Problems

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Solution: A circle radius \(y\) has a number of supermarkets \(X\) where \(X \sim Po(k \pi y^2)\). \[ \mathbb{P}(X = 0) = e^{-k\pi y^2} \frac{1}{0!} = e^{-k\pi y^2} \] The probability \(\mathbb{P}(Y < y) = 1-\mathbb{P}(Y \geq y) = 1-e^{-k\pi y^2}\), and in particular \(f_Y(y) = 2k\pi y e^{-k\pi y^2}\) (by differentiating). \begin{align*} && \mathbb{E}(Y) &= \int_0^\infty yf_Y(y) \d y \\ &&&= \int_0^\infty 2\pi y^2 k e^{-\pi k y^2} \d y \\ \sigma^2 = \frac{1}{2k\pi}:&&&= \pi k \sqrt{2 \pi}\sigma \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sigma }y^2 e^{-\frac12 \cdot 2\pi k y^2} \d y \\ &&&=\pi k \sqrt{2 \pi}\sigma \mathbb{E}\left (N(0, \sigma^2)^2 \right) \\ &&&= \pi k \sqrt{2 \pi}\sigma\sigma^2 \\ &&&= \pi k \sqrt{2 \pi} \frac{1}{(2k\pi)^{3/2}} \\ &&&= \frac{1}{2\sqrt{k}} \end{align*} \begin{align*} && \mathbb{E}(Y^2) &= \int_0^\infty y^2f_Y(y) \d y \\ &&&= \int_0^\infty 2\pi y^3 k e^{-\pi k y^2} \d y \\ &&&= \int_0^{\infty}y^2 2y \pi k e^{-\pi k y^2} \d y \\ \\ &&&= \left [-y^2 e^{-\pi k y^2}\right]_0^{\infty}+\int_0^\infty 2ye^{-\pi k y^2} \d y \\ &&&= \left [-\frac{1}{\pi k}e^{-\pi k y^2} \right]_0^{\infty} \\ &&&= \frac{1}{\pi k} \\ \Rightarrow && \textrm{Var}(Y) &= \mathbb{E}(Y^2) - \left [ \mathbb{E}(Y)\right]^2 \\ &&&= \frac{1}{\pi k} - \frac{1}{4k} \\ &&&= \frac{4 - \pi}{4\pi k} \end{align*}

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Solution: First notice that \begin{align*} && 1 &= \int_{-a}^2 f(x) \d x \\ &&&= 2ka + k\pi \\ \Rightarrow && k &= (\pi + 2a)^{-1} \end{align*}

1. \(\,\) \begin{align*} && \E[X] &= \int_{-a}^2 x f(x) \d x\\ &&&= \int_{-a}^0 2kx \d x + k\int_0^{2} x\sqrt{4-x^2} \d x\\ &&&= \left [kx^2 \right]_{-a}^0 +k \left [-\frac13(4-x^2)^{\frac32} \right]_0^2 \\ &&&= -ka^2 + \frac83k \\ &&&= \frac{\frac83-a^2}{\pi + 2a} \end{align*}
2. Consider \(\mathbb{P}(X < 0)\) then \(d < 0 \Leftrightarrow \mathbb{P}(X < 0) > \frac{9}{10}\) \begin{align*} && \frac{9}{10} &< \mathbb{P}(X < 0) \\ &&&= \int_{-a}^0 2k \d x \\ &&&= \frac{2a}{\pi+2a} \\ \Leftrightarrow && 9\pi &< 2a \\ \\ && \frac{9}{10} &= \int_{-a}^d 2k \d x \\ &&&= \frac{2(d+a)}{\pi + 2a} \\ \Rightarrow && 9\pi &= 2a + 20d \\ \Rightarrow && d &= \frac{2a-9\pi}{20} \end{align*}
3. Suppose \(d=\sqrt 2\) then \begin{align*} && \frac1{10} &= \int_{\sqrt{2}}^2 f(x) \d x \\ &&&= \int_{\sqrt{2}}^2 k\sqrt{4-x^2} \d x \\ &&&= k\left [ 2 \sin^{-1} \tfrac12 x + \tfrac12 x \sqrt{4-x^2}\right]_{\sqrt{2}}^2 \\ &&&= k\left (\pi -\frac{\pi}{2} - 1 \right) \\ \Rightarrow && \pi + 2a &= 5\pi - 10 \\ \Rightarrow && a &= 2\pi-5 \end{align*}

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Solution: Skewness is a measure of the symmetry (specifically the lack-thereof) in the distribution. How much mass is there on one side rather than another.

1. \(\,\) \begin{align*} && \gamma &= \frac{\E \left [ (X - \mu)^3 \right ]}{\sigma^3} \\ &&&= \frac{\E \left [ X^3 - 3\mu X^2 + 3\mu^2 X - \mu^3 \right ]}{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu \E[X^2] + 3\mu^2 \E[X] - \mu^3 }{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu (\mu^2 + \sigma^2) + 3\mu^2\cdot \mu- \mu^3 }{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu \sigma^2 - \mu^3 }{\sigma^3} \\ \end{align*} \begin{align*} && f(x) &= \begin{cases} 2x & \text{for } 0\le x\le 1\,, \\[2mm] 0 & \text{otherwise}\,. \end{cases} \\ && \E[X] &= \int_0^1 2x^2 \d x \\ &&&= \frac23 \\ && \E[X^2] &= \int_0^1 2x^3 \d x \\ &&&= \frac12 \\ && \E[X^3] &= \int_0^1 2x^4 \d x \\ &&&= \frac25 \\ \\ && \mu &= \frac23 \\ && \sigma^2 &= \frac12 - \frac49 = \frac{1}{18} \\ && \gamma &= \frac{\frac25 - 3 \cdot \frac23 \cdot \frac1{18} - \frac8{27}}{\frac{1}{54\sqrt2}} \\ &&&= -\frac{2\sqrt2}{5} \end{align*}
2. First note that \(\displaystyle F(x) = \int_0^x 2t \d t = x^2\) for \(x \in [0,1]\). In particular, \(F^{-1}(x) = \sqrt{x}\), so \begin{align*} && D &= \frac { {\rm F}^{-1}(\frac9{10}) - 2{\rm F} ^{-1}(\frac12) + {\rm F}^{-1} (\frac1{10}) } {{\rm F}^{-1}(\frac9{10}) - {\rm F} ^{-1} (\frac1{10})} \\ &&&= \frac{\sqrt{\frac9{10}} - 2 \sqrt{\frac5{10}} + \sqrt{\frac1{10}}}{\sqrt{\frac9{10}}-\sqrt{\frac1{10}}} \\ &&&= \frac{3-2\sqrt5+1}{3 - 1} \\ &&&= \frac{4-2\sqrt5}{2} = 2-\sqrt5 \end{align*} \begin{align*} && P &= \frac{3(\mu - M)}{\sigma} \\ &&&= \frac{3(\frac23 - \sqrt{\frac12})}{\frac{1}{3\sqrt2}} \\ &&&= 6 \sqrt2 - 9 \end{align*} First we compare \(P\) and \(D\), \(6\sqrt2-9\) and \(2-\sqrt5\) \begin{align*} && D & > P \\ \Leftrightarrow && 2-\sqrt5 &> 6\sqrt2 - 9 \\ \Leftrightarrow && 11 -6\sqrt2 &> \sqrt 5 \\ \Leftrightarrow && (121 + 72 - 132\sqrt2) & > 5 \\ \Leftrightarrow && 188 & > 132\sqrt2 \\ \Leftrightarrow && 47 & > 33 \sqrt 2\\ \Leftrightarrow && 2209 & > 2178 \end{align*} also \begin{align*} && P &> \gamma \\ \Leftrightarrow && 6\sqrt2 - 9 &> -\frac{2\sqrt2}{5} \\ \Leftrightarrow && 30\sqrt2 - 45 & > -2\sqrt2 \\ \Leftrightarrow && 32 \sqrt 2 &> 45 \\ \Leftrightarrow && 2048 &> 2025 \end{align*}

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Solution:

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2. Let \(Y \sim N(0,\frac14)\), then: \begin{align*} &&\int_0^\infty \frac{1}{\sqrt{2 \pi \cdot \frac14}} e^{-2x^2} \, dx &= \frac12\\ \Rightarrow && \int_0^\infty e^{-2x^2} &= \frac{\sqrt{\pi}}{2 \sqrt{2}} \\ \Rightarrow && k &= \boxed{\frac{2\sqrt{2}}{\sqrt{\pi}}} \end{align*}
3. \begin{align*} \mathbb{E}[X] &= \int_0^\infty x f(x) \, dx \\ &= \frac{2\sqrt{2}}{\sqrt{\pi}}\int_0^\infty x e^{-2x^2}\, dx \\ &= \frac{2\sqrt{2}}{\sqrt{\pi}} \left [-\frac{1}{4}e^{-2x^2} \right]_0^\infty \\ &= \frac{1}{\sqrt{2\pi}} \\ \end{align*} In order to calculate \(\mathbb{E}(X^2)\) it is useful to consider the related computation \(\mathbb{E}(Y^2)\). In fact, by symmetry, these will be the same values. Therefore \(\mathbb{E}(X^2) = \mathbb{E}(Y^2) = \mathrm{Var}(Y) = \frac{1}{4}\) (since \(\mathbb{E}(Y) = 0\)). Therefore \(\mathrm{Var}(Y) = \mathbb{E}(Y^2) - \mathbb{E}(Y)^2 = \frac14 - \frac{1}{2\pi}\)
4. \begin{align*} && \mathbb{P}(X < x) &= \frac12 \\ \Leftrightarrow && 2\mathbb{P}(0 \leq Y < x) &= \frac12 \\ \Leftrightarrow && 2\l \mathbb{P}(Y < x) - \frac12 \r &= \frac12 \\ \Leftrightarrow && \mathbb{P}(Y < x)&= \frac34 \\ \Leftrightarrow && \mathbb{P}(\frac{Y-0}{1/2} < \frac{x}{1/2})&= \frac34 \\ \Leftrightarrow && \mathbb{P}(Z < \frac{x}{1/2})&= \frac34 \\ \Leftrightarrow && \Phi(2x)&= \frac34 \\ \Leftrightarrow && 2x &= 0.6744895\cdots \\ \Leftrightarrow && x &= 0.3372\cdots \\ \Leftrightarrow && &= 0.337 \, (3 \text{sf}) \\ \end{align*}

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Solution:

1. \(X = \cos \theta\) \(\theta \sim U(0, 2\pi)\). Noting that \(\mathbb{P}(X \geq t ) = \frac{2}{2\pi}\cos^{-1} t\) so \(f_X(t) = \frac{1}{\pi} \frac{1}{\sqrt{1-x^2}}\) \begin{align*} && \E[X] &= 0 \tag{by symmetry} \\ && \E[X^2] &= \int_0^{2\pi} \cos^2 \theta \frac{1}{2 \pi} \d \theta \\ &&&= \frac{1}{2} \cdot 2\pi \cdot \frac{1}{2\pi} \\ &&&= \frac12 \\ \Rightarrow & &\var[X] &= \frac12 \\ \\ && \E[XY] &= \int_0^{2\pi} \cos \theta \sin \theta \frac{1}{2 \pi} \d \theta \\ &&&= \frac{1}{4\pi} \int_0^{2\pi} \sin 2\theta \d \theta \\ &&& =0 = \E[X]\E[Y] \end{align*} But note that clearly \(X\) and \(Y\) are not independent, since given \(X\) there are only two possible values of \(Y\).
2. \(\,\) \begin{align*} && \E \left [ XY \right] &= \E \left [ \left ( \frac1n \sum_{i=1}^n X_i \right)\left ( \frac1n \sum_{i=1}^n Y_i\right) \right] \\ &&&= \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n \E [X_i Y_j] \\ &&&= 0 = \E[X] \E[Y] \end{align*} Therefore \(X\) and \(Y\) are uncorrelated. Note that \(\E[X_i] = 0, \var[X_i] = \frac12\) so we can apply the central limit theorem to see that \(X \approx N(0, \frac{1}{2n})\), in particular \begin{align*} && 0.95 &\approx \mathbb{P}(|Z| < 2) \\ &&&= \mathbb{P} \left ( \Big |\frac{X}{\sqrt{\frac{1}{2n}}} \Big | < 2 \right ) \\ &&&= \mathbb{P}\left (|X| < \sqrt{\frac{2}{n}} \right) \end{align*}

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Solution:

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2. Since \(f(x)\) is continuous, \(a -bx\) joins \((2k, \ln k)\) and \((4k ,0)\). ie it has a gradient of \(\frac{-\ln k}{2k}\) and is zero at \(4k\), hence \(\displaystyle b = -\frac{\ln k}{2k}, a = 2\ln k\). The \(3\) sections have areas \(\int_1^k \ln x \d x = k \ln k -k +1\), \(k \ln k, k \ln k\). Therefore \begin{align*} &&1&= 3k\ln k - k +1 \\ \Rightarrow &&0 &= k(3\ln k - 1) \\ \Rightarrow &&\ln k &= \frac13 \\ \Rightarrow &&k &= e^{1/3} \\ && a &= \frac23 \\ && b&= -\frac16e^{-1/3} \end{align*}
3. Clearly \(1 > k \ln k > \frac{1}{3}\), therefore the median must lie between \(k\) and \(2k\). So we need, \(\frac12\) to be the area of the rectangle + the triangle, ie: \begin{align*} && \frac12 &= k \ln k + (2k-M) \ln k \\ &&&= \frac13 k + \frac13 (2k - M) \\ \Rightarrow && M &= 3k - \frac32 \\ \Rightarrow && M &= 3e^{1/3} - \frac32 \end{align*}

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Solution:

1. \begin{align*} 1 &= \int_{1/4}^{1/2} \frac{\lambda}{x\ln x} \, dx \\ &= \lambda\left [ \ln |\ln x| \right ]_{1/4}^{1/2} \\ &= \lambda \l \ln |-\ln 2| - \ln |-2 \ln 2| \r \\ &= \lambda (-\ln 2) \end{align*} So \(\lambda = -\frac{1}{\ln 2} = \frac{1}{\ln \frac12}\)
2. \begin{align*} 1 &= \int_{a}^{b} \frac{1}{x\ln x} \, dx \\ &= \left [ \ln |\ln x| \right ]_{a}^{b} \\ &= \l \ln \ln b - \ln \ln a \r \\ &= \ln \l \frac{\ln b}{\ln a} \r \\ \end{align*} So \(b = e^{a}\)
3. If \(\lambda = 1, a = e, b = e^e\), then \begin{align*} \P(\e^{3/2}\le X \le \e^2) &= \int_{e^{3/2}}^{e^2} \frac{1}{x \ln x} \, dx \\ &= \left [ \ln \ln x \right]_{e^{3/2}}^{e^2} \\ &= \ln 2 - \ln \frac{3}{2} \\ &= \ln \frac{4}{3} \\ &= \ln \l 1 + \frac{1}{3} \r \\ &\approx \frac{1}{3} - \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} - \frac{1}{4 \cdot 3^4} \\ &= \frac{31}{108} \end{align*}
4. Note that \(2 > e^{\frac12} > 1\) so \(a = e^{\frac12}, b = e^{\frac{e}2}\). Since \(3 > e \Rightarrow e^{3/2} > e^{\frac{e}{2}}\) this probability is out of range, therefore \(\P(\e^{3/2}\le X \le \e^2) = 0\)

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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*}

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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*} && f(x) &= \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \\ \Rightarrow && 1 &= \int_0^{\infty} f(x) \d x \\ &&&= \int_0^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ &&&= Ck^{a+1} \int_0^{\infty} \frac{x^a}{(x+k)^{2a+2} }\d x \\ &&&= Ck^{a+1} \frac{a!(2a-a)!}{(2a+1)!k^{2a-a+1}} \\ &&&= C \frac{a!a!}{(2a+1)!} \\ \Rightarrow && C &= \frac{(2a+1)!}{a!a!} \end{align*} \begin{align*} && I &= \int_0^v \frac{x^a}{(x+k)^{2a+2}} \d x\\ u = k^2/x, \d x = -k^2u^{-2} \d u: &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a}u^{-a}}{(k^2u^{-1} +k)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a-2a-2}u^{2a+2-a}}{(k +u)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{ k^2/v}^{\infty} \frac{u^{a}}{(k +u)^{2a+2}} \d u \\ \end{align*} At the median we want a value \(M\) such that \(M = k^2/M\) ie \(M = k\) \begin{align*} && \mathbb{E}(V) &= \int_0^{\infty} x f(x) \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \int_0^{\infty} \frac{x^{a+1}}{(x+k)^{2a+2}} \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \frac{(a+1)!(2a-(a+1))!}{(2a+1)!k^{2a-(a+1)+1}}\\ &&&= \frac{k^{a+1}}{a!} \frac{(a+1)(a-1)!}{k^{a}} \\ &&&= \frac{k(a+1)}{a} = \frac{a+1}a k \end{align*} \begin{align*} && \mathbb{P}(T < t) &= \mathbb{P}(\frac{s}{V} < t) \\ &&&= \mathbb{P}(V > \frac{s}{t}) \\ &&&= \int_{s/t}^{\infty} f(x) \d x \\ &&&= \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ \\ \Rightarrow && f_T(t) &= \frac{\d}{\d t} \left ( \mathbb{P}(T < t)\right) \\ &&&= \frac{\d}{\d t} \left ( \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \right) \\ &&&= - \frac{C \, k^{a+1} \, \left ( \frac{s}{t} \right)^a}{(\frac{s}{t}+k)^{2a+2}} \cdot \left (-\frac{s}{t^2} \right) \\ &&&= \frac{Ck^{a+1}s^{a+1}t^{2a+2}}{t^{a+2}(s+kt)^{2a+2}} \\ &&&= \frac{C(ks)^{a+1}t^a}{(s+kt)^{2a+2}} \\ &&&= \frac{C(\frac{s}{k})^{a+1}t^a}{(\frac{s}{k}+t)^{2a+2}} \end{align*} Therefore \(T\) follows the same distribution, but with parameter \(s/k\) rather than \(k\). In particular it has median \(s/k\) (and the product of the medians is \(s\)). However, the product of the expected time and expected speed is \(\frac{a+1}{a} k \frac{a+1}{a} \frac{s}{k} = \left ( \frac{a+1}{a} \right)^2s > s\)

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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: \begin{align*} && \E[X] &= \frac{a}{2}\tag{by symmetry} \\ &&\E[X^2] &= \int_0^a \frac{1}{a} x^2 \d x \\ &&&= \frac{a^3}{3a} = \frac{a^2}{3} \\ \Rightarrow && \var[X] &= \frac{a^2}{3} - \frac{a^2}{4} = \frac{a^2}{12} \\ \end{align*} \begin{align*} && V &=\frac{1}{2} \left ( \left ( X_1 - \frac{X_1+X_2}{2} \right )^2+\left ( X_2- \frac{X_1+X_2}{2} \right )^2 \right ) \\ &&&= \frac{1}{8} ((X_1 - X_2)^2 + (X_2 - X_1)^2 ) \\ &&&= \frac14 (X_1-X_2)^2 \\ \\ && \E[2V] &= \E \left [ \frac12 (X_1 - X_2)^2 \right] \\ &&&= \frac12 \E[X_1^2] - \E[X_1X_2] + \frac12 \E[X_2^2] \\ &&&= \frac{a^2}{3} - \frac{a^2}{4} = \frac{a^2}{12} \end{align*} Therefore \(2V\) is an unbiased estimator of the variance of \(X\).

We need \(|X_1 - X_2| < \frac{a}{\sqrt{3}}\) We are interested in the blue area, which is \(a^2 - a^2(1- \frac{1}{\sqrt{3}})^2 = a^2 \left ( \frac{2}{\sqrt{3}} - \frac13 \right)\) ie the probability is \(\frac{2\sqrt{3}-1}{3} \approx 0.8\)

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Solution: Let \(X \sim U[0, \tfrac12]\) be the shorter piece, so \(R = \frac{X}{1-X}\), and \begin{align*} && \mathbb{P}(R \leq r) &= \mathbb{P}(\tfrac{X}{1-X} \leq r) \\ &&&= \mathbb{P}(X \leq r - rX) \\ &&&= \mathbb{P}((1+r)X \leq r) \\ &&&= \mathbb{P}(X \leq \tfrac{r}{1+r} ) \\ &&&= \begin{cases} 0 & r < 0 \\ \frac{2r}{1+r} & 0 \leq r \leq 1 \\ 1 & r > 1 \end{cases} \\ \\ && f_R(r) &= \begin{cases} \frac{2}{(1+r)^2} & 0 \leq r \leq 1 \\ 0 & \text{otherwise} \end{cases} \end{align*} Let \(Y \sim U[\tfrac12, 1]\) be the longer piece, then \(R = \frac{1-Y}{Y} = Y^{-1} - 1\) and \begin{align*} \E[R] &= \int_{\frac12}^1 (y^{-1}-1) 2 \d y \\ &= 2\left [\ln y - y \right]_{\frac12}^1 \\ &= -2 + 2\ln2 +2\frac12 \\ &= 2\ln2 -1 \\ \\ \E[R^2] &= \int_{\frac12}^1 (y^{-1}-1)^2 2 \d y\\ &= 2\left [-y^{-1} -2\ln y + 1 \right]_{\frac12}^1 \\ &= 2 \left ( 2 - 2\ln 2+\frac12\right) \\ &= 3-4\ln 2 \\ \var[R] &= 3 - 4 \ln 2 -(2\ln 2-1)^2 \\ &= 2 - 4(\ln 2)^2 \end{align*}

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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.