# S5 2003 6 Edexcel

(a)

We must have G_X(1) = 1, so:

\displaystyle G_X(1) = k(1 + 1 + 3)^2 = 25k = 1

giving us:

\displaystyle k = \frac 1 {25}

(b)

We are interested in the coefficient of the t^2 term in:

\displaystyle G_X(t) = \frac 1 {25} (1 + t + 3t^2)(1 + t + 3t^2)

We need to look at what “contributes” to the t^2 term. To get t^2, we can multiply the constant term of the first bracket with the t^2 term from the second, the t term in the first bracket with the t term in the second, or the t^2 term in the first bracket with the constant term in the second. These are the only three possibilities. Adding these up, this means the coefficient of the t^2 term will be:

\displaystyle \frac 1 {25} \left(1 \times 3 + 1 \times 1 + 3 \times 1\right) = \frac 7 {25}

so:

\displaystyle \mathrm P (X = 2) = \frac 7 {25}

Alternatively, we can group and expand, ie.

\begin{align*}\frac 1 {25} (1 + t + 3t^2)^2 & = \frac 1 {25}((1 + t) + 3t^2)^2 \\ & = \frac 1 {25} \left((1 + t)^2 + 2 \times 3 t^2(1 + t) + (3t)^2\right) \\ & = \frac 1 {25} \left(t^2 + 2t + 1 + 6t^2 + 6t^3 + 9t^4\right)\end{align*}

of which we can see the t^2 term to be:

\displaystyle \frac {6 + 1} {25} = \frac 7 {25}

Expanding as in the second method for (b) we have:

\displaystyle G_X(t) = \frac 9 {25} t^4 + \frac 6 {25} t^3 + \frac 7 {25} t^2 + \frac 2 {25} t + \frac 1 {25}

Differentiating, we have:

\begin{align*}G'_X(t) &= \frac {9 \times 4} {25} t^3 + \frac {6 \times 3} {25} t^2 + \frac {7 \times 2} {25} t + \frac 2 {25} \\ & = \frac {36} {25} t^3 + \frac {18} {25} t^2 + \frac {14} {25} t + \frac 2 {25}\end{align*}

So:

\begin{align*}\mathrm E(X) &= G'_X(1) \\ &= \frac {36} {25} + \frac {18} {25} + \frac {14} {25} + \frac 2 {25} \\ &= \frac {14} 5\end{align*}

Differentiating again we have:

\begin{align*}G''_X(t) & = \frac {36 \times 3} {25} t^2 + \frac {18 \times 2} {25} t + \frac {14} {25} \\ & = \frac {108} {25} t^2 + \frac {36} {25} t + \frac {14} {25}\end{align*}

so:

\begin{align*}G''_X(1) & = \frac {108} {25} + \frac {36} {25} + \frac {14} {25} \\ & = \frac {158} {25}\end{align*}

We know:

\displaystyle \mathrm{var}(X) = G'_X(1) + G''_X(1) - (G'_X(1))^2

so:

\begin{align*}\mathrm {var}(X) & = \frac {158} {25} + \frac {14} 5 - \left(\frac {14} 5\right)^2 \\ & = \frac {32} {25}\end{align*}

Alternatively, we could have used the chain rule.

(d)

We know that:

\displaystyle G_{2X + 1}(t) = t G_X(t^2) = \frac t {25} (1 + t^2 + 3t^4)^2