Year: 2021
Paper: 3
Question Number: 11
Course: LFM Stats And Pure
Section: Continuous Probability Distributions and Random Variables
The total entry was a marginal increase from that of 2019, that of 2020 having been artificially reduced. Comfortably more than 90% attempted one of the questions, four others were very popular, and a sixth was attempted by 70%. Every question was attempted by at least 10% of the candidature. 85% of candidates attempted no more than 7 questions, though very nearly all the candidates made genuine attempts on at most six questions (the extra attempts being at times no more than labelling a page or writing only the first line or two). Generally, candidates should be aware that when asked to "Show that" they must provide enough working to fully substantiate their working, and that they should follow the instructions in a question, so if it says "Hence", they should be using the previous work in the question in order to complete the next part. Likewise, candidates should be careful when dividing or multiplying, that things are positive, or at other times non-zero.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The continuous random variable $X$ has probability density function
\[
f(x) = \begin{cases} \lambda e^{-\lambda x} & \text{for } x \geqslant 0, \\ 0 & \text{otherwise,} \end{cases}
\]
where $\lambda$ is a positive constant.
The random variable $Y$ is the greatest integer less than or equal to $X$, and $Z = X - Y$.
\begin{questionparts}
\item Show that, for any non-negative integer $n$,
\[
\mathrm{P}(Y = n) = (1 - e^{-\lambda})\,e^{-n\lambda}.
\]
\item Show that
\[
\mathrm{P}(Z < z) = \frac{1 - e^{-\lambda z}}{1 - e^{-\lambda}} \qquad \text{for } 0 \leqslant z \leqslant 1.
\]
\item Evaluate $\mathrm{E}(Z)$.
\item Obtain an expression for
\[
\mathrm{P}(Y = n \text{ and } z_1 < Z < z_2),
\]
where $0 \leqslant z_1 < z_2 \leqslant 1$ and $n$ is a non-negative integer.
Determine whether $Y$ and $Z$ are independent.
\end{questionparts}
\begin{questionparts}
\item $\,$ \begin{align*}
&& \mathbb{P}(Y = n) &= \mathbb{P}(X \in [n, n+1)) \\
&&&= \int_n^{n+1} \lambda e^{-\lambda x} \d x \\
&&&= \left [-e^{-\lambda x} \right]_n^{n+1} \\
&&&= e^{-\lambda n} - e^{-\lambda(n+1)} \\
&&&= e^{-\lambda n}(1- e^{-\lambda})
\end{align*}
\item $,$ \begin{align*}
&& \mathbb{P}(Z < z) &= \sum_{i=0}^{\infty} \mathbb{P}(X \in (n, n+z)) \\
&&&= \sum_{i=0}^{\infty} \int_{n}^{n+z} \lambda e^{-\lambda x} \d x \\
&&&= \sum_{i=0}^{\infty} [-e^{-\lambda x}]_{n}^{n+z} \\
&&&= \sum_{i=0}^{\infty} (1-e^{-\lambda x})e^{-\lambda n} \\
&&&= \frac{1-e^{-\lambda x}}{1-e^{-\lambda}}
\end{align*}
\item Give the cdf of $Z$, we see that $f_Z(z) = \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}}$ so
\begin{align*}
&& \E[Z] &= \int_0^1 z \frac{\lambda e^{-\lambda z}}{1-e^{-\lambda}} \d z \\
&&&= \frac{\lambda}{1-e^{-\lambda}} \int_0^1 ze^{-\lambda z} \d z \\
&&&= \frac{\lambda}{1-e^{-\lambda}} \left ( \left [-\frac{1}{\lambda} ze^{-\lambda z} \right]_0^1+\int_0^1 \frac{1}{\lambda} e^{-\lambda z} \d z \right) \\
&&&= \frac{\lambda}{1-e^{-\lambda}} \left ( -\frac{e^{-\lambda}}{\lambda} + \frac{1-e^{-\lambda}}{\lambda^2} \right) \\
&&&= \frac{1-e^{-\lambda}(1+\lambda)}{\lambda (1-e^{-\lambda})}
\end{align*}
\item $\,$ \begin{align*}
&& \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2)&= \mathbb{P}(X \in (n+z_1, n+z_2) ) \\
&&&= \int_{n+z_1}^{n+z_2} \lambda e^{-\lambda x} \d x \\
&&&= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2})
\end{align*}
Note that $\mathbb{P}(z_1 < Z < z_2) = \mathbb{P}( Z < z_2) -\mathbb{P}(Z< z_1) =\frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}}$
Therefore \begin{align*}
&& \mathbb{P}(Y = n \text{ and }z_1 < Z < z_2) &= e^{-n\lambda}(e^{-\lambda z_1} - e^{-\lambda z_2}) \\
&&&= e^{-\lambda n}(1-e^{-\lambda}) \frac{e^{-\lambda z_1} - e^{-\lambda z_2}}{1-e^{-\lambda}} \\
&&&= \mathbb{P}(Y=n) \mathbb{P}(z_1 < Z < z_2)
\end{align*}
So they are independent, which is to be expected from the memorylessness property of the exponential distribution.
\end{questionparts}
Comfortably the most popular applied question on the paper attracting slightly more than a third of candidates, it was the second most successful on the whole paper with a mean of 11/20. The quality of attempts for this question was high, with many candidates scoring full or close to full marks. Almost all candidates attempting it dealt with part (i) successfully. However, in part (ii) candidates often made incorrect conditioning arguments. The most common errors were computing P(Z < z|Y = n) rather than P(Z < z) and confusing P(Z < z|Y = n) with P(Z < z and Y = n). In part (iii), most candidates suitably obtained a probability density function for Z, but there were several computational mistakes in the integration by parts to evaluate the expectation. The independence argument in part (iv) was largely well executed, even when candidates had been unsuccessful in answering parts (ii) and (iii) of the question.