1 problem found
Five independent timers time a runner as she runs four laps of a track. Four of the timers measure the individual lap times, the results of the measurements being the random variables \(T_1\) to \(T_4\), each of which has variance \(\sigma^2\) and expectation equal to the true time for the lap. The fifth timer measures the total time for the race, the result of the measurement being the random variable \(T\) which has variance \(\sigma^2\) and expectation equal to the true race time (which is equal to the sum of the four true lap times). Find a random variable \(X\) of the form \(aT+b(T_1+T_2+T_3+T_4)\), where \(a\) and \(b\) are constants independent of the true lap times, with the two properties:
Solution: Let the expected total time for the race be \(\mu\). Let \(X = aT + b(T_1 + T_2+T_3+T_4)\) then \(\E[X] = a\E[T] + b\E[T_1+\cdots+T_4] = a \mu + b \mu = (a+b)\mu\). So \(a+b=1\). \begin{align*} && \var[X] &= a^2\var[T] + b^2(\var[T_1] + \var[T_2] + \var[T_3] + \var[T_4]) \\ &&&= a^2\sigma^2 + 4b^2 \sigma^2 \\ &&& = \sigma^2 (a^2 + 4(1-a)^2 ) \\ &&&= \sigma^2 (5a^2 - 8a + 4) \\ &&&= \sigma^2 \left ( 5 \left ( a - \frac45 \right)^2 - \frac{16}{5}+4 \right)\\ &&&= \sigma^2 \left ( 5 \left ( a - \frac45 \right)^2 + \frac{4}{5}\right) \end{align*} Therefore variance is minimised when \(a = \frac45, b = \frac15\). Let \(Y = cT + d(T_1 + T_2+T_3+T_4)\) then \begin{align*} && \E[Y^2] &= \E \left [c^2T^2 + 2cd T(T_1+T_2+T_3+T_4) + d^2(T_1+T_2+T_3+T_4)^2 \right] \\ &&&= c^2 (\mu^2 + \sigma^2) + 2cd \mu^2 + d^2 (\var[T_1 + \cdots + T_4] + \mu^2) \\ &&&= c^2(\mu^2+\sigma^2) + 2cd \mu^2 + d^2(4\sigma^2 + \mu^2) \\ &&&= (c^2 + 2cd + d^2) \mu^2 + (c^2+4d^2) \sigma^2 \\ &&&= (c+d)^2 \mu^2 + (c^2+4d^2) \sigma^2 \\ \\ \Rightarrow && d &= -c \\ && 1 &= c^2 + 4d^2 \\ \Rightarrow && c &= \pm \frac{1}{\sqrt5} \\ && d &= \mp \frac{1}{\sqrt5} \end{align*} Given our results, our best estimate for \(\mu\) is \(\frac45 \cdot 220 + \frac15 220.5 = 220.1\). Our estimate for \(\sigma^2 = \left( \frac{1}{\sqrt{5}}(220.5-220) \right)^2 = \frac{1}{20}\). Note that \(\var[X] = \frac45\sigma^2 \approx \frac{1}{25}\) so we are looking at an interval \((220.1 - 0.4, 220.1 + 0.4) = (219.7, 220.5)\) using an interval of two standard errors.