| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Derive MGF from PDF |
| Difficulty | Challenging +1.8 This is a substantial Further Maths Statistics question requiring multiple MGF techniques: deriving MGF from a non-standard PDF via integration, differentiating MGFs for moments, applying the independence theorem, performing a linear transformation of MGF, and using series expansion to prove asymptotic normality. While each individual step follows standard procedures, the multi-part structure, algebraic manipulation with the series expansion, and the conceptual understanding of limiting distributions make this significantly harder than typical A-level questions but not exceptionally difficult for S4 level. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M_Y(\theta) = E(e^{\theta Y}) = \int_0^\infty e^{\theta y} \frac{1}{\sqrt{2\pi y}} e^{-\frac{1}{2}y} dy\) | M1, A1 | |
| \(= \int_0^\infty \frac{1}{\sqrt{2\pi y}} e^{-\frac{1}{2}y(1-2\theta)} dy\) | A1 | |
| Substitute \(u = y(1-2\theta)\), \(du = dy(1-2\theta)\) | M1 | |
| \(\int_0^\infty \frac{\sqrt{1-2\theta}}{\sqrt{2\pi u}} e^{-\frac{1}{2}u} \frac{1}{1-2\theta} du\) | A1 | |
| \(= (1-2\theta)^{-1/2} \int_0^\infty \frac{1}{\sqrt{2\pi u}} e^{-\frac{1}{2}u} du\) | B1 | |
| \(= (1-2\theta)^{-1/2}\) | [6] | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either expand the mgf as a power series: \((1-2\theta)^{-1/2} = 1 + \theta + 3\frac{\theta^2}{2!} + \ldots\) | M1A1 | |
| \(E(Y) = 1\) (coefficient of \(\theta\)) | B1 | |
| \(E(Y^2) = 3\) (coefficient of \(\theta^2/2!\)) | B1 | |
| Hence \(\text{Var}(Y) = 2\) | B1 [5] | |
| Or differentiate the mgf: 1st derivative simplifies to \((1-2\theta)^{-3/2}\) | M1 | |
| Putting \(\theta=0\) gives \(E(Y)=1\) | A1, B1 | |
| 2nd derivative simplifies to \(3(1-2\theta)^{-5/2}\) | A1 | |
| Putting \(\theta=0\) gives \(E(Y^2)=3\), hence \(\text{Var}(Y)=2\) | B1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For independent rvs \(X\) and \(Y\), \(M_{X+Y}(\theta) = M_X(\theta)M_Y(\theta)\) | B1 | |
| Hence \(M_U(\theta) = (1-2\theta)^{-n/2}\) | B1 | |
| \(E(U) = n\) | B1 | |
| \(\text{Var}(U) = 2n\) | B1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M_W(\theta) = E\left(\exp\left(\theta\left(\frac{U-n}{\sqrt{2n}}\right)\right)\right)\) | B1 | |
| \(= \exp\left(-\frac{\theta n}{\sqrt{2n}}\right) E\left(\exp\left(\frac{\theta}{\sqrt{2n}}U\right)\right)\) | B1 | Or by use of general linear transformation result |
| \(= \exp\left(-\frac{\theta n}{\sqrt{2n}}\right) M_U\left(\frac{\theta}{\sqrt{2n}}\right)\) | B1 | |
| \(= \exp\left(-\frac{\theta n}{\sqrt{2n}}\right)\left(1 - \frac{2\theta}{\sqrt{2n}}\right)^{-n/2}\) | B1 | |
| Expanding \(\ln(M_W(\theta))\) gives \(-\sqrt{\frac{n}{2}}\,\theta + \frac{n}{2}\left(\sqrt{\frac{2}{n}}\,\theta + \frac{1}{2}\left(\sqrt{\frac{2}{n}}\,\theta\right)^2\right) + \text{terms of order } n^{-1/2}\) | M1A1 | Award max 3 marks from here if no account taken of terms beyond \(\theta^2\) |
| Convincing simplification to \(\frac{1}{2}\theta^2 + \text{terms of order } n^{-1/2}\) | B1 | Answer given |
| Hence tends to \(\frac{1}{2}\theta^2\) | ||
| The mgf therefore tends to \(\exp(\frac{1}{2}\theta^2)\) | B1 | Uniqueness of mgfs may be implied |
| That is, the distribution of \(W\) tends to the standard Normal | B1 [9] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M_Y(\theta) = E(e^{\theta Y}) = \int_0^\infty e^{\theta y} \frac{1}{\sqrt{2\pi y}} e^{-\frac{1}{2}y} dy$ | M1, A1 | |
| $= \int_0^\infty \frac{1}{\sqrt{2\pi y}} e^{-\frac{1}{2}y(1-2\theta)} dy$ | A1 | |
| Substitute $u = y(1-2\theta)$, $du = dy(1-2\theta)$ | M1 | |
| $\int_0^\infty \frac{\sqrt{1-2\theta}}{\sqrt{2\pi u}} e^{-\frac{1}{2}u} \frac{1}{1-2\theta} du$ | A1 | |
| $= (1-2\theta)^{-1/2} \int_0^\infty \frac{1}{\sqrt{2\pi u}} e^{-\frac{1}{2}u} du$ | B1 | |
| $= (1-2\theta)^{-1/2}$ | [6] | Answer given |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Either expand the mgf as a power series: $(1-2\theta)^{-1/2} = 1 + \theta + 3\frac{\theta^2}{2!} + \ldots$ | M1A1 | |
| $E(Y) = 1$ (coefficient of $\theta$) | B1 | |
| $E(Y^2) = 3$ (coefficient of $\theta^2/2!$) | B1 | |
| Hence $\text{Var}(Y) = 2$ | B1 [5] | |
| Or differentiate the mgf: 1st derivative simplifies to $(1-2\theta)^{-3/2}$ | M1 | |
| Putting $\theta=0$ gives $E(Y)=1$ | A1, B1 | |
| 2nd derivative simplifies to $3(1-2\theta)^{-5/2}$ | A1 | |
| Putting $\theta=0$ gives $E(Y^2)=3$, hence $\text{Var}(Y)=2$ | B1 [5] | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| For independent rvs $X$ and $Y$, $M_{X+Y}(\theta) = M_X(\theta)M_Y(\theta)$ | B1 | |
| Hence $M_U(\theta) = (1-2\theta)^{-n/2}$ | B1 | |
| $E(U) = n$ | B1 | |
| $\text{Var}(U) = 2n$ | B1 [4] | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M_W(\theta) = E\left(\exp\left(\theta\left(\frac{U-n}{\sqrt{2n}}\right)\right)\right)$ | B1 | |
| $= \exp\left(-\frac{\theta n}{\sqrt{2n}}\right) E\left(\exp\left(\frac{\theta}{\sqrt{2n}}U\right)\right)$ | B1 | Or by use of general linear transformation result |
| $= \exp\left(-\frac{\theta n}{\sqrt{2n}}\right) M_U\left(\frac{\theta}{\sqrt{2n}}\right)$ | B1 | |
| $= \exp\left(-\frac{\theta n}{\sqrt{2n}}\right)\left(1 - \frac{2\theta}{\sqrt{2n}}\right)^{-n/2}$ | B1 | |
| Expanding $\ln(M_W(\theta))$ gives $-\sqrt{\frac{n}{2}}\,\theta + \frac{n}{2}\left(\sqrt{\frac{2}{n}}\,\theta + \frac{1}{2}\left(\sqrt{\frac{2}{n}}\,\theta\right)^2\right) + \text{terms of order } n^{-1/2}$ | M1A1 | Award max 3 marks from here if no account taken of terms beyond $\theta^2$ |
| Convincing simplification to $\frac{1}{2}\theta^2 + \text{terms of order } n^{-1/2}$ | B1 | Answer given |
| Hence tends to $\frac{1}{2}\theta^2$ | | |
| The mgf therefore tends to $\exp(\frac{1}{2}\theta^2)$ | B1 | Uniqueness of mgfs may be implied |
| That is, the distribution of $W$ tends to the standard Normal | B1 [9] | |
---2 The random variable $Z$ has the standard Normal distribution. The random variable $Y$ is defined by $Y = Z ^ { 2 }$.\\
You are given that $Y$ has the following probability density function.
$$\mathrm { f } ( y ) = \frac { 1 } { \sqrt { 2 \pi y } } \mathrm { e } ^ { - \frac { 1 } { 2 } y } , \quad y > 0$$
(i) Show that the moment generating function (mgf) of $Y$ is given by
$$\mathrm { M } _ { Y } ( \theta ) = ( 1 - 2 \theta ) ^ { - \frac { 1 } { 2 } }$$
(ii) Use the mgf to obtain $\mathrm { E } ( Y )$ and $\operatorname { Var } ( Y )$.
The random variable $U$ is defined by
$$U = Z _ { 1 } ^ { 2 } + Z _ { 2 } ^ { 2 } + \ldots + Z _ { n } ^ { 2 } ,$$
where $Z _ { 1 } , Z _ { 2 } , \ldots , Z _ { n }$ are independent standard Normal random variables.\\
(iii) State an appropriate general theorem for mgfs and hence write down the mgf of $U$. State the values of $\mathrm { E } ( U )$ and $\operatorname { Var } ( U )$.
The random variable $W$ is defined by
$$W = \frac { U - n } { \sqrt { 2 n } }$$
(iv) Show that the logarithm of the $\operatorname { mgf }$ of $W$ is
$$- \sqrt { \frac { n } { 2 } } \theta - \frac { n } { 2 } \ln \left( 1 - \sqrt { \frac { 2 } { n } } \theta \right) .$$
Use the series expansion of $\ln ( 1 - t )$ to show that, as $n \rightarrow \infty$, this expression tends to $\frac { 1 } { 2 } \theta ^ { 2 }$.\\
State what this implies about the distribution of $W$ for large $n$.
\hfill \mbox{\textit{OCR MEI S4 2015 Q2 [24]}}