| 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 | Verify PDF integrates to 1 |
| Difficulty | Standard +0.3 This is a standard Further Maths Statistics question covering routine techniques: verifying a PDF integrates to 1 using substitution, finding moments, deriving an MLE by differentiation, and constructing a confidence interval. All parts follow textbook procedures with no novel insight required. The integration in part (i) is straightforward with the substitution u = x²/a, and the MLE derivation is mechanical. Slightly easier than average due to the step-by-step guidance and explicit 'show that' structure. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The integral is \(-\exp(-x^2/a)\), limits 0 and infinity. Evaluates to \(0-(-1)=1\) | B1 [1] | Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X^2) = \int_0^\infty \frac{2}{a} x^3 \exp\left(-\frac{x^2}{a}\right)dx\) or \(\int_0^\infty x^2 \frac{2}{a} x \exp\left(-\frac{x^2}{a}\right)dx\) | M1 | |
| \(= \left[x^2\left(-\exp\left(-\frac{x^2}{a}\right)\right)\right]_0^\infty + \int_0^\infty 2x\exp\left(-\frac{x^2}{a}\right)dx\) | M1A1 | M1 parts |
| \(= 0 + a = a\) | A1 | Cao |
| \(E(X^4) = \int_0^\infty \frac{2}{a} x^5 \exp\left(-\frac{x^2}{a}\right)dx\) or \(\int_0^\infty x^4 \frac{2}{a} x \exp\left(-\frac{x^2}{a}\right)dx\) | M1 | |
| \(= \left[x^4\left(-\exp\left(-\frac{x^2}{a}\right)\right)\right]_0^\infty + \int_0^\infty 4x^3\exp\left(-\frac{x^2}{a}\right)dx\) | A1 | Answer given so working must be convincing |
| \(= 0 + 2a \times a = 2a^2\) | A1 [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Likelihood is \(\left(\frac{2}{a}\right)^n \prod X_i \exp\left(-\frac{1}{a}\sum X_i^2\right)\) | M1A1A1 | M1 for recognisable attempt to obtain likelihood. A1 for \(\prod X_i\). A1 for the exp fn. |
| (log likelihood as given) Differentiate log likelihood | M1 | |
| to obtain \(-\frac{n}{a} + \frac{1}{a^2}\sum X_i^2\) | A1 | |
| Set to zero to obtain \(\text{MLE}(a) = \frac{1}{n}\sum X_i^2\) | M1A1 [7] | Justification of maximum not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(\text{MLE}(a)) = \frac{1}{n}\sum E(X_i^2) = \frac{1}{n}na = a\) | B1 | |
| That is, the MLE is unbiased | E1 | Seen or very clearly implied |
| \(\text{Var}(\text{MLE}(a)) = \frac{1}{n^2}\sum\text{Var}(X_i^2) = \frac{1}{n^2}\sum\left(E(X_i^4) - \left(E(X_i^2)\right)^2\right)\) | M1M1 | |
| \(= \frac{1}{n^2}n(2a^2 - a^2) = \frac{a^2}{n}\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum likelihood estimate (also unbiased) of \(a\) is 1.471 | B1 | Explanation not required if calculations are correct |
| with estimated standard error \((a/\sqrt{n})\) 0.1471 | B1 | |
| 95% CI is \(1.471 \pm 1.960 \times 0.1471 = 1.471 \pm 0.288\) | M1A1 [4] | Accept 2 for 1.960 |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| The integral is $-\exp(-x^2/a)$, limits 0 and infinity. Evaluates to $0-(-1)=1$ | B1 [1] | Answer given |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X^2) = \int_0^\infty \frac{2}{a} x^3 \exp\left(-\frac{x^2}{a}\right)dx$ or $\int_0^\infty x^2 \frac{2}{a} x \exp\left(-\frac{x^2}{a}\right)dx$ | M1 | |
| $= \left[x^2\left(-\exp\left(-\frac{x^2}{a}\right)\right)\right]_0^\infty + \int_0^\infty 2x\exp\left(-\frac{x^2}{a}\right)dx$ | M1A1 | M1 parts |
| $= 0 + a = a$ | A1 | Cao |
| $E(X^4) = \int_0^\infty \frac{2}{a} x^5 \exp\left(-\frac{x^2}{a}\right)dx$ or $\int_0^\infty x^4 \frac{2}{a} x \exp\left(-\frac{x^2}{a}\right)dx$ | M1 | |
| $= \left[x^4\left(-\exp\left(-\frac{x^2}{a}\right)\right)\right]_0^\infty + \int_0^\infty 4x^3\exp\left(-\frac{x^2}{a}\right)dx$ | A1 | Answer given so working must be convincing |
| $= 0 + 2a \times a = 2a^2$ | A1 [7] | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Likelihood is $\left(\frac{2}{a}\right)^n \prod X_i \exp\left(-\frac{1}{a}\sum X_i^2\right)$ | M1A1A1 | M1 for recognisable attempt to obtain likelihood. A1 for $\prod X_i$. A1 for the exp fn. |
| (log likelihood as given) Differentiate log likelihood | M1 | |
| to obtain $-\frac{n}{a} + \frac{1}{a^2}\sum X_i^2$ | A1 | |
| Set to zero to obtain $\text{MLE}(a) = \frac{1}{n}\sum X_i^2$ | M1A1 [7] | Justification of maximum not required |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(\text{MLE}(a)) = \frac{1}{n}\sum E(X_i^2) = \frac{1}{n}na = a$ | B1 | |
| That is, the MLE is unbiased | E1 | Seen or very clearly implied |
| $\text{Var}(\text{MLE}(a)) = \frac{1}{n^2}\sum\text{Var}(X_i^2) = \frac{1}{n^2}\sum\left(E(X_i^4) - \left(E(X_i^2)\right)^2\right)$ | M1M1 | |
| $= \frac{1}{n^2}n(2a^2 - a^2) = \frac{a^2}{n}$ | A1 [5] | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum likelihood estimate (also unbiased) of $a$ is 1.471 | B1 | Explanation not required if calculations are correct |
| with estimated standard error $(a/\sqrt{n})$ 0.1471 | B1 | |
| 95% CI is $1.471 \pm 1.960 \times 0.1471 = 1.471 \pm 0.288$ | M1A1 [4] | Accept 2 for 1.960 |
---1 The random variable $X$ has the following probability density function, in which $a$ is a (positive) parameter.
$$\mathrm { f } ( x ) = \frac { 2 } { a } x \mathrm { e } ^ { - x ^ { 2 } / a } , \quad x \geqslant 0 .$$
(i) Verify that $\int _ { 0 } ^ { \infty } \mathrm { f } ( x ) \mathrm { d } x = 1$.\\
(ii) Show that $\mathrm { E } \left( X ^ { 2 } \right) = a$ and $\mathrm { E } \left( X ^ { 4 } \right) = 2 a ^ { 2 }$.
The parameter $a$ is to be estimated by maximum likelihood based on an independent random sample from the distribution, $X _ { 1 } , X _ { 2 } , \ldots , X _ { n }$.\\
(iii) Show that the logarithm of the likelihood function is
$$n \ln 2 - n \ln a + \sum _ { i = 1 } ^ { n } \ln X _ { i } - \frac { 1 } { a } \sum _ { i = 1 } ^ { n } X _ { i } ^ { 2 }$$
Hence obtain the maximum likelihood estimator, $\hat { a }$, for $a$.\\[0pt]
[You are not required to verify that any turning point you find is a maximum.]\\
(iv) Using the results from part (ii), show that $\hat { a }$ is unbiased for $a$ and find the variance of $\hat { a }$.\\
(v) In a particular random sample from this distribution, $n = 100$ and $\sum x _ { i } ^ { 2 } = 147.1$. Obtain an approximate 95\% confidence interval for $a$. (You may assume that the Central Limit Theorem holds in this case.)
Option 2: Generating Functions\\
\hfill \mbox{\textit{OCR MEI S4 2015 Q1 [24]}}