# Question 2:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $M_X(\theta) = E\ e^{\theta x} = \int_0^\infty \frac{e^{\theta x - \frac{x}{\phi}} x^{k-1}}{\phi^k\ k-1\ !} dx$ | M1 | Correct integral with attempt to integrate by parts. |
| $= \frac{1}{\phi^k\ k-1\ !}\left[\left[x^{k-1} \frac{e^{\theta x - \frac{x}{\phi}}}{\theta - \frac{1}{\phi}}\right]_0^\infty - \int_0^\infty \frac{e^{\theta x - \frac{x}{\phi}}}{\theta - \frac{1}{\phi}}\ k-1\ x^{k-2} dx\right]$ | A1 B1 A1 | A1 for first term in integral by parts. B1 for sight of first term with limits. A1 for second term in integral by parts. |
| $= \frac{1}{\phi^k\ k-1\ !} \cdot \frac{\phi\ k-1}{1-\phi\theta} \int_0^\infty e^{\theta x - \frac{x}{\phi}} x^{k-2} dx$ | M1 M1 | M1 for re-arrangement into form that re-creates same form as original integral with $k-2$ instead of $k-1$. M1 for attempt at "reduction formula" method. |
| $= \frac{1}{\phi^k\ k-1\ !} \cdot \frac{\phi^2\ k-1\ k-2}{(1-\phi\theta)^2} \int_0^\infty e^{\theta x - \frac{x}{\phi}} x^{k-3} dx$ | | |
| $= \frac{1}{\phi^k\ k-1\ !} \cdot \frac{\phi^{k-1}\ k-1\ k-2\ \ldots 1}{(1-\phi\theta)^{k-1}} \int_0^\infty e^{\theta x - \frac{x}{\phi}} x^0 dx$ | A1 A1 A1 | A1 for power $k-1$ of $\phi$ in numerator. A1 for $(k-1)(k-2)\ldots 1$ in numerator. A1 for power $k-1$ of $(1-\phi\theta)$ in denominator. Final integral and integration explicit for following marks (AG). |
| The integral on the previous line is $\left[\frac{e^{\theta x - \frac{x}{\phi}}}{\theta - \frac{1}{\phi}}\right]_0^\infty$ | M1 A1 | M1 for attempt to deal with the $k=0$ term. A1 for $\frac{\phi}{1-\phi\theta}$ (may be implicit). |
| $= \frac{1}{\phi^k\ k-1\ !} \cdot \frac{\phi^k\ k-1\ !}{(1-\phi\theta)^k} = (1-\phi\theta)^{-k}$ | A1 | Beware printed answer. |
| **[12]** | | |

## Part (ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $M_W(\theta) = (1-\phi\theta)^{-nk}$ | B1 | |
| **[1]** | | |

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $M_Y(\theta) = \left(1 - \frac{2\phi\theta}{\phi}\right)^{-nk} = (1-2\theta)^{-nk}$ | B1 | |
| This is mgf of $\chi^2_{2nk}$, so by inversion theorem/uniqueness property, distribution of $Y$ is $\chi^2_{2nk}$. | M1 B1 | For some allusion to inversion theorem/uniqueness property. May be awarded for correct conclusion even if preceding M1 was not awarded. |
| **[3]** | | |

## Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| We have $\frac{2W}{\phi} \sim \chi^2_{2nk}$, and $0.95 = P\left(l < \chi^2_{2nk} < u\right)$ | | Must link 0.95, $\chi^2_{2nk}$ to $l$ and $u$, thence to $Y = \frac{2W}{\phi}$. (AG) |
| $\therefore 0.95 = P\left(l < \frac{2W}{\phi} < u\right)$ | | |
| Inserting an observed value $w$ of $W$ gives $2w/u < \phi$ and $2w/l > \phi$ so that a 95% confidence interval for $\phi$ is $(2w/u,\ 2w/l)$. | B2,1,0 | Final CI must not be embedded in a probability statement. |
| **[2]** | | |

## Part (v)

| Answer | Marks | Guidance |
|--------|-------|----------|
| We have $\chi^2_{40}$. | B1 | Follow-through from here if not 40 d.f. |
| Confidence interval is $\left(\frac{2w}{59.34},\ \frac{2w}{24.43}\right)$ | A1 | Both critical points correct, using candidate's d.f. |
| We need $E(W)$. Mgf of $W$ is $M_W(\theta) = (1-\phi\theta)^{-20}$ (see part (ii)). | M1 | For realising $E(W)$ is needed. $nk = 20$ may be seen below. |
| $E(W) = M'(0)$. | M1 | For attempt to find first derivative at zero, or by expanding the series. |
| $M'(\theta) = -20(1-\phi\theta)^{-21}(-\phi)$ $\therefore M'(0) = 20\phi$ | A1 | |
| $\therefore E(\text{length of CI}) = \frac{2 \times 20\phi}{24.43} - \frac{2 \times 20\phi}{59.34}$ | A1 | cao (no follow-through from candidate's incorrect d.f.) |
| $= 40\phi(0.040933 - 0.016852) = 40\phi(0.0240812) = 0.963(25)\phi$ | | Usual penalty for over-specification of final answer. (3 or 4 sf, awrt 0.963) |
| **[6]** | | |