Distribution of a branching process martingale at the limit

Question

Source: Probability with Martingale Williams, Page 11. In an attempt to get the distribution of $M_{\infty}$ for the case $\mu >1$ the author writes, you can easily check that, for $\lambda>0$ $$L(\lambda):=Eexp(-\lambda M_{\infty})=\frac{p\lambda+q-p}{q\lambda+q-p}=\pi e^{-\lambda.0}+\int_{0}^{\infty}(1-\pi)^2e^{-\lambda x}e^{-(1-\pi)x}dx$$ I can derive the first equality but do not understand the second equality derivation. I can probably check it backwards as stated, but how would one come up with it in a deductive fashion? This is used to conclude that $P(M_{\infty}>x)=(1-\pi)e^{-(1-\pi)x}$ for $x>0$.

Answer 1

One knows that $\int_0^\infty e^{-(\lambda+A)x} dx = {1\over \lambda+A}$ (for $A>0$), so presented with $$ {a\lambda+b\over c\lambda+d} $$ one is tempted to write $$ {a\lambda+b\over c\lambda+d}={(a/c)\lambda+(b/c)\over \lambda+(d/c)}={a\over c}+{(b/c)- (ad/c^2)\over \lambda+(d/c)} $$ thereby deducing that $$ \eqalign{ {a\lambda+b\over c\lambda+d} &={a\over c}+[(b/c)-(ad/c^2)]\int_0^\infty e^{-\lambda x}e^{-(d/c)x} dx\cr &=\pi+(1-\pi)^2 \int_0^\infty e^{-\lambda x}e^{(1-\pi)x} dx.\cr } $$ ($a=p$, $b=d=q-p$, $c=q$, and I presume that $\pi = p/q$.) Notice that $\pi=\lim_{\lambda\to\infty}E[\exp(-\lambda M_\infty)]=P[M_\infty=0]$.

We recognize $$ \pi+(1-\pi)^2 \int_0^\infty e^{-\lambda x}e^{(1-\pi)x} dx=\pi+(1-\pi) \int_0^\infty e^{-\lambda x}(1-\pi)e^{(1-\pi)x} dx $$ as the Laplace transform of the mixture of a point mass at $0$ (with weight $\pi$) and an exponential distribution with parameter $1-\pi$ (with weight $1-\pi$). In particular, $P[M_\infty>x]=(1-\pi)e^{-(1-\pi)x}$.