(a)     Consider the random variable \(X\) for which \({\text{E}}(X) = a\lambda  + b\), where \(a\) and \(b\)are constants and \(\lambda \) is a parameter.

Show that \(\frac{{X - b}}{a}\) is an unbiased estimator for \(\lambda \).

(b)     The continuous random variable Y has probability density function

\(f(y) = \left\{ \begin{array}{r}{\textstyle{2 \over 9}}(3 + y - \lambda ),\\0,\end{array} \right.\begin{array}{*{20}{l}}{{\rm{ for}}\, \lambda  - 3 \le y \le \lambda }\\{{\rm{ otherwise}}}\end{array}\)

where \(\lambda \) is a parameter.

          (i)     Verify that \(f(y)\) is a probability density function for all values of \(\lambda \).

          (ii)     Determine \({\text{E}}(Y)\).

          (iii)     Write down an unbiased estimator for \(\lambda \).

(a)     \({\text{E}}\left( {\frac{{X - b}}{a}} \right) = \frac{{a\lambda  + b - b}}{a}\)     M1A1

\( = \lambda \)     A1

(Therefore \(\frac{{X - b}}{a}\) is an unbiased estimator for \(\lambda \))     AG

[3 marks]

 

(b)     (i)     \(f(y) \geqslant 0\)     R1

 

Note:     Only award R1 if this statement is made explicitly.

 

          recognition or showing that integral of f is 1 (seen anywhere) R1

          EITHER

          \(\int_{\lambda  - 3}^\lambda  {\frac{2}{9}(3 + y - \lambda ){\text{d}}y} \)     M1

          \( = \frac{2}{9}\left[ {(3 - \lambda )y + \frac{1}{2}{y^2}} \right]_{\lambda  - 3}^\lambda \)     A1

          \( = \frac{2}{9}\left( {\lambda (3 - \lambda ) + \frac{1}{2}{\lambda ^2} - (3 - \lambda )(\lambda  - 3) - \frac{1}{2}{{(\lambda  - 3)}^2}} \right)\) or equivalent     A1

          \( = 1\)

          OR

          the graph of the probability density is a triangle with base length 3 and height \(\frac{2}{3}\)     M1A1

          its area is therefore \(\frac{1}{2} \times 3 \times \frac{2}{3}\)     A1

          \( = 1\)

          (ii)     \({\text{E}}(Y) = \int_{\lambda  - 3}^\lambda  {\frac{2}{9}y(3 + y - \lambda ){\text{d}}y} \)     M1

          \( = \frac{2}{9}\left[ {(3 - \lambda )\frac{1}{2}{y^2} + \frac{1}{3}{y^3}} \right]_{\lambda  - 3}^\lambda \)     A1

          \( = \frac{2}{9}\left( {(3 - \lambda )\frac{1}{2}\left( {{\lambda ^2} - {{(\lambda  - 3)}^2}} \right) + \frac{1}{3}\left( {{\lambda ^3} - {{(\lambda  - 3)}^3}} \right)} \right)\)     M1

          \( = \lambda  - 1\)     A1A1

 

Note:     Award 3 marks for noting that the mean is \(\frac{2}{3}{\text{rds}}\) the way along the base and then A1A1 for \(\lambda  - 1\).

 

Note:     Award A1 for \(\lambda \) and A1 for –1.

 

          (iii)     unbiased estimator: \(Y + 1\)     A1

 

Note:     Accept \(\bar Y + 1\).

     Follow through their \({\text{E}}(Y)\) if linear.

 

[11 marks]

 

Total [14 marks]

[N/A]