problem 59

5.2.7 problem 59

(a) Length biased sampling

L_{1} + L_{2} + L_{3} = 2 π

E [L_{1}] = E [L_{2}] = E [L_{3}] = \frac{2 π}{3}

But our point is more likely to be a part of the longest arc. If there was a $\frac{1}{3}$ chance of the point being in any one of the three points then $E [L] = \frac{2 π}{3}$

(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) & (5.42) = 1 - {\frac{2 π - y}{2 π}}^{3} & (5.43) \end{matrix}

PDF,

\begin{matrix} f (y) & = \frac{d}{dy} F (y) & (5.44) = \frac{3}{2 π} {(1 - \frac{y}{2 π})}^{2} & (5.45) \end{matrix}

(c)

\begin{matrix} E [L] & = 2 E [L_{1}] & (5.46) = 2 \int_{0}^{2 π} y \frac{3}{2 π} {(1 - \frac{y}{2 π})}^{2} dy & (5.47) = \frac{3}{π} \int_{0}^{2 π} y + \frac{y^{3}}{4 π^{2}} - \frac{y^{2}}{π} dy & (5.48) = \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}] & (5.49) = π & (5.50) \end{matrix}

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