Year: 2022
Paper: 2
Question Number: 12
Course: LFM Stats And Pure
Section: Continuous Probability Distributions and Random Variables
Candidates appeared to be generally well prepared for most topics within the examination, but there were a few situations in questions where some did not appear to be as proficient in standard techniques as needed. In particular, the method for finding invariant lines required in question 8 and the manipulation of trigonometric functions that were needed in question 10 caused considerable difficulties for some candidates. An additional issue that occurred at numerous points in the paper relates to the direction in which a deduction is required. It is important that candidates make sure that they know which statement is the one that they should start from as they deduce the other and that it is clear in their solution that the logic has gone in the correct direction. Clarity of solution is also an issue that candidates should be aware of, especially in the situations where the result to be reached has been given. It is important to check that there are no special cases that need to be considered separately, and when dividing by a function it is necessary to confirm that the function cannot be equal to 0 (and in the case of inequalities that the function always has the same sign). When drawing diagrams and sketching graphs it is useful if significant points that need to be clear are not drawn over the lines on the page as these can be difficult to interpret during the marking process.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The random variable $X$ has probability density function
\[\mathrm{f}(x) = \begin{cases} kx^n(1-x) & 0 \leqslant x \leqslant 1\,,\\ 0 & \text{otherwise}\,,\end{cases}\]
where $n$ is an integer greater than 1.
\begin{questionparts}
\item Show that $k = (n+1)(n+2)$ and find $\mu$, where $\mu = \mathrm{E}(X)$.
\item Show that $\mu$ is less than the median of $X$ if
\[6 - \frac{8}{n+3} < \left(1 + \frac{2}{n+1}\right)^{n+1}.\]
By considering the first four terms of the expansion of the right-hand side of this inequality, or otherwise, show that the median of $X$ is greater than $\mu$.
\item You are given that, for positive $x$, $\left(1 + \dfrac{1}{x}\right)^{x+1}$ is a decreasing function of $x$.
Show that the mode of $X$ is greater than its median.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& 1 &= \int_0^1 kx^n(1-x) \d x \\
&&&= k\left [\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^1 \\
&&&= \frac{k}{(n+1)(n+2)} \\
\Rightarrow && k &= (n+1)(n+2) \\
\\
&& \E[X] &= \int_0^1 kx^{n+1}(1-x) \d x \\
&&&= k\left [ \frac{x^{n+2}}{n+2} - \frac{x^{n+3}}{n+3} \right]_0^1 \\
&&&= \frac{(n+1)(n+2)}{(n+2)(n+3)} \\
&&&= \frac{n+1}{n+3}
\end{align*}
\item If $\mu$ is less than the median then \begin{align*}
&& \frac12 &> \int_0^{\mu} kx^{n}(1-x) \d x \\
&&&= \left [k \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^\mu \\
&&&= \mu^{n+1}\left ((n+2) - (n+1) \frac{n+1}{n+3} \right) \\
\Rightarrow && \left ( 1 + \frac{2}{n+1} \right)^{n+1} &> 2 \frac{(n+2)(n+3) - (n+1)^2}{n+3} \\
&&&= \frac{6n+10}{n+3} = 6 - \frac{8}{n+3}
\end{align*}
\begin{align*}
&& \left ( 1 + \frac{2}{n+1} \right)^{n+1} &= 1 + (n+1) \frac{2}{n+1} + \frac{(n+1)n}{2} \frac{4}{(n+1)^2} + \frac{(n+1)n(n-1)}{6} \frac{8}{(n+1)^3} + \cdots \\
&&&= 1 + 2 + \frac{2n}{n+1} + \frac{4n(n-1)}{3(n+1)^2} + \cdots \\
&&&= 5 - \frac{2}{n+1} + \frac{4((n+1)^2-3(n+1)+3)}{3(n+1)^2} \\
&&&= 6 + \frac13 - \frac{6}{n+1} - \frac{4}{3(n+1)^2} \\
&&&> 6 - \frac{8}{n+3}
\end{align*} as required.
\item To find the mode, we need to find the maximum of $x^n(1-x)$ which occurs when $nx^{n-1}(1-x) - x^n = x^{n-1}(n-(n+1)x)$ ie where $x = \frac{n}{n+1}$, therefore we need to look at:
\begin{align*}
&& \mathbb{P}(X \leq \tfrac{n}{n+1}) &= \int_0^{n/(n+1)} k x^n(1-x) \d x \\
&&&= \frac{n^{n+1}}{(n+1)^{n+1}} \left ( (n+2) - (n+1) \frac{n}{n+1} \right) \\
&&&= 2\frac{n^{n+1}}{(n+1)^{n+1}} \\
&&&= 2 \left ( \frac{1}{1+\frac1n} \right)^{n+1} \\
&&&\geq 2 \frac{1}{(1 + \frac11)^2} = \frac12
\end{align*} as required
\end{questionparts}
Part (i) of the question was generally well attempted, although some candidates opted to integrate by parts rather than noticing that the integral was simple once the brackets were expanded. A small number of candidates failed to simplify the mean fully. Part (ii) was found to be difficult. Most candidates attempted to compute an expression for the median rather than comparing the cumulative density function of the mean to ½. Those who did follow the intended approach were generally able to work through the algebra well. Many candidates were also confused about the direction of the logic in this question and instead showed the converse of the required result. In many cases the argument provided was reversible so many of the marks could still be awarded. Expansion of the binomial expression was generally done well, but the algebra of the part that followed proved difficult, with most candidates either giving up early on or making mistakes that either rendered the conclusion trivial or impossible to obtain. Only a handful of candidates successfully reached the correct condition and convincingly showed it to be true. Many candidates did not attempt part (iii). Those who had been successful in part (ii) almost always realised that they needed to consider the cumulative density function of the mode and compare it to ½ and almost all these candidates managed to deduce the argument completely and gain full credit.