1 problem found
Let \(X\) be a standard normal random variable. If \(M\) is any real number, the random variable \(X_{M}\) is defined in terms of \(X\) by \[ X_{M}=\begin{cases} X & \mbox{if }X < M,\\ M & \mbox{if }X\geqslant M. \end{cases} \] Show that the expectation of \(X_{M}\) is given by \[ \mathrm{E}(X_{M})=-\phi(M)+M(1-\Phi(M)), \] where \(\phi\) is the probability density function, and \(\Phi\) is the cumulative distribution function of \(X\). Fifty times a year, 1024 tourists disembark from a cruise liner at the port of Slaka. From there they must travel to the capital either by taxi or by bus. Officials of HOGPo are equally likely to direct a tourist to the bus station or to the taxi rank. Each bus of the bus coorperative holds 31 passengers, and the coorperative currently runs 16 buses. The bus coorperative makes a profit of 1 vloska for each passenger carried. It carries all the passengers it can, with any excess being (eventually) transported by taxi. What is the largest annual bribe the bus coorperative should consider paying to HOGPo in order to be allowed to run an extra bus?
Solution: Let \(X \sim N(0,1)\), and $\displaystyle X_{M}=\begin{cases} X & \text{if }X < M,\\ M & \text{if }X\geqslant M. \end{cases} $. Then we can calculate: \begin{align*} \mathbb{E}[X_M] &= \int_{-\infty}^M xf_X(x)\,dx + M\mathbb{P}(X \geq M) \\ &= \int_{-\infty}^M x \frac1{\sqrt{2\pi}}e^{-\frac12x^2}\,dx + M\mathbb{P}(X \geq M) \\ &= \left [ -\frac{1}{\sqrt{2\pi}}e^{-\frac12x^2} \right ]_{-\infty}^M + M (1-\mathbb{P}(X < M)) \\ &= -\phi(M) + M(1-\Phi(M)) \end{align*} Let \(B \sim B\left (1024, \frac12 \right)\) be the number of potential bus passengers. Then \(B \approx N(512, 256) = N(512, 16^2)\) which is a good approximation since both \(np\) and \(nq\) are large. The question is asking us, how much additional profit would the bus company get if they ran an additional bus. Currently each week they is (on average) \(512\) passengers worth of demand, but they can only supply \(496\) seats, so we should expect that there is demand for another bus. The question is how much that demand is worth. Using the first part of the question, we can see that their profit is something like a `capped normal', \(X_M\), except we are scaled and with a different cap. So we are interested in $\displaystyle Y_{M}=\begin{cases} B & \mbox{if }B< M,\\ M & \mbox{if }B\geqslant M. \end{cases}\(, but since \)B \approx N\left (512,16^2\right)$ this is similar to \begin{align*} Y_{M}&=\begin{cases} 16X+512 & \mbox{if }16X+512< M,\\ M & \mbox{if }16X+512\geq M. \end{cases} \\ &= \begin{cases} 16X+512 & \mbox{if }X< \frac{M-512}{16},\\ M & \mbox{if }X \geq \frac{M-512}{16}. \end{cases} \\ &= 16X_{\frac{M-512}{16}} + 512\end{align*} We are interested in \(\mathbb{E}[Y_{16\times31}]\) and \(\mathbb{E}[Y_{17\times31}]\), which are \(16\mathbb{E}[X_{-1}]+512\) and \(16\mathbb{E}[Y_{\frac{15}{16}}]+512\) Since \(\frac{15}{16} \approx 1\), lets look at \(16(\mathbb{E}[X_1] - \mathbb{E}[X_{-1}])\) \begin{align*} \mathbb{E}[X_1] - \mathbb{E}[X_{-1}] &= \left ( -\phi(1) + 1-\Phi(1)\right) - \left ( - \phi(-1) -(1 - \Phi(-1)) \right ) \\ &= -\phi(1) + \phi(-1) + 1-\Phi(1) + 1 - \Phi(-1) \\ &= 1 - \Phi(1) + \Phi(1) \\ &= 1 \end{align*} Therefore the extra \(31\) will fill roughly \(16\) of them. (This is a slight overestimate, which is worth bearing in mind). A better approximation might be that \(\mathbb{E}[X_t] - \mathbb{E}[X_{-1}] = \frac{t +1}{2}\) for \(t \approx 1\), (since we want something increasing). This would give us an approximation of \(15.5\), which is very close to the `true' answer. Therefore, over \(50\) bus runs, we should earn roughly \(800\) vloska extra from an additional bus. (Again an overestimate, and with an uncertain pay-off, they should consider offering maybe \(600\)). Since this is the future, we can quite easily calculate the exact values using the binomial distribution on a computer. This gives the true value as \(15.833\), and so they should pay up to \(791\)