Year: 2011
Paper: 2
Question Number: 13
Course: UFM Statistics
Section: Cumulative distribution functions
There were just under 1000 entries for paper II this year, almost exactly the same number as last year. After the relatively easy time candidates experienced on last year's paper, this year's questions had been toughened up significantly, with particular attention made to ensure that candidates had to be prepared to invest more thought at the start of each question – last year saw far too many attempts from the weaker brethren at little more than the first part of up to ten questions, when the idea is that they should devote 25-40 minutes on four to six complete questions in order to present work of a substantial nature. It was also the intention to toughen up the final "quarter" of questions, so that a complete, or nearly-complete, conclusion to any question represented a significant (and, hopefully, satisfying) mathematical achievement. Although such matters are always best assessed with the benefit of hindsight, our efforts in these areas seem to have proved entirely successful, with the vast majority of candidates concentrating their efforts on four to six questions, as planned. Moreover, marks really did have to be earned: only around 20 candidates managed to gain or exceed a score of 100, and only a third of the entry managed to hit the half-way mark of 60. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were attempted by almost all candidates; 3 and 4 by around three-quarters of them; 6, 7 and 9 by around half; the remaining questions were less popular, and some received almost no "hits". Overall, the highest scoring questions (averaging over half-marks) were 1, 2 and 9, along with 13 (very few attempts, but those who braved it scored very well). This at least is indicative that candidates are being careful in exercising some degree of thought when choosing (at least the first four) 'good' questions for themselves, although finding six successful questions then turned out to be a key discriminating factor of candidates' abilities from the examining team's perspective. Each of questions 4-8, 11 & 12 were rather poorly scored on, with average scores of only 5.5 to 6.6.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
What property of a distribution is measured by its \textit{skewness}?
\begin{questionparts}
\item One measure of skewness, $\gamma$, is given by
\[
\displaystyle
\gamma=
\frac{ \E\big((X-\mu)^3\big)}{\sigma^3}\,,
\]
where $\mu$ and $\sigma^2$ are the mean and variance of the random variable $X$.
Show that
\[
\gamma = \frac{ \E(X^3) -3\mu \sigma^2 - \mu^3}{\sigma^3}\,.
\]
The continuous random variable $X$ has probability density function $\f$ where
\[
\f(x)
= \begin{cases}
2x & \text{for } 0\le x\le 1\,, \\[2mm]
0 & \text{otherwise}\,.
\end{cases}
\]
Show that for this distribution $\gamma= -\dfrac{2\sqrt2}{5}$.
\item The \textit{decile skewness}, $D$, of a distribution is defined by
\[D=
\frac { {\rm F}^{-1}(\frac9{10}) - 2{\rm F} ^{-1}(\frac12) + {\rm F}^{-1} (\frac1{10}) }
{{\rm F}^{-1}(\frac9{10}) - {\rm F} ^{-1} (\frac1{10})}\,,
\]
where ${\rm F}^{-1}$ is the inverse of the cumulative distribution function.
Show that, for the above distribution, $ D= 2 -\sqrt5\,.$
The \textit{Pearson skewness}, $P$, of a distribution is defined by
\[
P = \frac{3(\mu-M)}{\sigma}
\,,\]
where $M$ is the median.
Find $P$ for the above distribution and show that $D > P > \gamma\,$.
\end{questionparts}Skewness is a measure of the symmetry (specifically the lack-thereof) in the distribution. How much mass is there on one side rather than another.
\begin{questionparts}
\item $\,$ \begin{align*}
&& \gamma &= \frac{\E \left [ (X - \mu)^3 \right ]}{\sigma^3} \\
&&&= \frac{\E \left [ X^3 - 3\mu X^2 + 3\mu^2 X - \mu^3 \right ]}{\sigma^3} \\
&&&= \frac{\E [ X^3 ]- 3\mu \E[X^2] + 3\mu^2 \E[X] - \mu^3 }{\sigma^3} \\
&&&= \frac{\E [ X^3 ]- 3\mu (\mu^2 + \sigma^2) + 3\mu^2\cdot \mu- \mu^3 }{\sigma^3} \\
&&&= \frac{\E [ X^3 ]- 3\mu \sigma^2 - \mu^3 }{\sigma^3} \\
\end{align*}
\begin{align*}
&& f(x) &= \begin{cases}
2x & \text{for } 0\le x\le 1\,, \\[2mm]
0 & \text{otherwise}\,.
\end{cases} \\
&& \E[X] &= \int_0^1 2x^2 \d x \\
&&&= \frac23 \\
&& \E[X^2] &= \int_0^1 2x^3 \d x \\
&&&= \frac12 \\
&& \E[X^3] &= \int_0^1 2x^4 \d x \\
&&&= \frac25 \\
\\
&& \mu &= \frac23 \\
&& \sigma^2 &= \frac12 - \frac49 = \frac{1}{18} \\
&& \gamma &= \frac{\frac25 - 3 \cdot \frac23 \cdot \frac1{18} - \frac8{27}}{\frac{1}{54\sqrt2}} \\
&&&= -\frac{2\sqrt2}{5}
\end{align*}
\item First note that $\displaystyle F(x) = \int_0^x 2t \d t = x^2$ for $x \in [0,1]$. In particular, $F^{-1}(x) = \sqrt{x}$, so
\begin{align*}
&& D &=
\frac { {\rm F}^{-1}(\frac9{10}) - 2{\rm F} ^{-1}(\frac12) + {\rm F}^{-1} (\frac1{10}) }
{{\rm F}^{-1}(\frac9{10}) - {\rm F} ^{-1} (\frac1{10})} \\
&&&= \frac{\sqrt{\frac9{10}} - 2 \sqrt{\frac5{10}} + \sqrt{\frac1{10}}}{\sqrt{\frac9{10}}-\sqrt{\frac1{10}}} \\
&&&= \frac{3-2\sqrt5+1}{3 - 1} \\
&&&= \frac{4-2\sqrt5}{2} = 2-\sqrt5
\end{align*}
\begin{align*}
&& P &= \frac{3(\mu - M)}{\sigma} \\
&&&= \frac{3(\frac23 - \sqrt{\frac12})}{\frac{1}{3\sqrt2}} \\
&&&= 6 \sqrt2 - 9
\end{align*}
First we compare $P$ and $D$, $6\sqrt2-9$ and $2-\sqrt5$
\begin{align*}
&& D & > P \\
\Leftrightarrow && 2-\sqrt5 &> 6\sqrt2 - 9 \\
\Leftrightarrow && 11 -6\sqrt2 &> \sqrt 5 \\
\Leftrightarrow && (121 + 72 - 132\sqrt2) & > 5 \\
\Leftrightarrow && 188 & > 132\sqrt2 \\
\Leftrightarrow && 47 & > 33 \sqrt 2\\
\Leftrightarrow && 2209 & > 2178
\end{align*}
also
\begin{align*}
&& P &> \gamma \\
\Leftrightarrow && 6\sqrt2 - 9 &> -\frac{2\sqrt2}{5} \\
\Leftrightarrow && 30\sqrt2 - 45 & > -2\sqrt2 \\
\Leftrightarrow && 32 \sqrt 2 &> 45 \\
\Leftrightarrow && 2048 &> 2025
\end{align*}
\end{questionparts}
This question was almost as unpopular as question 11, receiving under 70 attempts, very few of which ventured an opening opinion as to what skewness might measure. Those who could handle expectations lived up to them and scored well; the rest just found the question a little too overwhelming in its demands.