Find expressions for ${G’_X}(t)$ and ${G’’_X}(t)$.

Hence show that ${\text{Var}}(X) = \lambda$.

By considering the probability generating function, ${G_{X + Y}}(t)$, of $X + Y$, show that $X + Y$ follows a Poisson distribution with mean $\lambda  + \mu$.

Show that ${\text{P}}(X = x|X + Y = n) = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {1 - \frac{\lambda }{{\lambda + \mu }}} \right)^{n - x}}$, where $n$, $x$ are non-negative integers and $n \geqslant x$.

Identify the probability distribution given in part (c)(i) and state its parameters.

${G’_X}(t) = \lambda {{\text{e}}^{\lambda (t - 1)}}$     A1

${G’’_X}(t) = {\lambda ^2}{{\text{e}}^{\lambda (t - 1)}}$     A1

[2 marks]

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

${G’_X}(1) = \lambda$ and ${G’’_X}(1) = {\lambda ^2}$     (A1)

${\text{Var}}(X) = {\lambda ^2} + \lambda  - {\lambda ^2}$     A1

$= \lambda$     AG

[3 marks]

${G_{X + Y}}(t) = {{\text{e}}^{\lambda (t - 1)}} \times {{\text{e}}^{\mu (t - 1)}}$     M1

Note:     The M1 is for knowing to multiply pgfs.

$= {{\text{e}}^{(\lambda  + \mu )(t - 1)}}$     A1

which is the pgf for a Poisson distribution with mean $\lambda  + \mu$     R1AG

Note:     Line 3 identifying the Poisson pgf must be seen.

[3 marks]

${\text{P}}(X = x|X + Y = n) = \frac{{{\text{P}}(X = x \cap Y = n - x)}}{{{\text{P}}(X + Y = n)}}$     (M1)

$= \left( {\frac{{{{\text{e}}^{ - \lambda }}{\lambda ^x}}}{{x!}}} \right)\left( {\frac{{{{\text{e}}^{ - \mu }}{\mu ^{n - x}}}}{{(n - x)!}}} \right)\left( {\frac{{n!}}{{{{\text{e}}^{ - (\lambda  + \mu )}}{{(\lambda  + \mu )}^n}}}} \right)$ (or equivalent)     M1A1

$= \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right)\frac{{{\lambda ^x}{\mu ^{n - x}}}}{{{{(\lambda + \mu )}^n}}}$     A1

$= \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {\frac{\mu }{{\lambda + \mu }}} \right)^{n - x}}$     A1

leading to ${\text{P}}(X = x|X + Y = n) = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {1 - \frac{\lambda }{{\lambda + \mu }}} \right)^{n - x}}$     AG

[5 marks]

${\text{B}}\left( {n,{\text{ }}\frac{\lambda }{{\lambda  + \mu }}} \right)$     A1A1

Note:     Award A1 for stating binomial and A1 for stating correct parameters.

[2 marks]