2 problems found
Three points, \(A\), \(B\) and \(C\), lie in a horizontal plane, but are not collinear. The point \(O\) lies above the plane. Let \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\) and \(\overrightarrow{OC} = \mathbf{c}\). \(P\) is a point with \(\overrightarrow{OP} = \alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c}\), where \(\alpha\), \(\beta\) and \(\gamma\) are all positive and \(\alpha + \beta + \gamma < 1\). Let \(k = 1 - (\alpha + \beta + \gamma)\).
For any random variables \(X_1\) and \(X_2\), state the relationship between \(\E(aX_1+bX_2)\) and \(\E(X_1)\) and \(\E(X_2)\), where \(a\) and \(b\) are constants. If \(X_1\) and \(X_2\) are independent, state the relationship between \(\E(X_1X_2)\) and \(\E(X_1)\) and \(\E(X_2)\). An industrial process produces rectangular plates. The length and the breadth of the plates are modelled by independent random variables \(X_1\) and \(X_2\) with non-zero means \(\mu_1\) and \(\mu_2\) and non-zero standard deviations \(\sigma_1\) and \(\sigma_2\), respectively. Using the results in the paragraph above, and without quoting a formula for \(\var(aX_1+bX_2)\), find the means and standard deviations of the perimeter \(P\) and area \(A\) of the plates. Show that \(P\) and \(A\) are not independent. The random variable \(Z\) is defined by \(Z=P-\alpha A\), where \(\alpha \) is a constant. Show that \(Z\) and \(A\) are not independent if \[ \alpha \ne \dfrac{2(\mu_1^{\vphantom2} \sigma_2^2 +\mu_2^{\vphantom2}\sigma_1^2)} { \mu_1^2 \sigma_2^2 +\mu_2^2\sigma_1^2 + \sigma_1^2\sigma_2^2 } \;. \] Given that \(X_1\) and \(X_2\) can each take values 1 and 3 only, and that they each take these values with probability \(\frac 12\), show that \(Z\) and \(A\) are not independent for any value of \(\alpha\).
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 \(\alpha = \frac23\)