Problems

Solution:

1. \(\,\) \begin{align*} && \mathbb{E}(Z| a < Z < b) &= \mathbb{E}(Z\mathbb{1}_{(a,b)}) /\mathbb{E}(\mathbb{1}_{(a,b)}) \\ &&&= \int_a^b z \phi(z) \d z \Big / (\Phi(b) - \Phi(a)) \\ &&&= \frac{\int_a^b \frac{1}{\sqrt{2 \pi}}z e^{-\frac12 z^2} \d z}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left [-e^{-\frac12 z^2} \right]_a^b}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left (e^{-\frac12 a^2}-e^{-\frac12 b^2} \right)}{\Phi(b) - \Phi(a)} \\ \end{align*}
2. \(\,\) \begin{align*} && \mathbb{E}(X |X > 0) &= \mathbb{E}(\mu + \sigma Z | \mu + \sigma Z > 0) \\ &&&= \mathbb{E}(\mu + \sigma Z | Z > -\tfrac{\mu}{\sigma}) \\ &&&= \mathbb{E}(\mu| Z > -\tfrac{\mu}{\sigma})+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ &&&= \mu+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ \end{align*} Hence \begin{align*} &&\mathbb{E}(|X|) &= \mathbb{E}(X | X > 0)\mathbb{P}(X > 0) - \mathbb{E}(X | X < 0)\mathbb{P}(X < 0) \\ &&&=\left ( \mu+ \sigma \mathbb{E}(Z | Z > -\mu /\sigma)\right)(1-\Phi(-\mu/\sigma)) - \left ( \mu+ \sigma \mathbb{E}(Z | Z < -\mu /\sigma)\right)\Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2\pi}(1-\Phi(-\mu/\sigma))}(1-\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2 \pi} \Phi(-\mu/\sigma)} \Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \sqrt{\frac{2}{\pi}} \exp(-\tfrac12 \mu^2/\sigma^2) \end{align*} Finally, \begin{align*} && \textrm{Var}(|X|) &= \mathbb{E}(|X|^2) - [\mathbb{E}(|X|)]^2 \\ &&&= \mu^2 + \sigma^2 - m^2 \end{align*}

Solution: Let \(Y \sim Po(\lambda)\) \begin{align*} && 1 &= \sum_{k=1}^\infty \mathbb{P}(X = k ) \\ &&&= \sum_{k=1}^\infty A \frac{\lambda^k e^{-\lambda}}{k!}\\ &&&= Ae^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} \\ &&&= Ae^{-\lambda} \left (e^{\lambda}-1 \right) \\ \Rightarrow && A &= (1-e^{-\lambda})^{-1} \\ \\ && \E[X] &= \sum_{k=1}^{\infty} k \cdot \mathbb{P}(X=k) \\ &&&= A\sum_{k=1}^{\infty} k \frac{\lambda^k e^{-\lambda}}{k!} \\ &&&= A\E[Y] = A\lambda = \lambda(1-e^{-\lambda})^{-1} \\ \\ && \var[X] &= \E[X^2] - (\E[X])^2 \\ &&&= A\sum_{k=1}^{\infty} k^2 \frac{\lambda^k e^{-\lambda}}{k!} - \mu^2 \\ &&&= A\E[Y^2] - \mu^2 \\ &&&= A(\var[Y]+\lambda^2) - \mu^2 \\ &&&= A(\lambda + \lambda^2) - \mu^2 \\ &&&= A\lambda(1+\lambda) - \mu^2 \\ &&&= \mu(1+\lambda - \mu) \end{align*} Since \(A > 1\) we must have \(\mu > \lambda\) and since \(\var[X] > 0\) we must have \(1 + \lambda > \mu\) as required. If \(\lambda = 100\), then \(A \approx 1\) and \(P(X=\lambda) \approx P(Y = \lambda)\) and \(Y \approx N(\lambda, \lambda)\) so the value is approximately \(\displaystyle \int_{-\frac12}^{\frac12} \frac{1}{\sqrt{2\pi \lambda}} e^{-\frac{x^2}{2\lambda}} \d x \approx \frac{1}{\sqrt{200\pi}} = \frac{1}{\sqrt{630.\ldots}} \approx \frac{1}{25} = 0.04 \)

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 \(791\)