Year: 2015
Paper: 3
Question Number: 13
Course: UFM Statistics
Section: Bivariate data
A very similar number of candidates to 2014 once again ensured that all questions received a decent number of attempts, with seven questions being very popular rather than five being so in 2014, but the most popular questions were attempted by percentages in the 80s rather than 90s. All but one question was answered perfectly at least once, the one exception receiving a number of very close to perfect solutions. About 70% attempted at least six questions, and in those cases where more than six were attempted, the extra attempts were usually fairly superficial.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Each of the two independent random variables $X$ and $Y$ is uniformly distributed on the interval~$[0,1]$.
\begin{questionparts}
\item By considering the lines $x+y =$ $\mathrm{constant}$ in the $x$-$y$ plane, find the cumulative distribution function of $X+Y$.
\item
Hence show that the probability density function $f$ of $(X+Y)^{-1}$
is given by
\[
\f(t) =
\begin{cases}
2t^{-2} -t^{-3} & \text{for $ \tfrac12 \le t \le 1$} \\
t^{-3} & \text{for $1\le t <\infty$}\\
0 & \text{otherwise}.
\end{cases}
\]
Evaluate $\E\Big(\dfrac1{X+Y}\Big)\,$.
\item Find the cumulative distribution function of $Y/X$ and use this result to find the probability density function of $\dfrac X {X+Y}$.
Write down $\E\Big( \dfrac X {X+Y}\Big)$ and verify your result by integration.
\end{questionparts}\begin{questionparts}
\item $\mathbb{P}(X + Y \leq c) $ is the area between the $x$-axis, $y$-axis and the line $x + y = c$. There are two cases for this:
\[\mathbb{P}(X + Y \leq c) = \begin{cases}
0 & \text{ if } c \leq 0 \\
\frac{c^2}{2} & \text{ if } c \leq 1 \\
1- \frac{(2-c)^2}{2} & \text{ if } 1 \leq c \leq 2 \\
1 & \text{ otherwise} \end{cases}\]
\item
\begin{align*}
&& \mathbb{P}((X + Y)^{-1} \leq t) &= 1- \mathbb{P}(X + Y \leq \frac1{t}) \\
\Rightarrow && f_{(X+Y)^{-1}}(t) &= 0 -\begin{cases}
0 & \text{ if } \frac1{t} \leq 0 \\
\frac{\d}{\d t}\frac{1}{2t^2} & \text{ if } \frac{1}{t} \leq 1 \\
\frac{\d}{\d t} \l 1- \frac{(2-\frac1t)^2}{2} \r & \text{ if } 1 \leq \frac{1}{t} \leq 2 \\
0 & \text{ otherwise}\end{cases} \\
&& &= \begin{cases}
t^{-3} & \text{ if } t \geq 1 \\
(2-\frac1t)t^{-2} & \text{ if } \frac12 \leq t \leq 1\\
0 & \text{ otherwise}\end{cases} \\
&& &= \begin{cases}
t^{-3} & \text{ if } t \geq 1 \\
2t^{-2}-t^{-3} & \text{ if } \frac12 \leq t \leq 1\\
0 & \text{ otherwise}\end{cases}
\end{align*}
Therefore,
\begin{align*}
\E \Big(\dfrac1{X+Y}\Big) &= \int_{\frac12}^{\infty} t f_{(X+Y)^{-1}}(t) \, \d t \\
&= \int_{\frac12}^{1} t f_{(X+Y)^{-1}}(t) \, \d t + \int_{1}^{\infty} t f_{(X+Y)^{-1}}(t) \d t\\
&= \int_{\frac12}^{1} \l 2t^{-1} - t^{-2} \r \, \d t + \int_{1}^{\infty} t^{-2} \d t\\
&= \left [ 2 \ln (t) + t^{-1} \right]_{\frac12}^{1} + \left [ -t^{-1} \right ]_{1}^{\infty} \\
&= 1 + 2 \ln 2 -2 + 1 \\
&= 2 \ln 2
\end{align*}
\item \begin{align*}
&&\mathbb{P} \l \frac{Y}{X} \leq c \r &= \mathbb{P}( Y \leq c X) \\
&&&= \begin{cases} 0 & \text{if } c \leq 0 \\
\frac{c}{2} & \text{if } 0 \leq c \leq 1 \\
1-\frac{1}{2c} & \text{if } 1 \leq c \end{cases} \\
\\
\Rightarrow && \mathbb{P} \l \frac{X}{X+Y} \leq t\r &= \mathbb{P} \l \frac{1}{1+\frac{Y}{X}} \leq t\r \\
&&&= \mathbb{P} \l \frac{1}{t} \leq 1+\frac{Y}{X}\r \\
&&&= \mathbb{P} \l \frac{1}{t} - 1\leq \frac{Y}{X}\r \\
&&&= 1- \mathbb{P} \l \frac{Y}{X} \leq \frac{1}{t} - 1\r \\
&&&= 1 - \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\
\frac{1}{2t} - \frac{1}{2} & \text{if } 0 \leq \frac1{t} \leq 1 \\
1-\frac{t}{2-2t} & \text{if } 1 \leq \frac1{t} \end{cases} \\
&& f_{\frac{X}{X+Y}}(t) &= \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\
\frac{1}{2t^2} & \text{if } t \geq 1 \\
\frac{1}{2(1-t)^2} & \text{if } 0 \leq t \leq 1 \end{cases} \\
\Rightarrow && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^\infty t f(t) \d t \\
&&&= \int_0^1 \frac{1}{2(1-t)^2} \d t + \int_1^\infty \frac{1}{t^2} \d t \\
&&& = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \\
\\
&& \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^1 \int_0^1 \frac{x}{x+y} \d y\d x \\
&&&= \int_0^1 \l x \ln (x+1) - x \ln x \r \d x \\
&&&= \left [\frac{x^2}2 \ln(x+1) - \frac{x^2}{2} \ln(x) \right]_0^1 -\int_0^1 \l \frac{x^2}{2(x+1)} - \frac{x}{2} \r \d x \\
&&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x^2-1+1}{2(x+1)}\d x \\
&&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x -1}{2} + \frac{1}{2(x+1)}\d x \\
&&&= \frac{\ln 2}{2} + \frac{1}{4} - \frac{1}{4} + \frac{1}{2} - \frac{\ln 2}{2} \\
&&&= \frac{1}{2}
\end{align*}
We can also notice that $1 = \mathbb{E} \l \frac{X+Y}{X+Y} \r = \mathbb{E} \l \frac{X}{X+Y} \r + \mathbb{E} \l \frac{Y}{X+Y} \r = 2 \mathbb{E} \l \frac{X}{X+Y} \r$ so it's clearly true as long as we can show that the integral converges.
\end{questionparts}
The large majority of attempts got almost no marks, and as a consequence this was the second worst scoring question. A lot failed to draw the right sort of graph to attempt the first part of (i) or, if they did, frequently miscalculated the area to find the required quantity in the case 1 ≤ x ≤ 2. The next major problem was an inability to see how to find the cumulative distribution function. A surprisingly large number failed to multiply by the variable before integrating to find the expectation. A handful of candidates got most of the question right although only one made it clear with a symmetry argument why they could write down the result.