Solution. \[G_X(t) = \sum _{x=1}^{\infty } t^{x}\,2^{-x} = \sum _{x=1}^{\infty }\left (\frac {t}{2}\right )^{x} = \frac {t/2}{1-t/2} = \frac {t}{2-t},\] a geometric series converging for \(\left |t/2\right |<1\), that is \(\left |t\right |<2\). Check: \(G_X(1) = 1/(2-1) = 1\).

Differentiating by the quotient rule, \[G_X'(t) = \frac {(2-t) + t}{(2-t)^{2}} = \frac {2}{(2-t)^{2}}, \qquad G_X'(1) = 2 ,\] \[G_X''(t) = \frac {4}{(2-t)^{3}},\qquad G_X''(1) = 4 .\] Hence \(E(X) = 2\) and \[\Var (X) = G_X''(1) + G_X'(1) - \left [G_X'(1)\right ]^{2} = 4 + 2 - 4 = 2 .\] This is the geometric distribution with success probability \(\tfrac 12\) counted from one, for which the mean is \(1/p = 2\) and the variance \((1-p)/p^{2} = 2\) — agreeing with both.

Solution. \[G_X(t) = \sum _{x=0}^{n}t^{x}\binom {n}{x}\theta ^{x}(1-\theta )^{n-x} = \sum _{x=0}^{n}\binom {n}{x}\left (\theta t\right )^{x}(1-\theta )^{n-x} = \left (1-\theta +\theta t\right )^{n}\] by the binomial theorem. For independent \(X\sim \) BIN\((m,\theta )\) and \(Y\sim \) BIN\((n,\theta )\), \[G_{X+Y}(t) = \left (1-\theta +\theta t\right )^{m}\left (1-\theta +\theta t\right )^{n} = \left (1-\theta +\theta t\right )^{m+n},\] which is the generating function of BIN\((m+n,\theta )\), so by uniqueness that is the distribution.

If the success probabilities differ the product is \(\left (1-\theta _1+\theta _1t\right )^{m}\left (1-\theta _2+\theta _2t\right )^{n}\), which is not of the form \(\left (1-\theta +\theta t\right )^{k}\) for any single \(\theta \): the two bases are different polynomials in \(t\) and cannot be combined. The sum is then not binomial at all.

Solution. The GAM\((\alpha ,\beta )\) distribution with shape \(\alpha \) and scale \(\beta \) has \(M(t) = (1-\beta t)^{-\alpha }\), so matching gives \(\alpha =2\) and \(\beta =3\): by uniqueness \(X\sim \) GAM\((2,3)\).

Using the cumulant generating function is quicker than differentiating \(M\) twice: \[K_X(t) = \log M_X(t) = -2\log (1-3t),\] \[K_X'(t) = \frac {6}{1-3t},\qquad K_X''(t) = \frac {18}{(1-3t)^{2}},\qquad K_X'''(t) = \frac {108}{(1-3t)^{3}} .\] Evaluating at \(t=0\), \[\kappa _1 = E(X) = 6,\qquad \kappa _2 = \Var (X) = 18,\qquad \kappa _3 = 108 .\] These agree with \(\alpha \beta = 6\) and \(\alpha \beta ^{2} = 18\) for the gamma, and the third cumulant \(2\alpha \beta ^{3} = 108\).