= Solution
The moment generating <function> is $M_X(\theta)=\mathbb E(e^{\theta X})$ where finite. The <continuity> theorem states that, if moment generating functions are finite on a common open interval containing zero and converge there to the moment generating <function> of a probability law, the corresponding distributions converge weakly to that law.
Integrating the gamma density and using the gamma integral gives
$$
M_X(\theta)=\frac{\lambda^\alpha}{\Gamma(\alpha)}\int_0^\infty x^{\alpha-1}e^{-(\lambda-\theta)x}\,dx
=\left(\frac{\lambda}{\lambda-\theta}\right)^\alpha,\qquad\theta<\lambda.
$$
For a geometric waiting time, on $1,2,\ldots$, summing the geometric series gives
$$
M_Y(\theta)=\sum_{j\ge1}p(1-p)^{j-1}e^{j\theta}=
\frac{pe^\theta}{1-(1-p)e^\theta},\qquad \theta<-\log(1-p)
$$
when $0<p<1$. For $p=1$, $Y=1$ deterministically. The waiting times between successive heads are independent and identically distributed, so $N=Y_1+\cdots+Y_k$ and $M_N=M_Y^k$. Therefore
$$
M_{2pN}(\theta)=\left\{\frac{pe^{2p\theta}}{1-(1-p)e^{2p\theta}}\right\}^k
\longrightarrow (1-2\theta)^{-k}.
$$
Indeed the denominator is $p(1-2\theta)+O(p^2)$ and the numerator is $p+O(p^2)$. On any sufficiently small fixed neighbourhood of zero these moment generating functions exist for all sufficiently small $p$. The limit is the moment generating <function> of a gamma law with \b[shape $k$ and rate $1/2$], so
$$
\boxed{2pN\ \xrightarrow{d}\ \operatorname{Gamma}(k,1/2)}.
$$
Back to article page
