problem 59

5.2.7 problem 59

(a) Let $θ_{1}$ , $θ_{2}$ and $θ_{3}$ be the angles corresponding to points A, B and C respectively. From the statement of the problem, those angles are i.i.d. $Unif (0, 2 π)$ .

The angles $θ_{1}$ , $θ_{2}$ and $θ_{3}$ divide the $[0, 2 π)$ range in four successive sub-intervals. The argument is wrong because the problem is not symmetric with respect to the three arcs, but rather with respect to the four angle sub-intervals. The average length of each sub-interval is $2 π ∕ 4 = π ∕ 2$ , by symmetry. The length of the arc that contains the point (1,0) is the sum of the first and fourth sub-intervals, so it is twice as long as the other arcs on average.

(b)

θ_{1} = Unif (0, 2 π)

θ_{2} = Unif (0, 2 π)

θ_{3} = Unif (0, 2 π)

L_{1} = min (θ_{1}, θ_{2}, θ_{3})

CDF,

\begin{matrix} F (y) & = 1 - P (min (θ_{1}, θ_{2}, θ_{3}) > y) = 1 - {(\frac{2 π - y}{2 π})}^{3}, for 0 \leq y < 2 π \end{matrix}

PDF,

\begin{matrix} f (y) & = \frac{d}{dy} F (y) = \frac{3}{2 π} {(1 - \frac{y}{2 π})}^{2}, for 0 \leq y < 2 π \end{matrix}

(c)

\begin{matrix} E [L] & = 2 E [L_{1}] = 2 \int_{0}^{2 π} y \frac{3}{2 π} {(1 - \frac{y}{2 π})}^{2} dy = \frac{3}{π} \int_{0}^{2 π} (y + \frac{y^{3}}{4 π^{2}} - \frac{y^{2}}{π}) dy = \frac{3}{π} [\frac{4 π^{2}}{2} + \frac{1}{4 π^{2}} \frac{16 π^{4}}{4} - \frac{1}{π} \frac{8 π^{3}}{3}] = π \end{matrix}

We can reach the same result from the qualitative explanation given in part (a): since $L$ is the sum of the first and fourth sub-intervals, where the length of each sub-interval is $π ∕ 2$ on average, $E [L] = π ∕ 2 + π ∕ 2 = π$ .

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