(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]