Consider the random variable \(X \sim {\text{Geo}}(p)\).

(a)     State \({\text{P}}(X < 4)\).

(b)     Show that the probability generating function for X is given by \({G_X}(t) = \frac{{pt}}{{1 - qt}}\), where \(q = 1 - p\).

Let the random variable \(Y = 2X\).

(c)     (i)     Show that the probability generating function for Y is given by \({G_Y}(t) = {G_X}({t^2})\).

          (ii)     By considering \({G'_Y}(1)\), show that \({\text{E}}(Y) = 2{\text{E}}(X)\).

Let the random variable \(W = 2X + 1\).

(d)     (i)     Find the probability generating function for W in terms of the probability generating function of Y.

          (ii)     Hence, show that \({\text{E}}(W) = 2{\text{E}}(X) + 1\).

(a)     use of \({\text{P}}(X = n) = p{q^{n - 1}}{\text{ }}(q = 1 - p)\)     (M1)

\({\text{P}}(X < 4) = p + pq + p{q^2}{\text{ }}\left( { = 1 - {q^3}} \right){\text{ }}\left( { = 1 - {{(1 - p)}^3}} \right){\text{ }}( = 3p - 3{p^2} + {p^3})\)     A1

[2 marks]

 

(b)   \({G_X}(t) = {\text{P}}(X = 1)t + {\text{P}}(X = 2){t^2} +  \ldots \)     (M1)

\( = pt + pq{t^2} + p{q^2}{t^3} +  \ldots \)     A1

summing an infinite geometric series     M1

\( = \frac{{pt}}{{1 - qt}}\)     AG

[3 marks]

 

(c)     (i)     EITHER

          \({G_Y}(t) = {\text{P}}(Y = 1)t + {\text{P}}(Y = 2){t^2} +  \ldots \)     A1

          \( = 0 \times t + {\text{P}}(X = 1){t^2} + 0 \times {t^3} + {\text{P}}(X = 2){t^4} +  \ldots \)     M1A1

          \( = {G_X}({t^2})\)     AG

          OR

          \({G_Y}(t) = E({t^Y}) = E({t^{2X}})\)     M1A1

          \( = E\left( {{{({t^2})}^X}} \right)\)     A1

          \( = {G_X}({t^2})\)     AG

          (ii)     \({\text{E}}(Y) = {G'_Y}(1)\)     A1

          EITHER

          \( = 2t{G'_X}({t^2})\) evaluated at \(t = 1\)     M1A1

          \( = 2{\text{E}}(X)\)     AG

          OR

          \( = \frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{p{t^2}}}{{(1 - q{t^2})}}} \right) = \frac{{2pt(1 - q{t^2}) + 2pq{t^3}}}{{{{(1 - q{t^2})}^2}}}\) evaluated at \(t = 1\)     A1

          \( = 2 \times \frac{{p(1 - qt) + pqt}}{{{{(1 - qt)}^2}}}\) evaluated at \(t = 1{\text{ (or }}\frac{2}{p})\)     A1

          \( = 2{\text{E}}(X)\)     AG

[6 marks]

 

(d)     (i)     \({G_W}(t) = t{G_Y}(t)\) (or equivalent)     A2

          (ii)     attempt to evaluate \({G'_W}(t)\)     M1

          EITHER

          obtain \(1 \times {G_Y}(t) + t \times {G'_Y}(t)\)     A1

          substitute \(t = 1\) to obtain \(1 \times 1 + 1 \times {G'_Y}(1)\)     A1

          OR

          \( = \frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{p{t^3}}}{{(1 - q{t^2})}}} \right) = \frac{{3p{t^2}(1 - q{t^2}) + 2pq{t^4}}}{{{{(1 - q{t^2})}^2}}}\)     A1

          substitute \(t = 1\) to obtain \(1 + \frac{2}{p}\)     A1

          \( = 1 + 2{\text{E}}(X)\)     AG

[5 marks]

 

Total [16 marks]

[N/A]