problem 13

5.5.1 problem 13

Recall from problem 12 that the CDF of Y, the length of the longer piece, is $F (k) = 2 k - 1$ .

a. Let us find the CDF of $X ∕ Y$ .

P (X / Y < k) = P (\frac{1 - Y}{Y} < k) = P (Y > \frac{1}{k + 1}) = 1 - (\frac{2}{k + 1} - 1) = \frac{2 k}{k + 1}

To find the PDF, we derive the CDF using the quotient rule:

\frac{d}{dk} (\frac{2 k}{k + 1}) = 2 {(k + 1)}^{- 2}

b. Note that $X ∕ Y$ is minimized at when X is 0 and Y is 1 and maximized at 1 when X and Y are both 1/2. So, to find $E (X ∕ Y)$ , we must find $\int_{0}^{1} 2 k {(k + 1)}^{- 2} dk$ . This can be done with integration by parts, factoring out the constant 2, with $u = k$ , $dv = {(k + 1)}^{- 2} dk$ , $du = 1$ , and $v = - {(k + 1)}^{- 1}$ :

\int_{0}^{1} 2 k {(k + 1)}^{- 2} dk = 2 ((\frac{- k}{k + 1}) {∣ ∣ ∣}_{0}^{1} - \int_{0}^{1} - {(k + 1)}^{- 1} dk) = 2 (- \frac{1}{2} + ln (2)) = 2 ln (2) - 1

c. Following similar steps as in part A, the CDF of $Y ∕ X$ is $P (\frac{Y}{1 - Y} < k) = P (Y < \frac{k}{k + 1}) = \frac{k - 1}{k + 1}$ . Then the PDF using the quotient rule is $2 {(k + 1)}^{- 2}$ , the same as the PDF for X/Y!

Then, we need to evaluate the same integral as in part b to find $E (Y ∕ X)$ , but now with the limits set from 1 to infinity, since $Y ∕ X$ is minimized when X=Y and maximized when Y=1 and X=0:

\int_{1}^{\infty} 2 k {(k + 1)}^{- 2} dk = 2 ((\frac{- k}{k + 1}) {∣ ∣ ∣}_{1}^{\infty} - \int_{1}^{\infty} - {(k + 1)}^{- 1} dk) = 2 ((- 1 / 2) + ln (\infty) - ln (2)) = \infty

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