(i)     Prove that \({\text{E}}(X) = \lambda \).

(ii)     Prove that \({\text{Var}}(X) = \lambda \).

\(Y\) is a random variable, independent of \(X\), that also follows a Poisson distribution with mean \(\lambda \).

If \(S = 2X - Y\) find

(i)     \({\text{E}}(S)\);

(ii)     \({\text{Var}}(S)\).

Let \(T = \frac{Y}{2} + \frac{Y}{2}\).

(i)     Show that \(T\) is an unbiased estimator for \(\lambda \).

(ii)     Show that \(T\) is a more efficient unbiased estimator of \(\lambda \) than \(S\).

Could either \(S\) or \(T\) model a Poisson distribution? Justify your answer.

By consideration of the probability generating function, \({G_{X + Y}}(t)\), of \(X + Y\), prove that \(X + Y\) follows a Poisson distribution with mean \(2\lambda \).

Find

(i)     \({G_{X + Y}}(1)\);

(ii)     \({G_{X + Y}}( - 1)\).

Hence find the probability that \(X + Y\) is an even number.

(i)     \(G'(t) = \lambda {e^{\lambda (t - 1)}}\)     A1

\({\text{E}}(X) = G'(1)\)     M1

\( = \lambda \)     AG

(ii)     \(G''(t) = {\lambda ^2}{e^{\lambda (t - 1)}}\)     M1

\( \Rightarrow G''(1) = {\lambda ^2}\)     (A1)

\({\text{Var}}(X) = G''(1) + G'(1) - {\left( {G'(1)} \right)^2}\)     (M1)

\( = {\lambda ^2} + \lambda  - {\lambda ^2}\)     A1

\( = \lambda \)     AG

[6 marks]

(i)     \({\text{E}}(S) = 2\lambda  - \lambda  = \lambda \)     A1

(ii)     \({\text{Var}}(S) = 4\lambda  + \lambda  = 5\lambda \)     (A1)A1

Note:     First A1 can be awarded for either \(4\lambda \) or \(\lambda \).

[3 marks]

(i)     \({\text{E}}(T) = \frac{\lambda }{2} + \frac{\lambda }{2} = \lambda \;\;\;\)(so \(T\) is an unbiased estimator)     A1

(ii)     \({\text{Var}}(T) = \frac{1}{4}\lambda  + \frac{1}{4}\lambda  = \frac{1}{2}\lambda \)     A1

this is less than \({\text{Var}}(S)\), therefore \(T\) is the more efficient estimator     R1AG

Note:     Follow through their variances from (b)(ii) and (c)(ii).

[3 marks]

no, mean does not equal the variance     R1

[1 mark]

\({G_{X + Y}}(t) = {e^{\lambda (t - 1)}} \times {e^{\lambda (t - 1)}} = {e^{2\lambda (t - 1)}}\)     M1A1

which is the probability generating function for a Poisson with a mean of \(2\lambda \)     R1AG

[3 marks]

(i)     \({G_{X + Y}}(1) = 1\)     A1

(ii)     \({G_{X + Y}}( - 1) = {e^{ - 4\lambda }}\)     A1

[2 marks]

\({G_{X + Y}}(1) = p(0) + p(1) + p(2) + p(3) \ldots \)

\({G_{X + Y}}( - 1) = p(0) - p(1) + p(2) - p(3) \ldots \)

so \({\text{2P(even)}} = {G_{X + Y}}(1) + {G_{X + Y}}( - 1)\)     (M1)(A1)

\({\text{P(even)}} = \frac{1}{2}(1 + {e^{ - 4\lambda }})\)     A1

[3 marks]

Total [21 marks]

Solutions to the different parts of this question proved to be extremely variable in quality with some parts well answered by the majority of the candidates and other parts accessible to only a few candidates. Part (a) was well answered in general although the presentation was sometimes poor with some candidates doing the differentiation of \(G(t)\) and the substitution of \(t = 1\) simultaneously.

Part (b) was well answered in general, the most common error being to state that \({\text{Var}}(2X - Y) = {\text{Var}}(2X) - {\text{Var}}(Y)\).

Parts (c) and (d) were well answered by the majority of candidates.

Parts (c) and (d) were well answered by the majority of candidates.

Solutions to (e), however, were extremely disappointing with few candidates giving correct solutions. A common incorrect solution was the following:

\(\;\;\;{G_{X + Y}}(t) = {G_X}(t){G_Y}(t)\)

Differentiating,

\(\;\;\;{G'_{X + Y}}(t) = {G'_X}(t){G_Y}(t) + {G_X}(t){G'_Y}(t)\)

\(\;\;\;{\text{E}}(X + Y) = {G'_{X + Y}}(1) = {\text{E}}(X) \times 1 + {\text{E}}(Y) \times 1 = 2\lambda \)

This is correct mathematics but it does not show that \(X + Y\) is Poisson and it was given no credit. Even the majority of candidates who showed that \({G_{X + Y}}(t) = {{\text{e}}^{2\lambda (t - 1)}}\) failed to state that this result proved that \(X + Y\) is Poisson and they usually differentiated this function to show that \({\text{E}}(X + Y) = 2\lambda \).

In (f), most candidates stated that \({G_{X + Y}}(1) = 1\) even if they were unable to determine \({G_{X + Y}}(t)\) but many candidates were unable to evaluate \({G_{X + Y}}( - 1)\). Very few correct solutions were seen to (g) even if the candidates correctly evaluated \({G_{X + Y}}(1)\) and \({G_{X + Y}}( - 1)\).