TITLE: Show that the survival probability $\gamma$ satisfies $ \gamma = 1 - e^{-\lambda \gamma} $
QUESTION [1 upvotes]: The offspring distribution $X$ of a branching process is given by Poisson($\lambda$).
I need to show that the survival probability $\gamma$ satisfies the following equation:
$$ \gamma = 1 - e^{-\lambda \gamma}  $$
Now suppose $\kappa$ is the extinction probability, then it is the smallest solution in $[0,1]$ of the equation $$\kappa = G_X(\kappa) = \mathbb{E}[X^\kappa]$$ where $G_X$ is the probability generating function of $X$.
Since $\kappa = 1 - \gamma$ , therefore $$1- \gamma = G_X(1-\gamma)$$ $$ \space \space \space  = \mathbb{E}[X^{1-\gamma}]$$ $$= \sum_{i=0}^{\infty}i^{1-\gamma} \frac{e^{-\lambda}\lambda^i}{i !}$$
This implies, $$\gamma = 1 - e^{-\lambda} \sum_{i=0}^{\infty}i^{1-\gamma} \frac{\lambda^i}{i !}$$ Now, how do I show $$e^{-\lambda \gamma} = e^{-\lambda} \sum_{i=0}^{\infty}i^{1-\gamma} \frac{\lambda^i}{i !}$$ 
Is my approach corret or something wrong with it?

REPLY [1 votes]: The extinction probability P is the smallest solution $s = G(s)$, $s \in [0,1]$, and the survival probability is $\gamma = 1-P$.
For Poisson, $G(s) = e^{-\lambda(1-s)}$, thus
$s = e^{-\lambda(1-s)}$.
Since $P$ is the smallest solution of $s$, we make $P=s$ and substitute $\gamma$, leading to the final answer.
$P = e^{-\lambda(1-P)}$
$1-\gamma = e^{-\lambda(1-(1-\gamma))}$
$\gamma = 1-e^{-\lambda\gamma}$.
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