Proof.

(i)

Substituting \(t=0\) into the series, and using the convention \(0^{0}=1\), \[G_X(0) = 0^{0}P(X=0) + 0^{1}P(X=1) + 0^{2}P(X=2) + \cdots = P(X=0),\] since every term after the first has a factor \(0\).

(ii)

\[G_X(1) = \sum _{x=0}^{\infty }1^{x}P(X=x) = \sum _{x=0}^{\infty }P(X=x) = 1 .\] This is the standard check on any candidate generating function.

(iii)

Differentiating under the expectation, \[G_X'(t) = E\left (X\,t^{X-1}\right ) \quad \Longrightarrow \quad G_X'(1) = E(X).\]

(iv)

Differentiating once more, \[G_X''(t) = E\left (X(X-1)\,t^{X-2}\right ) \quad \Longrightarrow \quad G_X''(1) = E\left [X(X-1)\right ] = E\left (X^{2}\right ) - E(X).\] Hence \(E\left (X^{2}\right ) = G_X''(1) + G_X'(1)\), and \[\Var (X) = E\left (X^{2}\right ) - \left [E(X)\right ]^{2} = G_X''(1) + G_X'(1) - \left [G_X'(1)\right ]^{2}.\] □

Solution. \[G_X'(t) = (-4)(-0.9)(0.0001)\left (1-0.9t\right )^{-5} = 0.00036\left (1-0.9t\right )^{-5},\] and since \(1-0.9(1) = 0.1\), we have \(\left (0.1\right )^{-5} = 10^{5}\), so \[G_X'(1) = 0.00036 \times 10^{5} = 36 .\] Similarly \[G_X''(t) = (-4)(-5)(-0.9)^{2}(0.0001)\left (1-0.9t\right )^{-6} = 0.00162\left (1-0.9t\right )^{-6},\] \[G_X''(1) = 0.00162 \times 10^{6} = 1620 .\] Therefore \[E(X) = 36,\qquad \Var (X) = 1620 + 36 - 36^{2} = 1656 - 1296 = 360 .\]