S4 June 2017 Q6
6. The independent random variables \(X_1\) and \(X_2\) are each distributed \(\mathrm{B}(n, p)\), where \(n \gt 1\)
An unbiased estimator for \(p\) is given by
where \(a\) and \(b\) are constants.
[You may assume that if \(X_1\) and \(X_2\) are independent then \(\mathrm{E}(X_1X_2) = \mathrm{E}(X_1)\mathrm{E}(X_2)\)]
| Scheme | Marks |
|---|---|
| \(\mathrm{E}\left(\dfrac{aX_1 + bX_2}{n}\right) = \dfrac{anp + bnp}{n} = ap + bp = (a + b)p\) | M1 |
| \(a + b = 1\ *\) | A1* cso |
| (2) |
Notes
M1 Using \(\dfrac{a\mathrm{E}(X_1) + b\mathrm{E}(X_2)}{n}\) and subst \(\mathrm{E}(X_1) = np\) and \(\mathrm{E}(X_2) = np\)
Acso* Answer given. Need \(p(a + b) = p\) and statement \(a + b = 1\) and no errors
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}\left(\dfrac{aX_1 + bX_2}{n}\right) = \dfrac{1}{n^2}\left(a^2np(1 - p) + b^2np(1 - p)\right)\) | M1 A1 |
| \(= \dfrac{p(1 - p)\left(a^2 + b^2\right)}{n}\) | |
| \(= \dfrac{p(1 - p)\left(a^2 + (1 - a)^2\right)}{n}\) | M1d |
| \(= \dfrac{\left(2a^2 - 2a + 1\right)p(1 - p)}{n}\ \ *\) | A1* cso |
| (4) |
Notes
M1 Using \(\dfrac{a^2\mathrm{Var}(X_1) + b^2\mathrm{Var}(X_2)}{n^2}\) and subst \(\mathrm{Var}(X_1) = np(1 - p)\) – may be implied by \(\dfrac{1}{n^2}\left(a^2np(1 - p) + b^2np(1 - p)\right)\)
A1 correct answer in any form
M1d dep on 1st M1 Subst \(b = 1 - a\)
A1cso* method must be shown and no errors.
| Scheme | Marks |
|---|---|
| Min value when \(\dfrac{(4a - 2)p(1 - p)}{n} = 0\) \(\Rightarrow 4a - 2 = 0\) | M1A1 |
| \(a = \dfrac{1}{2},\ \ b = \dfrac{1}{2}\) | A1A1ft |
| \(\dfrac{\mathrm{d}^2\mathrm{Var}(\hat{p})}{\mathrm{d}a^2} = \dfrac{4p(1 - p)}{n} \gt 0\) or \(\because\) quadratic with positive \(x^2\) \(\therefore\) minimum point or sketch | B1 |
| (5) |
Notes
M1 \(\dfrac{\mathrm{d}}{\mathrm{d}a}(\mathrm{Var})\) (must differentiate with respect \(a\)) or attempt to complete the square
A1 correct diff \(= 0\) or \(2\left(a - \tfrac{1}{2}\right)^2 + \tfrac{1}{2}\)
A1 \(a = 0.5\)
A1 ft for \(b = 1 - a\)
B1 for a reason why minimum
| Scheme | Marks |
|---|---|
| (i) \(\mathrm{E}\left(\dfrac{aX_1 + bX_2}{n}\right)^2 = \mathrm{E}\left(\dfrac{a^2X_1^{\,2} + b^2X_2^{\,2} + 2abX_1X_2}{n^2}\right)\) | M1 |
| \(= \dfrac{1}{n^2}\left(a^2np(1 - p) + a^2n^2p^2 + b^2np(1 - p) + b^2n^2p^2 + 2abn^2p^2\right)\) | M1d |
| \(= \dfrac{(a^2 + b^2)np(1 - p) + (a + b)^2n^2p^2}{n^2}\) \(= \dfrac{(a^2 + b^2)p(1 - p)}{n} + p^2(a + b)^2\) | |
| \(= \dfrac{(a^2 + b^2)p(1 - p)}{n} + p^2\ \ ; \gt p^2\) since \(\dfrac{(a^2 + b^2)p(1 - p)}{n} \gt 0\) oe \(\therefore\) biased | A1;A1 |
| (ii) As \(n \to \infty\) \(\mathrm{E}\left(\hat{p}^2\right) \to p^2\) Therefore bias \(\to 0\) | B1 |
| (5) |
Notes
(i) M1 multiplying out and using \(\mathrm{E}(aX) = a\,\mathrm{E}(X)\) [may use their values of \(a\) and \(b\)]
M1d dependent on previous M being awarded Using \(\mathrm{E}(X^2) = \mathrm{Var}(X) + [\mathrm{E}(X)]^2\)
A1 \(\dfrac{(2a^2 - 2a + 1)p(1 - p)}{n} + p^2\) or \(\dfrac{(a^2 + b^2)p(1 - p)}{n} + p^2\) must be of the form \(p^2 +\) a single term
A1 for a reason why it is not equal \(p^2\) plus statement to say biased.
(ii) B1 Follow on from their expression \(p^2 + \ldots\) with \(a\) and \(b\).
| Scheme | Marks |
|---|---|
| \(\mathrm{E}\left(X_1(X_1 - 1)\right) = \mathrm{E}\left(X_1^{\,2}\right) - \mathrm{E}\left(X_1\right)\) \(= np(1 - p) + n^2p^2 - np\) | M1 |
| \(= np - np^2 + n^2p^2 - np\) \(= np^2(n - 1)\) | A1 |
| Unbiased estimator \(= \dfrac{X_1(X_1 - 1)}{n(n - 1)}\) | A1 |
| (3) | |
| (19 marks) |
Notes
M1 multiplying out correctly and subst \(np\) for \(\mathrm{E}(X)\) or using \(\mathrm{E}(\hat{p})^2 = \mathrm{Var}(\hat{p}) + \left[\mathrm{E}(\hat{p})\right]^2\)
Allow \(= \dfrac{(2a^2 - 2a + 1)p(1 - p)}{n} + p^2\)
A1 \(np^2(n - 1)\)
A1 \(\dfrac{X_1(X_1 - 1)}{n(n - 1)}\)
NB \(\dfrac{X_1(X_1 - 1)}{n(n - 1)}\) gains all 3 marks.