are paid a salary of \(A\) pounds per year. With a Teaching Assessment Exercise impending it is decided to try to lower the student failure rate by offering each lecturer an alternative salary of \(B/(1+X)\) pounds, where \(X\) is the number of his or her students who fail the end of year examination. Dr Doom has \(N\) students, each with independent probability \(p\) of failure. Show that she should accept the new salary scheme if $$A(N+1)p < B(1-(1-p)^{N+1}).$$ Under what circumstances could \(X\), for Dr Doom, be modelled by a Poisson random variable? What would Dr Doom's expected salary be under this model?
Solution: \begin{align*} && \E[\text{salary}] &= B\sum_{k=0}^N \frac{1}{1+k}\binom{N}{k}p^k(1-p)^{N-k} \\ \\ && (q+x)^N &= \sum_{k=0}^N \binom{N}{k}x^kq^{N-k} \\ \Rightarrow && \int_0^p(q+x)^N \d x &= \sum_{k=0}^N \binom{N}{k} \frac{p^{k+1}}{k+1}q^{N-k} \\ && \frac{(q+p)^{N+1}-q^{N+1}}{N+1} &= \frac{p}{B} \E[\text{salary}] \\ \Rightarrow && \E[\text{salary}] &= B\frac{1-q^{N+1}}{p(N+1)} \end{align*} Therefore if \(Ap(N+1) < B(1-(1-p)^{N+1})\) the expected value of the new salary is higher. (Whether or not the new salary is worth it in a risk adjusted sense is for the birds). We could model \(X\) by a Poisson random variable if \(N\) is large and \(Np = \lambda \) is small. Suppose \(X \approx Po(\lambda)\) then \begin{align*} \E \left [\frac{B}{1+X} \right] &= B\sum_{k=0}^\infty \frac{1}{1+k}\frac{e^{-\lambda}\lambda^k}{k!} \\ &= \frac{B}{\lambda} \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^{k+1}}{(k+1)!} \\ &= \frac{B}{\lambda}e^{-\lambda}(e^{\lambda}-1) \\ &= \frac{B(1-e^{-\lambda})}{\lambda} = B \frac{1-e^{-Np}}{Np} \end{align*}
Let $$ {\rm f}(x)=\sin^2x + 2 \cos x + 1 $$ for \(0 \le x \le 2\pi\). Sketch the curve \(y={\rm f}(x)\), giving the coordinates of the stationary points. Now let $$ \hspace{0.6in}{\rm g}(x)={a{\rm f}(x)+b \over c{\rm f}(x)+d} \hspace{0.8in} ad\neq bc\,,\; d\neq -3c\,,\; d\neq c\;. $$ Show that the stationary points of \(y={\rm g}(x)\) occur at the same values of \(x\) as those of \(y={\rm f}(x)\), and find the corresponding values of \({\rm g}(x)\). Explain why, if \(d/c <-3\) or \(d/c>1\), \(|{\rm g}(x)|\) cannot be arbitrarily large.
Let $$ {\rm I}(a,b) = \int_0^1 t^{a}(1-t)^{b} \, \d t \; \qquad (a\ge0,\ b\ge0) .$$
Solution:
The value \(V_N\) of a bond after \(N\) days is determined by the equation $$ V_{N+1} = (1+c) V_{N} -d \qquad (c>0, \ d>0), $$ where \(c\) and \(d\) are given constants. By looking for solutions of the form \(V_T= A k^T + B\) for some constants \(A,B\) and \(k\), or otherwise, find \(V_N\) in terms of \(V_0\). What is the solution for \(c=0\)? Show that this is the limit (for fixed \(N\)) as \(c\rightarrow 0\) of your solution for \(c>0\).
Solution: Suppose \(V_T = Ak^T + B\), then \begin{align*} && Ak^{T+1}+B &= (1+c)(Ak^T+B) - d \\ \Rightarrow && (k-1-c)Ak^T &= cB -d \\ \Rightarrow && k &= 1+c \\ && B &= \frac{d}{c} \\ && A &= V_0 - B \\ \Rightarrow && V_N &= (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \end{align*} When \(c = 0\), \(V_{N+1} = V_N - d \Rightarrow V_N = V_0 - Nd\). \begin{align*} \lim_{c \to 0} \left ( (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \right) &= \lim_{c \to 0} \left ( \frac{(V_0 c-d)(1+c)^N + d}{c} \right ) \\ &= \lim_{c \to 0} \left ( \frac{V_0c - d-Ncd+NV_0c^2 + o(c^2) + d}{c} \right ) \\ &= \lim_{c \to 0} \left ( V_0 - Nd + o(c) \right ) \\ &= V_0 - Nd \end{align*}
Show that the equation (in plane polar coordinates) \(r=\cos\theta\), for $-\frac{1}{2}\pi \le \theta \le \frac{1}{2}\pi$, represents a circle. Sketch the curve \(r=\cos2\theta\) for \(0\le\theta\le 2\pi\), and describe the curves \(r=\cos2n\theta\), where \(n\) is an integer. Show that the area enclosed by such a curve is independent of \(n\). Sketch also the curve \(r=\cos3\theta\) for \(0\le\theta\le 2\pi\).
The exponential of a square matrix \({\bf A}\) is defined to be $$ \exp ({\bf A}) = \sum_{r=0}^\infty {1\over r!} {\bf A}^r \,, $$ where \({\bf A}^0={\bf I}\) and \(\bf I\) is the identity matrix. Let $$ {\bf M}=\left(\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0 \end{array} \right) \,. $$ Show that \({\bf M}^2=-{\bf I}\) and hence express \(\exp({\theta {\bf M}})\) as a single \(2\times 2\) matrix, where \(\theta\) is a real number. Explain the geometrical significance of \(\exp({\theta {\bf M}})\). Let $$ {\bf N}=\left(\begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right) \,. $$ Express similarly \(\exp({s{\bf N}})\), where \(s\) is a real number, and explain the geometrical significance of \(\exp({s{\bf N}})\). For which values of \(\theta\) does $$ \exp({s{\bf N}})\; \exp({\theta {\bf M}})\, = \, \exp({\theta {\bf M}})\;\exp({s{\bf N}}) $$ for all \(s\)? Interpret this fact geometrically.
Solution: \begin{align*} \mathbf{M}^2 &= \begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}^2 \\ &= \begin{pmatrix} 0 \cdot 0 + (-1) \cdot 1 & 0 \cdot (-1) + (-1) \cdot 0 \\ 1 \cdot 0 + 0 \cdot 1 & 1 \cdot (-1) + 0 \cdot 0 \end{pmatrix} \\ &= \begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix} \\ &= - \mathbf{I} \end{align*} \begin{align*} \exp(\theta \mathbf{M}) &= \sum_{r=0}^\infty \frac1{r!} (\theta \mathbf{M})^r \\ &= \sum_{r=0}^\infty \frac{1}{r!} \theta^r \mathbf{M}^r \\ &= \cos \theta \mathbf{I} + \sin \theta \mathbf{M} \\ &= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{align*} This is a rotation of \(\theta\) degrees about the origin. \begin{align*} && \mathbf{N}^2 &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}^2 \\ && &= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \\ \Rightarrow && \exp(s\mathbf{N}) &= \sum_{r=0}^\infty \frac{1}{r!} (s\mathbf{N})^r \\ &&&= \mathbf{I} + s \mathbf{N} \\ &&&= \begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix} \end{align*} This is a shear, leaving the \(y\)-axis invariant, sending \((1,1)\) to \((1+s, 1)\). Suppose those matrices commute, for all \(s\), ie \begin{align*} && \begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} &= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix} \\ \Rightarrow && \begin{pmatrix} \cos \theta - s \sin \theta & -\sin \theta + s \cos \theta \\ \sin \theta & \cos \theta \end{pmatrix} &= \begin{pmatrix} \cos \theta & s \cos \theta - \sin \theta \\ \sin \theta & s \sin \theta + \cos \theta \end{pmatrix} \\ \Rightarrow && \sin \theta &= 0 \\ \Rightarrow && \theta &=n \pi, n \in \mathbb{Z} \end{align*} Clearly it doesn't matter when we do nothing. If we are rotating by \(\pi\) then it also doesn't matter which order we do it in as the stretch happens in both directions equally.
Sketch the graph of \({\rm f}(s)={ \e}^s(s-3)+3\) for \(0\le s < \infty\). Taking \({\e\approx 2.7}\), find the smallest positive integer, \(m\), such that \({\rm f}(m) > 0\). Now let $$ {\rm b}(x) = {x^3 \over \e^{x/T} -1} \, $$ where \(T\) is a positive constant. Show that \({\rm b}(x)\) has a single turning point in \(0 < x < \infty\). By considering the behaviour for small \(x\) and for large \(x\), sketch \({\rm b}(x)\) for \(0\le x < \infty\). Let $$ \int_0^\infty {\rm b}(x)\,\d x =B, $$ which may be assumed to be finite. Show that \(B = K T^n\) where \(K\) is a constant, and \(n\) is an integer which you should determine. Given that \(\displaystyle{B \approx 2 \int_0^{Tm} {\rm b}(x) {\,\rm d }x}\), use your graph of \({\rm b}(x)\) to find a rough estimate for \(K\).