OCR MEI S4 2010 June — Question 1 24 marks

Exam BoardOCR MEI
ModuleS4 (Statistics 4)
Year2010
SessionJune
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicThe Gamma Distribution
TypeUnbiased estimator verification
DifficultyStandard +0.8 This is a multi-part Further Maths Statistics question requiring integration to find E(X), verification of unbiased estimators, variance calculations, and relative efficiency comparison. While the techniques are standard for S4 (gamma distribution moments, estimator properties), it requires careful algebraic manipulation across multiple connected parts and understanding of asymptotic properties, placing it moderately above average difficulty.
Spec5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05b Unbiased estimates: of population mean and variance

1 The random variable \(X\) has probability density function $$\mathrm { f } ( x ) = \frac { x \mathrm { e } ^ { - x / \lambda } } { \lambda ^ { 2 } } \quad ( x > 0 )$$ where \(\lambda\) is a parameter \(( \lambda > 0 ) . X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\) are \(n\) independent observations on \(X\), and \(\bar { X } = \frac { 1 } { n } \sum _ { i = 1 } ^ { n } X _ { i }\) is their mean.
  1. Obtain \(\mathrm { E } ( X )\) and deduce that \(\hat { \lambda } = \frac { 1 } { 2 } \bar { X }\) is an unbiased estimator of \(\lambda\).
  2. \(\operatorname { Obtain } \operatorname { Var } ( \hat { \lambda } )\).
  3. Explain why the results in parts (i) and (ii) indicate that \(\hat { \lambda }\) is a good estimator of \(\lambda\) in large samples.
  4. Suppose that \(n = 3\) and consider the alternative estimator $$\tilde { \lambda } = \frac { 1 } { 8 } X _ { 1 } + \frac { 1 } { 4 } X _ { 2 } + \frac { 1 } { 8 } X _ { 3 } .$$ Show that \(\tilde { \lambda }\) is an unbiased estimator of \(\lambda\). Find the relative efficiency of \(\tilde { \lambda }\) compared with \(\hat { \lambda }\). Which estimator do you prefer in this case?

1 The random variable $X$ has probability density function

$$\mathrm { f } ( x ) = \frac { x \mathrm { e } ^ { - x / \lambda } } { \lambda ^ { 2 } } \quad ( x > 0 )$$

where $\lambda$ is a parameter $( \lambda > 0 ) . X _ { 1 } , X _ { 2 } , \ldots , X _ { n }$ are $n$ independent observations on $X$, and $\bar { X } = \frac { 1 } { n } \sum _ { i = 1 } ^ { n } X _ { i }$ is their mean.\\
(i) Obtain $\mathrm { E } ( X )$ and deduce that $\hat { \lambda } = \frac { 1 } { 2 } \bar { X }$ is an unbiased estimator of $\lambda$.\\
(ii) $\operatorname { Obtain } \operatorname { Var } ( \hat { \lambda } )$.\\
(iii) Explain why the results in parts (i) and (ii) indicate that $\hat { \lambda }$ is a good estimator of $\lambda$ in large samples.\\
(iv) Suppose that $n = 3$ and consider the alternative estimator

$$\tilde { \lambda } = \frac { 1 } { 8 } X _ { 1 } + \frac { 1 } { 4 } X _ { 2 } + \frac { 1 } { 8 } X _ { 3 } .$$

Show that $\tilde { \lambda }$ is an unbiased estimator of $\lambda$. Find the relative efficiency of $\tilde { \lambda }$ compared with $\hat { \lambda }$. Which estimator do you prefer in this case?

\hfill \mbox{\textit{OCR MEI S4 2010 Q1 [24]}}