Year: 2020
Paper: 3
Question Number: 11
Course: LFM Stats And Pure
Section: Continuous Probability Distributions and Random Variables
No solution available for this problem.
In spite of the change to criteria for entering the paper, there was still a very healthy number of candidates, and the vast majority handled the protocols for the online testing very well. Just over half the candidates attempted exactly six questions, and whilst about 10% attempted a seventh, hardly any did more than seven. With 20% attempting five questions, and 10% attempting only four, overall, there were very few candidates not attempting the target number. There was a spread of popularity across the questions, with no question attracting more than 90% of candidates and only one less than 10%, but every question received a good number of attempts. Likewise, there was a spread of success on the questions, though every question attracted at least one perfect solution.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The continuous random variable $X$ is uniformly distributed on $[a,b]$ where $0 < a < b$.
\begin{questionparts}
\item Let $\mathrm{f}$ be a function defined for all $x \in [a,b]$
\begin{itemize}
\item with $\mathrm{f}(a) = b$ and $\mathrm{f}(b) = a$,
\item which is strictly decreasing on $[a,b]$,
\item for which $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ for all $x \in [a,b]$.
\end{itemize}
The random variable $Y$ is defined by $Y = \mathrm{f}(X)$. Show that
\[ \mathrm{P}(Y \leqslant y) = \frac{b - \mathrm{f}(y)}{b - a} \quad \text{for } y \in [a,b]. \]
Find the probability density function for $Y$ and hence show that
\[ \mathrm{E}(Y^2) = -ab + \int_a^b \frac{2x\,\mathrm{f}(x)}{b-a} \; \mathrm{d}x. \]
\item The random variable $Z$ is defined by $\dfrac{1}{Z} + \dfrac{1}{X} = \dfrac{1}{c}$ where $\dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$. By finding the variance of $Z$, show that
\[ \ln\left(\frac{b-c}{a-c}\right) < \frac{b-a}{c}. \]
\end{questionparts}
Just one candidate more attempted this question than question 12, and with 20% attempting it, it was the most popular of the applied questions. Overall, there was only moderate success with the mean score just slightly better than 40%. However, there was a wide range of attempts, and although only a few obtained full marks, there were a number of strong attempts that just dropped a few marks in passing. The first part of the question was generally well attempted, with many candidates gaining full marks. However, some struggled with the initial justification, often by failing to properly use and justify the decreasing property of the function, whilst others were led astray by attempting to find an explicit form for the function, by attempting to sketch a graph instead of providing a proof, or by failing to notice the reversal of the inequality at all. Candidates had more difficulty with the second part of the question. Some failed to justify the use of the previous part, whilst others confused f(x) with the pdf of Z or Y. Many candidates correctly realised that they would need to use the strict positivity of the variance, but due to algebraic errors or other issues were unable to simplify to the required result. Finally, to receive full marks, candidates needed to ensure that relevant terms were positive in order to rearrange the inequality, which many failed to do.