| Date | May 2017 | Marks available | 3 | Reference code | 17M.3sp.hl.TZ0.4 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Hence and Find | Question number | 4 | Adapted from | N/A |
Question
The random variables \({X_1}\) and \({X_2}\) are a random sample from \({\text{N}}(\mu ,{\text{ 2}}{\sigma ^2})\). The random variables \({Y_1}\), \({Y_2}\) and \({Y_3}\) are a random sample from \({\text{N}}(2\mu ,{\text{ }}{\sigma ^2})\).
The estimator \(U\) is used to estimate \(\mu \) where \(U = a({X_1} + {X_2}) + b({Y_1} + {Y_2} + {Y_3})\) and \(a\), \(b\) are constants.
Given that \(U\) is unbiased, show that \(2a + 6b = 1\).
Show that \({\text{Var}}(U) = (39{b^2} - 12b + 1){\sigma ^2}\).
Hence find the value of \(a\) and the value of \(b\) which give the best unbiased estimator of this form, giving your answers as fractions.
Hence find the variance of this best unbiased estimator.
Markscheme
\({\text{E}}(U) = a\left( {{\text{E}}({X_1}) + {\text{E}}({X_2})} \right) + b\left( {{\text{E}}({Y_1}) + {\text{E}}({Y_2}) + {\text{E}}({Y_3})} \right)\) (M1)
\( = 2a\mu + 6b\mu \) A1
(for an unbiased estimator,) \({\text{E}}(U) = \mu \) R1
giving \(2a + 6b = 1\) AG
Note: Condone omission of E on LHS.
[3 marks]
\({\text{Var}}(U) = {a^2}\left( {{\text{Var}}({X_1}) + {\text{Var}}({X_2})} \right) + {b^2}\left( {{\text{Var}}({Y_1}) + {\text{Var}}({Y_2}) + {\text{Var}}({Y_3})} \right)\) (M1)
\( = 4{a^2}{\sigma ^2} + 3{b^2}{\sigma ^2}\) A1
\( = 4{\left( {\frac{{1 - 6b}}{2}} \right)^2}{\sigma ^2} + 3{b^2}{\sigma ^2}\) A1
\( = (39{b^2} - 12b + 1){\sigma ^2}\) AG
[3 marks]
the best unbiased estimator (of this form) will be found by minimising \({\text{Var}}(U)\) (R1)
For example, \(\frac{{\text{d}}}{{{\text{d}}b}}\left( {{\text{Var}}(U)} \right) = (78b - 12){\sigma ^2}\) (A1)
for a minimum, \(b = \frac{{12}}{{78}}\,\,\,\left( { = \frac{2}{{13}}} \right)\) so that \(a = \frac{3}{{78}}\,\,\,\left( { = \frac{1}{{26}}} \right)\) A1
[3 marks]
\({\text{Var}}U = \left( {39{{\left( {\frac{2}{{13}}} \right)}^2} - 12\left( {\frac{2}{{13}}} \right) + 1} \right){\sigma ^2}\)
\( = \frac{{{\sigma ^2}}}{{13}}\,\,\,(0.0769{\sigma ^2})\) A1
[1 mark]
