Problems

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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: \(\E(aX_1+bX_2) = a \E(X_1) + b\E(X_2)\) for any \(X_1, X_2\) \(\E(X_1X_2)=\E(X_1)\E(X_2)\). if \(X_1, X_2\) are independent. \begin{align*} && \E(P) &= \E(2(X_1+X_2)) = 2(\E[X_1]+\E[X_2]) \\ &&&= 2(\mu_1 + \mu_2) \\ && \var(P) &= \E[\left ( 2(X_1+X_2) \right)^2] - \E[2(X_1+X_2)]^2 \\ &&&= 4\E[X_1^2+2X_1X_2+X_2^2] -4(\mu_1 + \mu_2)^2 \\ &&&= 4(\mu_1^2 + \sigma_1^2 + 2\mu_1\mu_2 + \mu_2^2 + \sigma_2^2) - 4(\mu_1 + \mu_2)^2 \\ &&&= 4(\sigma_1^2+\sigma_2^2) \\ && \textrm{SD}(P) &= 2 \sqrt{\sigma_1^2+\sigma_2^2}\\ \\ && \E(A) &= \E[X_1X_2] = \E[X_1]\E[X_2] \\ &&&= \mu_1\mu_2 \\ && \var(A) &= \E[(X_1X_2)^2] - (\mu_1\mu_2)^2 \\ &&&= (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) - (\mu_1\mu_2)^2\\ &&&= \mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2 + \sigma_1^2 \sigma_2^2\\ && \textrm{SD}(A) &= \sqrt{\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2 + \sigma_1^2 \sigma_2^2} \end{align*} \begin{align*} \E[PA] &= \E[2(X_1+X_2)X_1X_2] \\ &= 2\E[X_1^2X_2] + 2\E[X_1X_2^2]\\ &= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2)\\ &\neq 2(\mu_1 + \mu_2)\mu_1\mu_2 \\ &= \E[P]\E[A] \end{align*} \begin{align*} && \E[Z] &= \E[P] - \alpha \E[A] \\ &&&= 2(\mu_1+\mu_2) - \alpha \mu_1 \mu_2 \\ \\ && \E[ZA] &= \E[PA - \alpha A^2] \\ &&&= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2) - \alpha \E[A^2] \\ &&&= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2) - \alpha \E[X_1^2]\E[X_2^2] \\ &&&= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2) - \alpha (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) \\ \text{if ind.} && \E[Z]\E[A] &= \E[ZA]\\ && (2(\mu_1+\mu_2) - \alpha \mu_1 \mu_2) \mu_1\mu_2 &= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2) - \alpha (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) \\ \Rightarrow && 2(\mu_1^2\mu_2+\mu_1\mu_2^2) - \alpha \mu_1^2\mu_2^2 &= 2(\mu_1^2\mu_2+\mu_1\mu_2^2) + 2\sigma_1^2\mu_2 + 2\sigma_2^2\mu_1 - \alpha (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) \\ \Rightarrow && \alpha ((\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) - \mu_1^2\mu_2^2) &= 2(\sigma_1^2\mu_2 + \sigma_2^2\mu_1) \\ \Rightarrow && \alpha &= \frac{ 2(\sigma_1^2\mu_2 + \sigma_2^2\mu_1) }{\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2 + \sigma_1^2 \sigma_2^2} \end{align*} Therefore if they are not independent if \(\alpha \neq \) the expression. \begin{array}{c|c|c|c|c|c} & X_1 & X_2 & A & P & Z \\ \hline 0.25 & 1 & 1 & 1 & 4 & 4-\alpha \\ 0.25 & 1 & 3 & 3 & 8 & 8-3\alpha \\ 0.25 & 3 & 1 & 3 & 8 & 8-3\alpha \\ 0.25 & 3 & 3 & 9 & 12 & 12-9\alpha \\ \end{array} If \(\mathbb{P}(A = 1, Z = 4-\alpha) = \mathbb{P}(A = 1)\mathbb{P}(Z = 4-\alpha)\) then \(\mathbb{P}(Z = 4-\alpha) = 1\), but that mean \(4-\alpha = 8-3\alpha = 12-9\alpha\) which is not a consistent set of equations as the first two are solved by \(\alpha = 2\) and the second by \(\alpha = \frac23\)