| Date | None Specimen | Marks available | 16 | Reference code | SPNone.2.hl.TZ0.7 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Determine, Find, and Show that | Question number | 7 | Adapted from | N/A |
Question
The random variable \(X\) has cumulative distribution function\[F(x) = \left\{ {\begin{array}{*{20}{c}}
0&{x < 0} \\
{{{\left( {\frac{x}{a}} \right)}^3}}&{0 \leqslant x \leqslant a} \\
1&{x > a}
\end{array}} \right.\]where \(a\) is an unknown parameter. You are given that the mean and variance of \(X\) are \(\frac{{3a}}{4}\) and \(\frac{{3{a^2}}}{{80}}\) respectively. To estimate the value of \(a\) , a random sample of \(n\) independent observations, \({X_1},{X_2}, \ldots {X_n}\) is taken from the distribution of \(X\) .
(i) Find an expression for \(c\) in terms of \(n\) such that \(U = c\sum\limits_{i = 1}^n {{X_i}} \) is an unbiased estimator for \(a\) .
(ii) Determine an expression for \({\text{Var}}(U)\) in this case.
(i) Show that \({\rm{P}}(Y \le y) = {\left( {\frac{y}{a}} \right)^{3n}},0 \le y \le a\) and deduce an expression for the probability density function of \(Y\) .
(ii) Find \({\rm{E}}(Y)\) .
(iii) Show that \({\rm{Var}}(Y) = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) .
(iv) Find an expression for \(d\) in terms of \(n\) such that \(V = dY\) is an unbiased estimator for \(a\) .
(v) Determine an expression for \({\rm{Var}}(V)\) in this case.
Show that \(\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{3n + 2}}{5}\) and hence state, with a reason, which of \(U\) or \(V\) is the more efficient estimator for \(a\) .
Markscheme
(i) \({\rm{E}}(U) = c \times n \times \frac{{3a}}{4} = a \Rightarrow c = \frac{4}{{3n}}\) M1A1
(ii) \(Var(U) = \frac{{16}}{{9{n^2}}} \times n \times \frac{{3{a^2}}}{{80}} = \frac{{{a^2}}}{{15n}}\) M1A1
[4 marks]
(i) \({\rm{P}}(Y \le y) = {\rm{P}}({\text{all }}Xs \le y)\) M1
\( = \left[ {\rm{P}} \right.{\left. {(X \le y)} \right]^n}\) (A1)
\( = {\left( {{{\left( {\frac{y}{a}} \right)}^3}} \right)^n}\) (A1)
Note: Only one of the two A1 marks above may be implied.
\( = {\left( {\frac{y}{a}} \right)^{3n}}\) AG
\(g(y) = \frac{{\rm{d}}}{{{\rm{d}}y}}{\left( {\frac{y}{a}} \right)^{3n}} = \frac{{3n{y^{3n - 1}}}}{{{a^{3n}}}},(0 < y < a)\) M1A1
(\(g(y) = 0\) otherwise)
(ii) \({\rm{E}}(Y) = \int_0^a {\frac{{3n{y^{3n}}}}{{{a^{3n}}}}} {\rm{d}}y\) M1
\( = \left[ {\frac{{3n{y^{3n + 1}}}}{{(3n + 1){a^{3n}}}}} \right]_0^a\) A1
\( = \frac{{3na}}{{3n + 1}}\) A1
(iii) \({\rm{Var}}(Y) = \int_0^a {\frac{{3n{y^{3n + 1}}}}{{{a^{3n}}}}} {\rm{d}}y - {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}\) M1
\( = \left[ {\frac{{3n{y^{3n + 2}}}}{{(3n + 2){a^{3n}}}}} \right]_0^a - {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}\) A1
\( = \frac{{3n{a^2}}}{{3n + 2}} - \frac{{9{n^2}{a^2}}}{{{{(3n + 1)}^2}}}\) M1
\( = \frac{{3n{a^2}(9{n^2} + 6n + 1) - 9{n^2}{a^2}(3n + 2)}}{{(3n + 2)(3n + 1)}}\) A1
\( = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) AG
(iv) \({\rm{E}}(V) = d \times \frac{{3na}}{{3n + 1}} = a \Rightarrow d = \frac{{3n + 1}}{{3n}}\) M1A1
(v) \(Var(V) = {\left( {\frac{{3n + 1}}{{3n}}} \right)^2} \times \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) M1
\( = \frac{{{a^2}}}{{3n(3n + 2)}}\) A1
[16 marks]
\(\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{\frac{{{a^2}}}{{15n}}}}{{\frac{{{a^2}}}{{3n(3n + 2)}}}}\) A1
\( = \frac{{3n + 2}}{5}\) AG
\(V\) is the more efficient estimator because \(3n + 2 > 5\) (for \(n > 1\) ) R1
[2 marks]
