Year: 2017
Paper: 3
Question Number: 12
Course: UFM Statistics
Section: Bivariate data
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.2
Banger Comparisons: 2
The discrete random variables $X$ and $Y$ can each take the values $1$, $\ldots\,$, $n$ (where $n\ge2$). Their joint probability distribution is given by
\[
\P(X=x, \ Y=y) = k(x+y) \,,
\]
where $k$ is a constant.
\begin{questionparts}
\item Show that \[
\P(X=x) = \dfrac{n+1+2x}{2n(n+1)}\,.
\]
Hence determine whether $X$ and $Y$ are independent.
\item Show that the covariance of $X$ and $Y$ is negative.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& \mathbb{P}(X = x) &= \sum_{y=1}^n \mathbb{P}(X=x,Y=y) \\
&&&= \sum_{y=1}^n k(x+y) \\
&&&= nkx + k\frac{n(n+1)}2 \\
\\
&& 1 &= \sum_{x=1}^n \mathbb{P}(X=x) \\
&&&= nk\frac{n(n+1)}{2} + kn\frac{n(n+1)}2 \\
&&&= kn^2(n+1) \\
\Rightarrow && k &= \frac{1}{n^2(n+1)} \\
\Rightarrow && \mathbb{P}(X = x) &= \frac{nx}{n^2(n+1)} + \frac{n(n+1)}{2n^2(n+1)} \\
&&&= \frac{n+1+2x}{2n(n+1)} \\
\\
&& \mathbb{P}(X=x)\mathbb{P}(Y=y) &= \frac{(n+1)^2+2(n+1)(x+y)+4xy}{4n^2(n+1)^2} \\
&&&\neq \frac{x+y}{n^2(n+1)}
\end{align*}
Therefore $X$ and $Y$ are not independent.
\item $\,$ \begin{align*}
&& \E[X] &= \sum_{x=1}^n x \mathbb{P}(X=x) \\
&&&= \sum_{x=1}^n x \mathbb{P}(X=x)\\
&&&= \sum_{x=1}^n x \frac{n+1+2x}{2n(n+1)} \\
&&&= \frac{1}{2n(n+1)} \left ( (n+1) \sum x + 2\sum x^2\right)\\
&&&= \frac{1}{2n(n+1)} \left ( \frac{n(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{3} \right) \\
&&&= \frac{1}{2} \left ( \frac{n+1}{2} + \frac{2n+1}{3} \right)\\
&&&= \frac{1}{2} \left ( \frac{7n+5}{6} \right)\\
&&&= \frac{7n+5}{12}
\\
\\
&& \textrm{Cov}(X,Y) &= \mathbb{E}\left[XY\right] - \E[X] \E[Y] \\
&&&= \sum_{x=1}^n \sum_{y=1}^n xy \frac{x+y}{n^2(n+1)} - \E[X]^2 \\
&&&= \frac{1}{n^2(n+1)} \sum \sum (x^2 y+xy^2) - \E[X]^2 \\
&&&= \frac{1}{n^2(n+1)} \left (\sum y \right )\left (\sum x^2\right ) - \E[X]^2 \\
&&&=\frac{(n+1)(2n+1)}{12} - \left ( \frac{7n+5}{12}\right)^2 \\
&&&= \frac1{144} \left (12(2n^2+3n+1) - (49n^2+70n+25) \right)\\
&&&= \frac{1}{144} \left (-25n^2-34n-13 \right) \\
&&& < 0
\end{align*}
since $\Delta = 34^2 - 4 \cdot 25 \cdot 13 = 4(17^2-25 \times 13) = -4 \cdot 36 < 0$
\end{questionparts}
The least popular question on the paper, it was still attempted by just over 10% of the entry achieving marks only very marginally less good than for question 3. It was fairly well done overall, though a few were completely confused, so the marks tended to either be very high, very low or about around half marks for some who did not do much on part (ii).