Problems

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Solution:

1. If \(n_3 = 9\) and we are looking for \(0 < n_1 < n_2 < n_3\) we can consider values for each \(n_2\). \begin{array}{clc|c} n_2 & \text{range} & \text{count} \\ \hline 6 & 4-5 & 2 \\ 7 & 3-6 & 4 \\ 8 & 2-7 & 6 \\ \hline & & 12 \end{array} When \(n_3 = 10\) \begin{array}{clc|c} n_2 & \text{range} & \text{count} \\ \hline 6 & 5 & 1 \\ 7 & 4-6 & 3 \\ 8 & 3-7 & 5 \\ 9 & 2-8 & 7 \\ \hline & & 16 \end{array} When \(n_3 = 2n+1\) we can have \(2 + 4 + \cdots + 2n-2 = n(n-1)\) When \(n_3 = 2n\) we can have \(1 + 3 + \cdots + 2n-3 = (n-1)^2\)
2. For the 3 rods to form a triangle, it suffices for the sum of the lengths of the shorter rods to be larger than \(N\). When \(N = 2n+1\) there are \(n(n-1)\) ways this can happen, out of \(\binom{2n}{2}\) ways to choos the numbers, ie \begin{align*} && P &= \frac{n(n-1)}{\frac{2n(2n-1)}{2}} \\ &&&= \frac{n-1}{2n-1} \end{align*} When \(N = 2n\) there are \((n-1)^2\) ways this can happen, out of \(\binom{2n-1}{2}\) ways, ie \begin{align*} && P &= \frac{(n-1)^2}{\frac{(2n-1)(2n-2)}{2}} \\ &&&= \frac{n-1}{2n-1} \end{align*}
3. The number of ways this can happen is: \begin{align*} C &= \sum_{k=3}^{2M+1} \# \{ \text{triangles where }k\text{ is largest} \} \\ &= \sum_{k=1}^{M} \# \{ \text{triangles where }2k+1\text{ is largest} \} +\sum_{k=1}^{M} \# \{ \text{triangles where }2k\text{ is largest} \}\\ &= \sum_{k=1}^{M} n(n-1)+\sum_{k=1}^{M} (n-1)^2\\ &= \sum_{k=1}^{M} (2n^2-3n+1)\\ &= \frac26M(M+1)(2M+1) - \frac32M(M+1) + M \\ &= \frac16 M(4M+1)(M-1) \end{align*} Therefore the probability is \begin{align*} && P &= \frac{M(4M+1)(M-1)}{6 \binom{2M+1}{3}} \\ &&&= \frac{M(4M+1)(M-1)}{(2M+1)2M(2M-1)} \\ &&&= \frac{(4M+1)(M-1)}{2(2M+1)(2M-1)} \end{align*}

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Solution:

1. \begin{align*} && \mathbb{P}(Y_k \leq y) &= \sum_{j=k}^n\mathbb{P}(\text{exactly }j \text{ values less than }y) \\ &&&= \sum_{j=k}^m \binom{m}{j} y^j(1-y)^{n-j} \end{align*}
2. This is the number of ways to choose a committee of \(m\) people with the chair from those \(m\) people. This can be done in two ways. First: choose the committee in \(\binom{n}{m}\) ways and choose the chair in \(m\) ways so \(m \binom{n}{m}\). Alternatively, choose the chain in \(n\) ways and choose the remaining \(m-1\) committee members in \(\binom{n-1}{m-1}\) ways. Therefore \(m \binom{n}{m} = n \binom{n-1}{m-1}\) \begin{align*} (n-m) \binom{n}{m} &= (n-m) \binom{n}{n-m} \\ &= n \binom{n-1}{n-m-1} \\ &= n \binom{n-1}{m} \end{align*} \begin{align*} f_{Y_k}(y) &= \frac{\d }{\d y} \l \sum^{n}_{m=k}\binom{n}{m}y^{m}\left(1-y\right)^{n-m} \r \\ &= \sum^{n}_{m=k} \l \binom{n}{m}my^{m-1}\left(1-y\right)^{n-m} -\binom{n}{m}(n-m)y^{m}\left(1-y\right)^{n-m-1} \r \\ &= \sum^{n}_{m=k} \l n \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n \binom{n-1}{m} y^{m}\left(1-y\right)^{n-m-1} \r \\ &= n\sum^{n}_{m=k} \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n\sum^{n+1}_{m=k+1} \binom{n-1}{m-1} y^{m-1}\left(1-y\right)^{n-m} \\ &= n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \end{align*} \begin{align*} &&1 &= \int_0^1 f_{Y_k}(y) \d y \\ &&&= \int_0^1 n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \d y \\ &&&= n \binom{n-1}{k-1} \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\ \Rightarrow && \frac{1}{n \binom{n-1}{k-1}} &= \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\ \end{align*}
3. \begin{align*} && \mathbb{E}(Y_k) &= \int_0^1 y f_{Y_k}(y) \d y \\ &&&= \int_0^1 n \binom{n-1}{k-1} y^{k}(1-y)^{n-k} \\ &&&= n \binom{n-1}{k-1}\int_0^1 y^{k}(1-y)^{n-k} \d y \\ &&&= n \binom{n-1}{k-1}\int_0^1 y^{k+1-1}(1-y)^{n+1-(k+1)} \d y \\ &&&= n \binom{n-1}{k-1} \frac{1}{(n+1) \binom{n}{k}}\\ &&&= \frac{n}{n+1} \cdot \frac{k}{n} \\ &&&= \frac{k}{n+1} \end{align*}

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Solution:

1. The probability no-one picks the winning number is \(\left ( 1 - \frac{1}{N}\right)^N \approx \frac1e\). \begin{align*} && \mathbb{E}(\text{profit}) &= Nc - (1-e^{-1})J \\ &&& < Nc -(1- \tfrac12 )J \\ &&& < Nc - \frac12 J \\ &&&= \frac{2Nc-J}{2} \end{align*} Therefore if \(J = 2Nc\) the expected profit is negative.
2. \(\,\) \begin{align*} && 1 &= \sum_{\text{all numbers}} \mathbb{P}(\text{pick }i) \\ &&&= \sum_{\text{popular numbers}} \mathbb{P}(\text{pick }i)+\sum_{\text{unpopular numbers}} \mathbb{P}(\text{pick }i) \\ &&&=\gamma N \frac{a}{N} + (1-\gamma)N \frac{b}{N} \\ &&&= \gamma a + (1-\gamma)b \end{align*} \begin{align*} && \mathbb{P}(\text{no-one picks winning number}) &= \mathbb{P}(\text{no-one picks winning number} | \text{winning number is popular})\mathbb{P})(\text{winning number is popular}) + \\ &&&\quad + \mathbb{P}(\text{no-one picks} | \text{unpopular})\mathbb{P}(\text{unpopular}) \\ &&&= \left (1 - \frac{a}{N} \right)^N \gamma + \left (1 - \frac{b}{N} \right)^N (1-\gamma) \\ &&&\approx \gamma e^{-a} + (1-\gamma)e^{-b} \\ \\ && \mathbb{E}(\text{profit}) &= Nc - (1-\gamma e^{-a} - (1-\gamma)e^{-b})J \\ &&&= Nc-J+J\gamma e^{-a} +J(1-\gamma)e^{-b} \end{align*} If \(\gamma = \frac18\) and \(a=9b\), then \(1=\frac18 a + \frac78b = 2b \Rightarrow b = \frac12, a = \frac92\) and \begin{align*} && \mathbb{E}(\text{profit}) &= Nc-J +J\tfrac18e^{-9/2}+J\tfrac78e^{-1/2} \\ &&&= Nc-J+\tfrac18Je^{-1/2}(e^{-4}+7) \end{align*} If we can show \(e^{-1/2}\frac{e^{-4}+7}{8} > \frac12\) we'd be done, so \begin{align*} && e^{-1/2}\frac{e^{-4}+7}{8} &> \frac12 \\ \Leftrightarrow && e^{-4}+7 &>4e^{1/2} \\ \Leftrightarrow && 49+14e^{-4}+e^{-8} &>16e \\ \end{align*} But clearly the LHS \(>49\) and the RHS \(<48\) so we're done

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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: \begin{align*} && \mathbb{P}(X > 4) &= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdot \frac{n-3}{n} \\ && \mathbb{P}(X > 3) &= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \\ \Rightarrow && \mathbb{P}(X =4) &= \mathbb{P}(X > 3) - \mathbb{P}(X > 4) \\ &&&= \frac{(n-1)(n-2)}{n^2} \left (1 - \frac{n-3}{n} \right) \\ &&&= \frac{3(n-1)(n-2)}{n^3} \end{align*}

1. Notice that \begin{align*} && \mathbb{P}(X > r) &= \frac{n-1}{n} \cdots \frac{n-r+1}{n} \\ \Rightarrow && \mathbb{P}(X = r) &= \frac{n-1}{n} \cdots \frac{n-r+2}{n} \left (1 - \frac{n-r+1}{n} \right) \\ &&&= \frac{(n-1)\cdots(n-r+2)(r-1)}{n^{r-1}} \\ &&&= \left (1 - \frac{1}n \right)\left (1 - \frac{2}{n} \right) \cdots \left (1 - \frac{r-2}{n} \right) \frac{r-1}{n} \\ \Rightarrow && 1 &= \sum \mathbb{P}(X = r) \\ &&&= \sum_{r=2}^{n+1} \mathbb{P}(X = r) \\ &&&= \frac 1n + \left(1-\frac1n\right) \frac 2 n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac3 n + \cdots \end{align*}
2. \(\,\) \begin{align*} \mathbb{E}(X) &= \sum_{r=2}^{n+1} r\cdot\mathbb{P}(X = r) \\ &= \frac 2n + \left(1-\frac1n\right) \frac {2\cdot3} n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac{3\cdot4} n + \cdots \end{align*}
3. \(\displaystyle \mathbb{P}(X \geq k) = \frac{n-1}{n} \cdots \frac{n-r+2}{n}\)
4. \(\,\) \begin{align*} && \mathbb{E}(Y) &= \sum_{r=1}^N r \cdot \mathbb{P}(Y = r) \\ &&&= \sum_{r=1}^N \sum_{j=1}^r \mathbb{P}(Y = r) \\ &&&= \sum_{j=1}^N \sum_{r=j}^N \mathbb{P}(Y=r) \\ &&&= \sum_{j=1}^N \mathbb{P}(Y \geq j) \end{align*} Let \(P_k = \left(1-\frac1n\right)\left(1-\frac2n\right) \cdots \left(1-\frac1n\right)\left(1-\frac{k}n\right) \) \begin{align*} && \mathbb{E}(X) &= P_1 \frac{1 \cdot 2 }{n} + P_2 \cdot \frac{2 \cdot 3}{n} + \cdots + P_k \cdot \frac{k(k+1)}{n} + \cdots \\ && &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + \sum_{k=1}^{n} \frac{k}{n}P_k \\ && \text{Using the identity } & \frac{k}{n}P_k = \frac{k}{n} \prod_{i=1}^{k-1} \left(1-\frac{i}{n}\right) = P_k - P_{k+1}: \\ && \sum_{k=1}^{n} \frac{k}{n}P_k &= (P_1 - P_2) + (P_2 - P_3) + \cdots + (P_n - P_{n+1}) \\ && &= P_1 - P_{n+1} = 1 - 0 = 1 \\ \\ \Rightarrow && \mathbb{E}(X) &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + 1 \\ && &= \mathbb{P}(X \geq 1) + \mathbb{P}(X \geq 2) + \mathbb{P}(X \geq 3) + \cdots \\ && &= 1 + P_1 + P_2 + P_3 + \cdots \\ && &= 1 + \sum_{k=1}^{n} P_k \\ \\ \Rightarrow && 1 + \sum_{k=1}^{n} P_k &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + 1 \\ \Rightarrow && \sum_{k=1}^{n} P_k &= \sum_{k=1}^{n} \frac{k^2}{n}P_k \\ \Rightarrow && 0 &= \sum_{k=1}^{n} P_k \left( 1 - \frac{k^2}{n} \right) \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:

1. Firstly, we consider the probability that all darts lie within a distance \(s\) from the centre, ie \begin{align*} \mathbb{P}(\text{all darts within }s) &= \prod_{k=1}^s \mathbb{P}(\text{dart within }s) \\ &= \left ( \frac{\pi s^2}{\pi} \right)^n \\ &= s^{2n} \end{align*} Therefore the pdf is \(2ns^{2n-1}\), and the expected area is \(\int_{s=0}^1 \pi s^2 \cdot 2n s^{2n-1} \d s = 2n \pi \frac{1}{2n+2} = \frac{n}{n+1} \pi\). \begin{align*} \mathbb{P}(\text{n-1 within }s) &= \underbrace{s^{2n}}_{\text{all within }s} + \underbrace{ns^{2n-2}(1-s^2)}_{\text{all but 1 within }s}\\ &= ns^{2n-2}-(n-1)s^{2n} \end{align*} Therefore the pdf is \(n(2n-2)s^{2n-3} - 2n(n-1)s^{2n-1} = 2n(n-1)(s^{2n-3}-s^{2n-1})\) and the expected area is: \begin{align*} \int \pi s^2 \cdot2n(n-1)(s^{2n-3}-s^{2n-1})\d s &= 2n(n-1) \pi \left ( \frac{1}{2n} - \frac{1}{2n+2} \right) \\ &= n(n-1)\pi \frac{2}{n(n+1)} \\ &= \frac{n-1}{n+1} \pi \end{align*}
2. Now consider a square of side-length \(s\), we must have \(\mathbb{P}(\text{all darts within square}) = \left ( \frac{s^2}{4} \right)^n\) and therefore the pdf is \(n \frac{s^{n-1}}{4^n}\). Therefore the expected area is \(\displaystyle \int_0^2 s^2 \cdot n \frac{s^{n-1}}{4^n} \d s = \frac{n}{n+1} \frac{2^{2n+1}}{2^{2n}} = \frac{4n}{n+1}\)
3. It is clearly larger as the square dartboard contains all of the circular dartboard, and there will be some probability that the darts land outside the circular dartboard, making the circle much larger.

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Solution: There are \(3\) possible numbers of points on the curved part of the perimeter. \(0\): The area is \(0\) \(1\):

The area of the triangle is \(\frac12 |x| \sin \theta\) Where \(X\) is the point along the diameter which is \(U[-1,1]\) and \(\theta \sim U(0, \pi)\) Therefore \begin{align*} \mathbb{E}(A|\text{one on diameter}) &= \int_{0}^\pi \frac{1}{\pi} \int_{-1}^1\frac{1}{2}\frac12 |x| \sin \theta \d x \d \theta \\ &= \frac{1}{2\pi}\frac12 \int_{0}^\pi \sin \theta \d \theta \cdot 2\int_{0}^1 x\d x \\ &=\frac{1}{2\pi}\cdot 2 \cdot \frac12 = \frac{1}{2\pi} \end{align*} \(2\): If both are on the curved section Then the area is \(\frac12 \sin \theta\) where \(\theta = |\theta_1 - \theta_2|\) and \(\theta_i \sim U[0, \pi]\) Therefore the area is \begin{align*} \mathbb{E}(A|\text{none on diameter}) &= \int_{0}^\pi\frac{1}{\pi} \int_{0}^\pi\frac{1}{\pi} \frac12 \sin |\theta_1 - \theta_2| \d \theta_1 \d \theta_2 \\ &= \frac{1}{\pi^2}\frac12 \int_{0}^\pi \left (\int_{0}^{\theta_2} \sin (\theta_2 - \theta_1) \d \theta_1-\int_{\theta_2}^{\pi} \sin (\theta_2 - \theta_1) \d \theta_1 \right)\d \theta_2 \\ &= \frac{1}{\pi^2}\frac12 \int_{0}^\pi \left [2\cos(\theta_2 - \theta_2)-\cos(\theta_2 - 0)-\cos(\theta_2 - \pi) \right]\d \theta_2 \\ &= \frac{1}{\pi} \end{align*} Therefore the expected area is: \begin{align*} \mathbb{E}(A ) &= \mathbb{E}(A|\text{one on diameter})\cdot \mathbb{P}(\text{one on diameter}) + \mathbb{E}(A|\text{none on diameter})\cdot \mathbb{P}(\text{none on diameter}) \\ &= \frac{1}{2\pi}\mathbb{P}(\text{one on diameter}) + \frac{1}{\pi}\cdot \mathbb{P}(\text{none on diameter}) \\ &= \frac{1}{2\pi} \cdot 2 \cdot \frac{\pi}{\pi + 2} \cdot \frac{2}{\pi + 2} + \frac1{\pi} \cdot \frac{\pi}{\pi + 2} \cdot \frac{\pi}{\pi+2} \\ &= \frac{2 + \pi}{(\pi+2)^2} \\ &= \frac{1}{\pi+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. \(\,\)
   

   + - ↺
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*} && \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*}