Problems

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Solution:

1. Note that \(\displaystyle F(y) = \mathbb{P}(X_i < y) = \int_0^y \lambda e^{-\lambda t} \d t = 1-e^{-\lambda y}\). Notice also that \begin{align*} G(y) &= \mathbb{P}(Y < y) \\ &= \mathbb{P}(\max_i(X_i) < y) \\ &= \mathbb{P}(X_i < y \text{ for all }i) \\ &= \prod_{i=1}^n \mathbb{P}(X_i < y) \\ &= \prod_{i=1}^n (1-e^{-\lambda y})\\ &= (1-e^{-\lambda y})^n \end{align*} as required.
2. \begin{align*} && \mathbb{P}(Y < L(\alpha)) &= \alpha \\ \Rightarrow && (1-e^{-\lambda L(\alpha)})^n &= \alpha \\ \Rightarrow && 1-e^{-\lambda L(\alpha)} &= \alpha^{\tfrac1n} \\ \Rightarrow && L(\alpha) &= -\frac{1}{\lambda}\ln \left (1-\alpha^{\tfrac1n} \right) \end{align*} Notice also: \begin{align*} && \mathbb{P}(Y > U(\alpha)) &= \alpha \\ \Rightarrow && 1 - (1-e^{-\lambda U(\alpha)})^n &= \alpha \\ \Rightarrow && U(\alpha) &= -\frac{1}{\lambda}\ln \left ( 1-(1-\alpha)^{\tfrac1n} \right) \end{align*}
3. \begin{align*} \lambda L(\alpha) &= -\ln \left (1-\alpha^{\tfrac1n} \right) \\ &= -\ln \left (1-e^{\tfrac1n \ln \alpha} \right) \\ &\approx - \ln \left ( 1 - 1 - \frac1n \ln \alpha\right) \tag{\(e^t \approx 1 + t\)} \\ &= -\ln \left ( \frac{1}{n} \ln \frac{1}\alpha \right) \\ &= - \ln \frac{1}{n} - \ln \left ( \ln \frac{1}{\alpha} \right )\\ &= \ln n - \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \end{align*} since if \(n\) is large, \(\frac{\ln \alpha}{n}\) is small.
4. The median is the value where \(\mathbb{P}(Y < M) = \frac12\), or in other words \(L(\frac12)\), but this is \(\approx \frac{\ln n - \ln (\ln 2)}{\lambda} \to \infty\). \begin{align*} && \lambda U(\alpha) &\approx \ln n - \ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right) \\ \Rightarrow && \lambda(U(\alpha) - L(\alpha)) &\approx -\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right)+ \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \\ \Rightarrow && U(\alpha) - L(\alpha) &\to \frac{1}{\lambda} \left ( \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right)-\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right ) \right) \end{align*} which doesn't depend on \(n\).
5. Suppose \(\alpha = \frac{1}{20}\) then \begin{align*} U(\alpha) - L(\alpha) &\approx \frac{1}{\lambda} \left (\ln \ln 20 - \ln \ln \frac{20}{19} \right) \\ &= \lambda^{-1} \left (\ln \ln 20 - \ln \ln (1 + \frac{1}{19}) \right) \\ &\approx \lambda^{-1} \left (\ln 3 - \ln \frac{1}{19} \right) \tag{\(\ln(1+t) \approx t\)} \\ &\approx \lambda^{-1} \ln 3 \cdot 19 \\ &\approx \lambda^{-1} (1 + 3) \\ &\approx 4\lambda^{-1} \end{align*} [Note that \(\ln \ln 20 - \ln \ln \frac{20}{19} = 4.0673\ldots\)]

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(Y = n) &= \mathbb{P}(X \in [n, n+1)) \\ &&&= \int_n^{n+1} \lambda e^{-\lambda x} \d x \\ &&&= \left [-e^{-\lambda x} \right]_n^{n+1} \\ &&&= e^{-\lambda n} - e^{-\lambda(n+1)} \\ &&&= e^{-\lambda n}(1- e^{-\lambda}) \end{align*}
2. \(,\) \begin{align*} && \mathbb{P}(Z < z) &= \sum_{i=0}^{\infty} \mathbb{P}(X \in (n, n+z)) \\ &&&= \sum_{i=0}^{\infty} \int_{n}^{n+z} \lambda e^{-\lambda x} \d x \\ &&&= \sum_{i=0}^{\infty} [-e^{-\lambda x}]_{n}^{n+z} \\ &&&= \sum_{i=0}^{\infty} (1-e^{-\lambda x})e^{-\lambda n} \\ &&&= \frac{1-e^{-\lambda x}}{1-e^{-\lambda}} \end{align*}
3. Give the cdf of \(Z\), we see that \(f_Z(z) = \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}}\) so \begin{align*} && \E[Z] &= \int_0^1 z \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}} \d z \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \int_0^1 ze^{-\lambda z} \d z \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \left ( \left [-\frac{1}{\lambda} ze^{-\lambda z} \right]_0^1+\int_0^1 \frac{1}{\lambda} e^{-\lambda z} \d z \right) \\ &&&= \frac{\lambda}{1-e^{-\lambda}} \left ( -\frac{e^{-\lambda}}{\lambda} + \frac{1-e^{-\lambda}}{\lambda^2} \right) \\ &&&= \frac{1-e^{-\lambda}(1+\lambda)}{\lambda (1-e^{-\lambda})} \end{align*}
4. \(\,\) \begin{align*} && \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2)&= \mathbb{P}(X \in (n+z_1, n+z_2) ) \\ &&&= \int_{n+z_1}^{n+z_2} \lambda e^{-\lambda x} \d x \\ &&&= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \end{align*} Note that \(\mathbb{P}(z_1 < Z < z_2) = \mathbb{P}( Z < z_2) -\mathbb{P}(Z< z_1) =\frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}}\) Therefore \begin{align*} && \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2) &= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \\ &&&= e^{-\lambda n}(1-e^{-\lambda}) \frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}} \\ &&&= \mathbb{P}(Y=n) \mathbb{P}(z_1 < Z < z_2) \end{align*} So they are independent, which is to be expected from the memorylessness property of the exponential distribution.

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Solution:

1. \(\mathbb{P}(X + Y \leq c) \) is the area between the \(x\)-axis, \(y\)-axis and the line \(x + y = c\). There are two cases for this: \[\mathbb{P}(X + Y \leq c) = \begin{cases} 0 & \text{ if } c \leq 0 \\ \frac{c^2}{2} & \text{ if } c \leq 1 \\ 1- \frac{(2-c)^2}{2} & \text{ if } 1 \leq c \leq 2 \\ 1 & \text{ otherwise} \end{cases}\]
2. \begin{align*} && \mathbb{P}((X + Y)^{-1} \leq t) &= 1- \mathbb{P}(X + Y \leq \frac1{t}) \\ \Rightarrow && f_{(X+Y)^{-1}}(t) &= 0 -\begin{cases} 0 & \text{ if } \frac1{t} \leq 0 \\ \frac{\d}{\d t}\frac{1}{2t^2} & \text{ if } \frac{1}{t} \leq 1 \\ \frac{\d}{\d t} \l 1- \frac{(2-\frac1t)^2}{2} \r & \text{ if } 1 \leq \frac{1}{t} \leq 2 \\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ (2-\frac1t)t^{-2} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ 2t^{-2}-t^{-3} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \end{align*} Therefore, \begin{align*} \E \Big(\dfrac1{X+Y}\Big) &= \int_{\frac12}^{\infty} t f_{(X+Y)^{-1}}(t) \, \d t \\ &= \int_{\frac12}^{1} t f_{(X+Y)^{-1}}(t) \, \d t + \int_{1}^{\infty} t f_{(X+Y)^{-1}}(t) \d t\\ &= \int_{\frac12}^{1} \l 2t^{-1} - t^{-2} \r \, \d t + \int_{1}^{\infty} t^{-2} \d t\\ &= \left [ 2 \ln (t) + t^{-1} \right]_{\frac12}^{1} + \left [ -t^{-1} \right ]_{1}^{\infty} \\ &= 1 + 2 \ln 2 -2 + 1 \\ &= 2 \ln 2 \end{align*}
3. \begin{align*} &&\mathbb{P} \l \frac{Y}{X} \leq c \r &= \mathbb{P}( Y \leq c X) \\ &&&= \begin{cases} 0 & \text{if } c \leq 0 \\ \frac{c}{2} & \text{if } 0 \leq c \leq 1 \\ 1-\frac{1}{2c} & \text{if } 1 \leq c \end{cases} \\ \\ \Rightarrow && \mathbb{P} \l \frac{X}{X+Y} \leq t\r &= \mathbb{P} \l \frac{1}{1+\frac{Y}{X}} \leq t\r \\ &&&= \mathbb{P} \l \frac{1}{t} \leq 1+\frac{Y}{X}\r \\ &&&= \mathbb{P} \l \frac{1}{t} - 1\leq \frac{Y}{X}\r \\ &&&= 1- \mathbb{P} \l \frac{Y}{X} \leq \frac{1}{t} - 1\r \\ &&&= 1 - \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t} - \frac{1}{2} & \text{if } 0 \leq \frac1{t} \leq 1 \\ 1-\frac{t}{2-2t} & \text{if } 1 \leq \frac1{t} \end{cases} \\ && f_{\frac{X}{X+Y}}(t) &= \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t^2} & \text{if } t \geq 1 \\ \frac{1}{2(1-t)^2} & \text{if } 0 \leq t \leq 1 \end{cases} \\ \Rightarrow && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^\infty t f(t) \d t \\ &&&= \int_0^1 \frac{1}{2(1-t)^2} \d t + \int_1^\infty \frac{1}{t^2} \d t \\ &&& = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \\ \\ && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^1 \int_0^1 \frac{x}{x+y} \d y\d x \\ &&&= \int_0^1 \l x \ln (x+1) - x \ln x \r \d x \\ &&&= \left [\frac{x^2}2 \ln(x+1) - \frac{x^2}{2} \ln(x) \right]_0^1 -\int_0^1 \l \frac{x^2}{2(x+1)} - \frac{x}{2} \r \d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x^2-1+1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x -1}{2} + \frac{1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \frac{1}{4} + \frac{1}{2} - \frac{\ln 2}{2} \\ &&&= \frac{1}{2} \end{align*} We can also notice that \(1 = \mathbb{E} \l \frac{X+Y}{X+Y} \r = \mathbb{E} \l \frac{X}{X+Y} \r + \mathbb{E} \l \frac{Y}{X+Y} \r = 2 \mathbb{E} \l \frac{X}{X+Y} \r\) so it's clearly true as long as we can show that the integral converges.

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(T > t + \delta t | T > t) &= \frac{\mathbb{P}(T < t + \delta t)}{\mathbb{P}(T > t )} \\ &&&= \frac{\int_t^{t+\delta t} f(s) \d s}{1-F(t)} \\ &&&\approx \frac{f(t)\delta t}{1-F(t)} \\ &&&= h(t) \delta t \end{align*}
2. If \(F(t) = t/a\) then \(f(t) = 1/a\) and \(h(t) = \frac{1/a}{1-t/a} = \frac{1}{a-t}\)
   

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3. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \frac{1}{t} \\ \Rightarrow && -\ln (1-F) &= \ln t + C\\ \Rightarrow && 1-F &= \frac{A}{t} \\ && F &= 1 - \frac{A}{t} \\ F(a) = 0: && F &= 1 - \frac{a}{t} \\ && f(t) &= \frac{a}{t^2} \end{align*}
4. (\(\Rightarrow\)) \begin{align*} && \frac{F'}{1-F} &= k \\ \Rightarrow && -\ln(1-F) &= kt+C \\ \Rightarrow && 1-F &= Ae^{-kt} \\ F(b) = 0: && 1 &= Ae^{-kb} \\ \Rightarrow && 1-F &= e^{-k(t-b)}\\ \Rightarrow && f &= ke^{-k(t-b)} \\ \end{align*} (\(\Leftarrow\)) \(f(t) = ke^{-k(t-b)} \Rightarrow F(t) = 1-e^{-k(t-b)}\) and the result is clear.
5. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \left ( \frac{\lambda}{\theta^{\lambda}} \right) t^{\lambda-1} \\ \Rightarrow && -\ln(1-F) &= \left ( \frac{t}{\theta} \right)^{\lambda} +C\\ \Rightarrow && F &= 1-A\exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ F(0) = 0: && 0 &= 1-A \\ \Rightarrow && F &= 1 - \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ \Rightarrow && f &= \lambda t^{\lambda -1} \theta^{-\lambda} \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \end{align*}

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Solution:

1. \begin{align*} && \frac12 &= \mathbb{P}(X \leq x_m) \\ \Leftrightarrow && \frac12 &= \mathbb{P}(e^X \leq e^{x_m} = y_m) \end{align*} Therefore the median is \(y_m = e^{x_m}\)
2. \begin{align*} && \mathbb{P}(Y \leq y) &= \mathbb{P}(e^X \leq y) \\ &&&= \mathbb{P}(X \leq \ln y) \\ &&&= F(\ln y) \\ \Rightarrow && f_Y(y) &= f(\ln y)/y \\ \\ && f'_Y(y) &= \frac{f'(\ln y) - f(\ln y)}{y^2} \end{align*} Therefore since the mode satisfies \(f'_Y = 0\) we must have \(f'(\ln \lambda ) = f(\ln \lambda)\)
3. This is the integral of the pdf of \(N(\mu + \sigma^2, \sigma^2)\) and therefore is clearly \(1\). \begin{align*} && \E[Y] &= \int_{-\infty}^{\infty} e^x \cdot \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)} \d x \\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (x - (x-\mu)^2/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp ((2x \sigma^2- (x-\mu)^2)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2+2\mu \sigma^2-\sigma^4)/(2\sigma^2)) \d x\\ &&&= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu+\sigma^2)^2)/(2\sigma^2)+\mu +\frac12\sigma^2) \d x\\ &&&= \e^{\mu +\frac12\sigma^2}\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp (-(x-\mu-\sigma^2)^2)/(2\sigma^2)) \d x\\ &&&= \e^{\mu +\frac12\sigma^2} \end{align*}
4. Notice that \(y_m = e^\mu < e^{\mu + \tfrac12 \sigma^2} = \E[Y]\), so it suffices to prove that \(\lambda < e^{\mu}\) Notice that \(f'(x) - f(x) = f(x)[-(x-\mu)/\sigma^2 - 1]\) and therefore \(\ln y - \mu = -\sigma^2\) so \(\lambda = e^{\mu - \sigma^2}\) which is clearly less than \(e^{\mu}\) as required.

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Solution:

1. 1. \(\,\) \begin{align*} && 0 &\leq f(t) &\leq M \\ \Rightarrow && \int_0^x 0 \d t &\leq \int_0^x f(t) \d t & \leq \int_0^x M \d x \\ \Rightarrow && 0 &\leq F(x) &\leq Mx \end{align*}
   2. \(\,\) \begin{align*} && \int_0^1 2g(x)F(x)f(x) \d x &= \left [ g(x) F(x)^2 \right] - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \\ &&&= g(1) - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \end{align*}
2. 1. \(\,\) \begin{align*} && 1 &= \int_0^1 kF(y)f(y) \d y \\ &&&= k\left [ \frac12 F(y)^2\right]_0^1 \\ &&&= \frac{k}{2} \\ \Rightarrow && k &= 2 \end{align*}
   2. \(\,\) \begin{align*} \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &\geq \int_0^1 y^n 2My f(y) \d y \\ &= 2M\int_0^1 y^{n+1} f(y) \d y \\ &= 2M \E[X^{n+1}] = 2M\mu_{n+1} \\ \\ \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &= 1 - \int_0^1 ny^{n-1} F(y)^2 \d y \\ &\geq 1 - \int_0^1 ny^{n-1}My F(y) \d y \\ &= 1 - M\int_0^1 ny^n F(y) \d y \\ &= 1 - M[\frac{n}{n+1}y^{n+1} F(y)]_0^1 + M\int_0^1\frac{n}{n+1} y^{n+1} f(y) \d y \\ &= 1 - \frac{nM}{n+1} + \frac{nM}{n+1} \mu_{n+1} \end{align*}
   3. Since \(\E[Y^{n-1}] \geq 0\) we must have \begin{align*} && 2M\mu_n \geq 1 + \frac{(n-1)M}{n}\mu_n - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \left (2M + \frac{(n-1)M}{n} \right) \geq 1 - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \frac{3Mn-M}{n} & \geq \frac{n-(n-1)M}{n} \\ \Rightarrow && \mu_n & \geq \frac{n-(n-1)M}{3Mn-M} \end{align*}

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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*}

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Solution: \begin{align*} && 1 &= \int_{-\infty}^{\infty} f(x) \d x \\ &&&= k[1 + \lambda] \\ \Rightarrow && k &= \frac{1}{1+\lambda} \\ \\ && \mu &= \int_{-\infty}^\infty x f(x) \d x \\ &&&= k \int_{-\infty}^\infty x \phi(x) \d x + k \lambda \int_{-\infty}^{\infty} x g(x) \d x \\ &&&= k \cdot 0 + k \lambda \cdot \frac{\lambda}{2} \\ &&&= \frac{\lambda^2}{2(1+\lambda)} \\ \\ && \E[X^2] &= \int_{-\infty}^\infty x^2 f(x) \d x \\ &&&= k \int_{-\infty}^\infty x^2 \phi(x) \d x + k \lambda \int_{-\infty}^{\infty} x^2 g(x) \d x \\ &&&= k \cdot 1 + k \lambda \int_0^{\lambda} \frac{x^2}{\lambda} \d \lambda \\ &&&= k + \frac{k \lambda^3}{3} \\ &&&= \frac{3+\lambda^3}{3(1+\lambda)} \\ && \var[X] &= \frac{3+\lambda^3}{3(1+\lambda)} - \frac{\lambda^4}{4(1+\lambda)^2} \\ &&& = \frac{(3+\lambda^3)4(1+\lambda) - 3\lambda^4}{12(1+\lambda)^2} \\ &&&= \frac{\lambda^4+4\lambda^3+12\lambda + 12}{12(1+\lambda)^2} \end{align*}

1. \(\,\)
   

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2. \(\,\) \begin{align*} && \mathbb{P}(X \leq x) &= \int_{-\infty}^x f(x) \d x \\ &&&= \begin{cases} \frac13 \Phi(x) & \text{if } x < 0 \\ \frac13\Phi(x) + \frac13x & \text{if } 0 \leq x \leq 2 \\ \frac13 \Phi(x) + \frac23 & \text{if } 2 < x \end{cases} \end{align*} When \(\lambda = 2\), \(\mu = \frac{4}{6} = \frac23\), \(\sigma^2 = \frac{16+32+24+12}{12 \cdot 9} = \frac{7}{9}\), so \(\mu + 2 \sigma = \frac23 + \frac{2\sqrt7}{3}>2\). Therefore \begin{align*} && \P(0 < X < \mu + 2\sigma) &= \frac13 \Phi\left (\frac{2+2\sqrt{7}}{3} \right) + \frac23 - \Phi(0) \\ &&&= \tfrac13 \cdot 0.9921 +\tfrac23 - \tfrac12 \\ &&&= 0.4974 \end{align*}

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Solution:

\begin{align*} && 1 &= \int_0^1 f(x) \d x \\ &&&= \int_0^1 kx e^{-ax^2} \d x \\ &&&= \left [-\frac{k}{2a}e^{-ax^2} \right]_0^1 \\ &&&= \frac{k(1-e^{-a})}{2a} \\ \Rightarrow && k &= \frac{2a}{1-e^{-a}} \end{align*} To find the mode, we want \(f'(x) = 0\), ie \begin{align*} && 0 &= f'(x) \\ &&&= -2kax^2e^{-ax^2} + k e^{-ax^2} \\ &&&= ke^{-ax^2} \left (1-2ax^2 \right)\\ \end{align*} So either \(m = \frac{1}{\sqrt{2a}}\) (if \(a > \frac12\)) or \(f(x)\) is increasing and the mode is \(m = 1\) (if \(a < \frac12\)). \begin{align*} && \frac12 &= \int_0^h f(x) \d x \\ &&&= \left [ -\frac{e^{-ax^2}}{1-e^{-a}} \right]_0^h \\ &&&= \frac{1-e^{-ah^2}}{1-e^{-a}} \\ \Rightarrow && e^{-ah^2}&= 1-\frac12(1-e^{-a}) \\ \Rightarrow && -a h^2 &= \ln \left ( \frac12(1+e^{-a}) \right) \\ \Rightarrow && h &= \sqrt{-\frac1a \ln (\tfrac12(1+e^{-a}))} \end{align*} \(h > m\) already means \(a > \frac12\) so \begin{align*} && h &> m \\ \Leftrightarrow &&\sqrt{-\frac1a \ln (\tfrac12(1+e^{-a}))} &> \frac{1}{\sqrt{2a}} \\ \Leftrightarrow && -\ln (\tfrac12(1+e^{-a})) &> \frac12 \\ \Leftrightarrow && e^{-1/2} & > \frac12(1+e^{-a}) \\ \Leftrightarrow && 2e^{-1/2}-1 &>e^{-a} \\ \Leftrightarrow && \ln(2e^{-1/2}-1) &>-a \\ \Leftrightarrow && a& > -\ln(2e^{-1/2}-1) \\ \end{align*} Noting that \begin{align*} && -\ln(2e^{-1/2} - 1) &= -\ln \left (\frac{2-\sqrt{e}}{e^{1/2}} \right) \\ &&&= \frac12 -\ln(\underbrace{2 - \sqrt{e}}_{<1}) \\ &&&> \frac12 \end{align*} If \(a > -\ln(2e^{-1/2}-1)\) then \begin{align*} && \mathbb{P}(X > m | X < h) &= \frac{\mathbb{P}(m < X < h)}{\mathbb{P}(X < h)} \\ &&&= \frac{e^{-am^2}-e^{-ah^2}}{1-e^{-ah^2}} \\ &&&= \frac{e^{-a\frac{1}{2a}}-e^{\ln \left ( \frac12(1+e^{-a}) \right)}}{1-e^{\ln \left ( \frac12(1+e^{-a}) \right)}} \\ &&&= \frac{e^{-1/2}-\frac12(1+e^{-a})}{1-\frac12(1+e^{-a})} \\ &&&= \frac{2e^{-1/2}-1-e^{-a}}{1-e^{-a}} \end{align*} as required.

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Solution: Let \(H_i\) be the random variable of how many heads the \(i\)th coin throws on a given day. Then \(H_i \sim B(M,p)\), and the probability that a given coin produces fewer than \(m\) heads is \(p_h = \P(H_i < m)\) Let \(C\) be the random variable the number of coins producing fewer than \(m\) heads, then \(C \sim B(L, p_h)\). The probability that more than \(l\) of the coins produce fewer than \(m\) heads is therefore \(\P(C > l)\). Finally, the probability that on exactly \(k\) days more than \(l\) of the coins will produce fewer than \(m\) heads is: \[ \binom{K}{k} \cdot \P(C > l)^k \cdot (1-\P(C > l))^{K-k} \] Let's start by assuming that all our Binomials can be approximated by a normal distribution. \(B(M,p) \approx N(Mp, Mp(1-p))\) and so: \begin{align*} p_h &= \P(H_i < m) \\ &\approx \P( \sqrt{Mp(1-p)}Z+Mp < m-\frac12) \\ &= \P \l Z < \frac{2m-2Mp-1}{2\sqrt{Mp(1-p)}} \r \\ &= \Phi\l\frac{2m-2Mp-1}{2\sqrt{Mp(1-p)}} \r \end{align*} \(B(L, p_h) \approx B \l L, \P \l Z < \frac{2m-2Mp-1}{2\sqrt{Mp(1-p)}} \r\r = B(L, \frac{h}{L}) \approx N(h, \frac{h(L-h)}{L})\) Therefore \begin{align*} \P(C > l) &= 1-\P(C \leq l) \\ &\approx 1- \P \l \sqrt{\frac{h(L-h)}{L}} Z + h \leq l+\frac12 \r \\ &= 1 - \P \l Z \leq \frac{2l-2h+1}{2\sqrt{\frac{h(L-h)}{L}}}\r \\ &= 1- \Phi\l \frac{2l-2h+1}{2\sqrt{\frac{h(L-h)}{L}}} \r \\ &= \Phi\l \frac{2h-2l-1}{2\sqrt{\frac{h(L-h)}{L}}} \r \end{align*} If we can approximate \(\sqrt{1-\frac{h}{L}}\) by \(1\) then we obtain the approximation in the question. Alternatively, \(B(L, \frac{h}{L}) \approx Po(h)\) and \(Po(h) \approx N(h,h)\) so we obtain: \begin{align*} \P(C > l) &= 1-\P(C \leq l) \\ &\approx 1 - \P(\sqrt{h} Z +h < l + \frac12) \\ &= 1 - \P \l Z < \frac{2l-2h+1}{2\sqrt{h}} \r \\ &= \Phi \l \frac{2h - 2l -1}{2\sqrt{h}}\r \end{align*} as required. [I think this is what the examiners expected]. Considering the case \(K=7\), \(k=2\), \(L=500\), \(l=4\), \(M=100\), \(m=48\) and \(p=0.6\), we have the first normal approximation depends on \(Mp\) and \(M(1-p)\) being large. They are \(60\) and \(40\) respectively, so this is likely a good approximation. The first approximation finds that \begin{align*} h &= 500 \cdot \Phi \l \frac{2 \cdot 48 - 2 \cdot 60 - 1}{2\sqrt{24}} \r \\ &= 500 \cdot \Phi \l \frac{2 \cdot 48 - 2 \cdot 60 - 1}{2\sqrt{24}} \r \\ &= 500 \cdot \Phi \l \frac{-25}{2 \sqrt{24}} \r \\ &\approx 500 \cdot \Phi (-2.5) \\ &= 500 \cdot 0.0062 \\ &\approx 3.1 \end{align*} The second binomial approximation will be good if \(500 \cdot \frac{3.1}{500} = 3.1\) is large, but this is quite small. Therefore, we shouldn't expect this to be a good approximation. However, since \(m = 48\) is far from the mean (in a normalised sense), we might expect the percentage error to be large. [Alternatively, using what I expect the desired approach] The approximation of \(B(L, \frac{h}{L}) \approx Po(h)\) is acceptable since \(n>50\) and \(h < 5\). The approximation of \(Po(h) \sim N(h,h)\) is not acceptable since \(h\) is small (in particular \(h < 15\)) Finally, we can compute all these values exactly using a modern calculator. \begin{array}{l|cc} & \text{correct} & \text{approx} \\ \hline p_h & 0.005760\ldots & 0.005362\ldots \\ \P(C > l) & 0.164522\ldots & 0.133319\ldots \\ \text{ans} & 0.231389\ldots & 0.182516\ldots \end{array} We can also see how the errors propagate, by doing the calculations assuming the previous steps are correct, and also including the Poisson step. \begin{array}{lccc} & \text{correct} & \text{approx} & \text{using approx } p_h \\ \hline p_h & 0.005760\ldots & 0.005362\ldots & - \\ \P(C > l)\quad [Po(h)] & 0.164522\ldots & 0.165044\ldots & 0.134293\ldots \\ \P(C > l)\quad [N(h,h)] & 0.164522\ldots & 0.169953\ldots & 0.133319\ldots \\ \P(C > l)\quad [N(h,h(1-\frac{h}{L})] & 0.164522\ldots & 0.169255\ldots & 0.132677\ldots \\ \text{ans} & 0.231389\ldots & 0.231389\ldots \end{array} By doing this, we discover that the largest errors are actually coming not from approximating the second approximation but from the small absolute (but large relative error) in the first approximation. This is, in fact, a coincidence; we can observe it by investigating the specific values being used. The first approximation looks as follows:

You might not be able to tell, but there's actually two plots on this chart. However, let's zoom in on the area we are worried about: We can see there are small differences, which could be large in percentage terms. (As we found when we computed them directly). First, we can immediately see that if we just look at the distribution of \(B(L, p_h)\) and \(B(L, p_{h_\text{approx}})\) we get quite different results, even before we do any approximations. If we plot the probability distribution of \(B(L, p_h)\) vs \(N(Lp_h, Lp_h(1-p_h))\) we find that it is not a great approximation. However, the CDF happens to be a very good approximation *just* for the value we care about. Very lucky, but not possible for someone sitting STEP to know at the time!

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Solution: \begin{align*} && G(a) &= \frac{F(a)}{2-F(a)}\\ &&&= 0 \tag{\(F(a)= 0\)}\\ \\ && G(b) &= \frac{F(b)}{2-F(b)} \\ &&&= \frac{1}{2-1} = 1 \tag{\(F(b)=1\)}\\ \\ && G'(y) &= \frac{F'(y)(2-F(y))+F(y)F'(y)}{(2-F(y))^2} \\ &&&= \frac{2F'(y)}{(2-F(y))^2} \geq 0 \tag{\(F'(y) \geq 0\)} \end{align*} \begin{align*} && 0 \leq F(y)\leq1\\ \Leftrightarrow&& 1\leq 2-F(y) \leq 2\\ \Leftrightarrow &&1 \leq (2-F(y))^2 \leq 4\\ \Leftrightarrow && 1 \geq \frac{1}{(2-F(y))^2} \geq \frac14 \\ \Leftrightarrow && 2 \geq \frac{2}{(2-F(y))^2} \geq\frac12 \end{align*} \begin{align*} && \mathbb{E}(Y) &= \int_a^b y G'(y) \d y \\ &&&= \int_a^b y F'(y) \underbrace{\frac{2}{(2-F(y))^2}}_{\in [\frac12, 2]} \d y \\ &&&\leq 2 \E[X] \\ &&&\geq \frac12 \E[X]\\ \\ && \E[Y^2] &\leq 2\E[X^2] \\ && \E[Y^2] &\geq \frac12\E[X^2] \\ \\ \Rightarrow && \var[Y] &= \E[Y^2]-\E[Y]^2 \\ &&& \leq 2 \E[X^2] - (\tfrac12\E[X])^2 \\ &&&= 2 \var[X] + \tfrac74(\E[X])^2 \end{align*}

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Solution:

1. Suppose the needle's center is \(x\) cm from the nearest line and makes an angle of \(\theta\). Then if \(\sin \theta > x\) it will cross the line, otherwise it will not. Given that \(\theta \sim U(0, \frac{\pi}{2})\), we can see that \begin{align*} && \mathbb{P}(\text{needle crosses}) &= \mathbb{P}(\sin \theta > x) \\ &&&= \mathbb{P}(\theta > \sin^{-1} x) \\ &&&= 1-\frac{2\sin^{-1} x}{\pi} \end{align*}
2. The length of the short segment is \(L = 1 - \frac{x}{\sin \theta}\) and \(\theta \sim U(\sin^{-1} x, \frac{\pi}{2})\). So \begin{align*} && F_L(l) &= \mathbb{P}(L < l) \\ &&&= \mathbb{P}\left (1 - \frac{x} {\sin \theta} < l\right) \\ &&&= \mathbb{P}\left ( \sin \theta < \frac{x}{1-l}\right) \\ &&&= \mathbb{P}\left (\theta < \sin^{-1} \frac{x}{1-l}\right) \\ &&&= \frac{ \sin^{-1} \frac{x}{1-l} - \sin^{-1} x }{\frac{\pi}{2} - \sin^{-1}x} \end{align*}
3. The needle (with probability \(1\)) cannot hit \(2\) lines, so let's only consider the line it's nearest too. The distance to this line is uniform on \([0,1]\), and the so we want to calculate. \begin{align*} && \mathbb{P}(\text{needle crosses}) &= \int_0^1 \left (1 - \frac{2\sin^{-1}x}{\pi} \right) \d x \\ &&&= 1 - \frac{2}{\pi} \int_0^1 \sin^{-1} x \d x\\ &&&= 1 - \frac{2}{\pi} \left ( \frac{\pi}{2} - 1 \right) \\ &&&= \frac{2}{\pi} \end{align*} Therefore the probability it misses is \(1 - \frac{\pi}{2}\)