Year: 2008
Paper: 1
Question Number: 12
Course: LFM Stats And Pure
Section: Continuous Uniform Random Variables
There were significantly more candidates attempting this paper this year (an increase of nearly 25%), but many found it to be very difficult and only achieved low scores. The mean score was significantly lower than last year, although a similar number of candidates achieved very high marks. This may be, in part, due to the phenomenon of students in the Lower Sixth (Year 12) being entered for the examination before attempting papers II and III in the Upper Sixth. This is a questionable practice, as while students have enough technical knowledge to answer the STEP I questions at this stage, they often still lack the mathematical maturity to be able to apply their knowledge to these challenging problems. Again, a key difficulty experienced by most candidates was a lack of the algebraic skill required by the questions. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many students were simply unable to progress on some questions because they did not know how to handle the algebra. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was also pleasing that one of the applied questions, question 13, attracted a very large number of attempts. However, the examiners were again left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers and other available resources to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1484.0
Banger Comparisons: 1
In this question, you may use without proof the results:
\[
\sum_{r=1}^n r = \tfrac12 n(n+1)
\qquad\text{and}\qquad
\sum_{r=1}^n r^2 = \tfrac1 6 n(n+1)(2n+1)\,.
\]
The independent random variables $X_1$ and $X_2$ each take values $1$, $2$, $\ldots$, $N$, each value being equally likely. The random variable $X$ is defined by
\[
X=
\begin{cases}
X_1 & \text { if } X_1\ge X_2\\
X_2 & \text { if } X_2\ge X_1\;.
\end{cases}
\]
\begin{questionparts}
\item Show that $\P(X=r) = \dfrac{2r-1}{N^2}\,$ for $r=1$, $2$, $\ldots$, $N$.
\item Find an expression for the expectation, $\mu$, of $X$ and show that $\mu=67.165$ in the case $N=100$.
\item The median, $m$, of $X$ is defined to be the integer such that
$\P(X\ge m) \ge \frac 12$ and $\P(X\le m)\ge \frac12$. Find an expression for $m$ in terms of $N$
and give an explicit value for $m$ in the case $N=100$.
\item Show that when $N$ is very large,
\[
\frac \mu m
\approx \frac {2\sqrt2}3\,.
\]
\end{questionparts}\begin{align*}
\P(X = r) &= \P(X_1 = r, X_2 \leq r) + \P(X_2 = r, X_1 < r) \\
&= \P(X_1 = r) \P(X_2 \leq r) + \P(X_2 = r)\P( X_1 < r) \\
&= \frac{1}{N} \frac{r}{N} + \frac{1}{N} \frac{r-1}{N} \\
&= \frac{2r-1}{N^2}
\end{align*}
\begin{align*}
\E(X) &= \sum_{r=1}^N r \P(X = r) \\
&= \sum_{r=1}^N \frac{2r^2 - r}{N^2} \\
&= \frac{1}{N^2} \l \frac{N(N+1)(2N+1)}{3} - \frac{N(N+1)}{2} \r \\
&= \frac{N+1}{N} \l \frac{4N-1}{6} \r
\end{align*}
When $N = 100$, this is equal to $\frac{101 \cdot 399}{6 \cdot 100} = \frac{101 \cdot 133}{200} = 67.165$
\begin{align*}
&&\frac12 &\leq \P(X \leq m) \\
&&&=\sum_{r=1}^m \P(X=r) \\
&&&=\sum_{r=1}^m \frac{2r-1}{N^2} \\
&&&= \frac{1}{N^2} \l m(m+1) - m \r \\
&&&= \frac{m^2}{N^2} \\
\Rightarrow && m^2 &\geq \frac{N^2}{2} \\
\Rightarrow && m &\geq \frac{N}{\sqrt{2}} \\
\Rightarrow && m &= \left \lceil \frac{N}{\sqrt{2}} \right \rceil
\end{align*}
When $N = 100$, $100/\sqrt{2} = \sqrt{2}50$.
$\sqrt{2} > 1.4 \Rightarrow 50\sqrt{2} > 70$
$\sqrt{2} < 1.42 \Rightarrow 50 \sqrt{2} < 71$, therefore $\displaystyle \left \lceil \frac{100}{\sqrt{2}} \right \rceil = 71$
\begin{align*}
\lim_{N \to \infty} \frac{\frac{(N+1)(4N-1)}{6N}}{ \left \lceil\frac{N}{\sqrt{2}} \right \rceil} &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l \frac{4N^2 +3N - 1}{2N^2} \r \tag{since the floor will be irrelevant}\\
&= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l 2 + \frac{3}{2N} - \frac{1}{N^2} \r \\
&= \lim_{N \to \infty} \frac{2\sqrt{2}}{3}
\end{align*}
This was by far the least popular question on the paper, as is often the case with Probability and Statistics questions. Of the candidates who attempted it, most successfully answered part (i), and a significant number were also confident in the manipulation of sums required for part (ii). Success was clearly dependent upon taking great care to ascertain the meaning of the event X = r in terms of X₁ and X₂. Part (iii) proved much more problematic, as almost no-one made use of both defining inequalities for the median; one inequality on its own may appear to give the correct answer, but is insufficient to gain the marks. The handful of candidates who attempted part (iv) were generally successful in their attempts.