| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2016 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find PGF from probability distribution |
| Difficulty | Hard +2.3 This is a challenging Further Maths Statistics question requiring deep understanding of maximum likelihood estimation with a non-standard distribution (Cauchy). Part (iii) demands algebraic manipulation of log-likelihood derivatives to obtain a cubic equation, part (iv) requires analyzing roots using the discriminant and understanding MLE theory, and part (v) involves evaluating multiple candidates. The multi-step nature, proof elements, and requirement to work with an unfamiliar distribution place this well above typical A-level questions. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Graph: bell-shaped curve centred on \(x = m\), symmetric, tails approaching zero | G2 | |
| Mode: \(f(x)\) takes maximum when \(x - m = 0\) (or differentiate and set to zero) | M1E1 | Derivative is \(-\frac{1}{\pi}\frac{2(x-m)}{(1+(x-m)^2)^2}\) |
| Median: integral is \(\frac{1}{\pi}\arctan(x-m)\). Evaluated between \(-\infty\) and \(m\) or \(m\) and \(\infty\) this evaluates to 0.5 | M1A1 | "symmetry" SC B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Likelihood function is \(f(x_1)\) | B1 | |
| Differentiate wrt \(m\) and set to zero: \(\frac{1}{\pi}\frac{2(x_1-m)}{(1+(x_1-m)^2)^2}=0\) | M1 | |
| Giving \(x_1 = m\) | A1 | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Likelihood function is \(f(x_1)f(x_2)\) | B1 | |
| The derivative of the log likelihood wrt \(m\) is \(\frac{x_1-m}{1+(x_1-m)^2}+\frac{x_2-m}{1+(x_2-m)^2}\) | M1A1 | |
| Setting to zero and convincing algebra to given result | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\hat{m}=\frac{1}{2}(x_1+x_2)\) | B1 | |
| \(\hat{m}=\frac{1}{2}(x_1+x_2)\pm\frac{1}{2}\sqrt{(x_1-x_2)^2-4}\) | M1A1 | Or equivalent |
| Convincing argument that if \( | x_1-x_2 | <2\) then only 1 root is real |
| Convincing argument that the ML function is positive, continuous and tends to zero at \(\pm\infty\). Hence the only turning point is a maximum | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The three values of \(\hat{m}\) are: \(0, \pm\sqrt{3}\) | B1B1 | B1 for 0, B1 both |
| Likelihood at zero is \(\frac{1}{25\pi^2}\) or \(\frac{0.04}{\pi^2}\) (approx. 0.0040) | B1 | Accept approx. values |
| Likelihood at \(\pm\sqrt{3}\) is \(\frac{1}{16\pi^2}\) or \(\frac{0.0625}{\pi^2}\) (approx. 0.0063) | B1 | |
| So \(\pm\sqrt{3}\) are joint MLE, 0 is not MLE | E1 |
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph: bell-shaped curve centred on $x = m$, symmetric, tails approaching zero | G2 | |
| Mode: $f(x)$ takes maximum when $x - m = 0$ (or differentiate and set to zero) | M1E1 | Derivative is $-\frac{1}{\pi}\frac{2(x-m)}{(1+(x-m)^2)^2}$ |
| Median: integral is $\frac{1}{\pi}\arctan(x-m)$. Evaluated between $-\infty$ and $m$ or $m$ and $\infty$ this evaluates to 0.5 | M1A1 | "symmetry" SC B1 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Likelihood function is $f(x_1)$ | B1 | |
| Differentiate wrt $m$ and set to zero: $\frac{1}{\pi}\frac{2(x_1-m)}{(1+(x_1-m)^2)^2}=0$ | M1 | |
| Giving $x_1 = m$ | A1 | Answer given |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Likelihood function is $f(x_1)f(x_2)$ | B1 | |
| The derivative of the log likelihood wrt $m$ is $\frac{x_1-m}{1+(x_1-m)^2}+\frac{x_2-m}{1+(x_2-m)^2}$ | M1A1 | |
| Setting to zero and convincing algebra to given result | M1A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\hat{m}=\frac{1}{2}(x_1+x_2)$ | B1 | |
| $\hat{m}=\frac{1}{2}(x_1+x_2)\pm\frac{1}{2}\sqrt{(x_1-x_2)^2-4}$ | M1A1 | Or equivalent |
| Convincing argument that if $|x_1-x_2|<2$ then only 1 root is real | E1 | |
| Convincing argument that the ML function is positive, continuous and tends to zero at $\pm\infty$. Hence the only turning point is a maximum | E1 | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| The three values of $\hat{m}$ are: $0, \pm\sqrt{3}$ | B1B1 | B1 for 0, B1 both |
| Likelihood at zero is $\frac{1}{25\pi^2}$ or $\frac{0.04}{\pi^2}$ (approx. 0.0040) | B1 | Accept approx. values |
| Likelihood at $\pm\sqrt{3}$ is $\frac{1}{16\pi^2}$ or $\frac{0.0625}{\pi^2}$ (approx. 0.0063) | B1 | |
| So $\pm\sqrt{3}$ are joint MLE, 0 is not MLE | E1 | |
---1 The random variable $X$ has a Cauchy distribution centred on $m$. Its probability density function ( pdf ) is $\mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = \frac { 1 } { \pi } \frac { 1 } { 1 + ( x - m ) ^ { 2 } } , \quad \text { for } - \infty < x < \infty$$
\begin{enumerate}[label=(\roman*)]
\item Sketch the pdf. Show that the mode and median are at $x = m$.
\item A sample of size 1 , consisting of the observation $x _ { 1 }$, is taken from this distribution. Show that the maximum likelihood estimate (MLE) of $m$ is $x _ { 1 }$.
\item Now suppose that a sample of size 2 , consisting of observations $x _ { 1 }$ and $x _ { 2 }$, is taken from the distribution. By considering the logarithm of the likelihood function or otherwise, show that the MLE, $\hat { m }$, satisfies the cubic equation
$$\left( 2 \hat { m } - \left( x _ { 1 } + x _ { 2 } \right) \right) \left( \hat { m } ^ { 2 } - \left( x _ { 1 } + x _ { 2 } \right) \hat { m } + 1 + x _ { 1 } x _ { 2 } \right) = 0$$
\item Obtain expressions for the three roots of this equation. Show that if $\left| x _ { 1 } - x _ { 2 } \right| < 2$ then only one root is real. How do you know, without doing further calculations, that in this case the real root will be the MLE of $m$ ?
\item Obtain the three possible values of $\hat { m }$ in the case $x _ { 1 } = - 2$ and $x _ { 2 } = 2$. Evaluate the likelihood function for each value of $\hat { m }$ and comment on your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S4 2016 Q1 [24]}}