# Conditional probability branching process

Consider a discrete time branching process $X_{n}$ with $X_{0}=1.$ Establish the simple inequality $$P\{X_{n}>L\ \textrm{for some}\ 0\leq n\leq m\ |\ X_{m}=0 \}\leq [P\{X_{m}=0\}]^L$$

Note: This issue is the second edition of the book “The First Course in Stochastic Processes” Samuel Karlin author. Could anyone help me to solve it? I have no idea to answer it! I thank you!

#### Solutions Collecting From Web of "Conditional probability branching process"

Let $T=\inf\{n:X_n>L\}$. Define $\mathbb{E}_x[\cdot]=\mathbb{E}[\cdot|X_0=x]$, $\mathcal{F}_n=\sigma(X_0,\dots,X_n)$, and $\mathcal{F}_T=\{A:A\bigcap\{T=n\}\in\mathcal{F}_n\text{ for all$n$}\}$.

Then since

$$\{X_n>L\text{ for some 0\le n\le m }\}=\{\max_{0\le n\le m}X_n>L\}=\{T\le m\}$$

we have

$$P_1\{\max_{0\le n\le m}X_n>L,X_m=0\}=P_1\{T\le m,X_m=0\}$$
$$=\mathbb{E}_1[1\{T\le m\}1\{X_m=0\}]$$
$$=\mathbb{E}_1[1\{T\le m\}\mathbb{E}_1[1\{X_{(m-T)+T}=0\}|\mathcal{F}_T]]$$
$$=^1\mathbb{E}_1[1\{T\le m\}\mathbb{E}_{X_T}[1\{X_{m-T}=0\}]$$
$$=^2\mathbb{E}_1[1\{T\le m\}(P_1\{X_{m-T}=0\})^{X_T}]]$$
$$\le^3 \mathbb{E}_1[1\{T\le m\}(P_1\{X_m=0\})^{L+1}]]\le (P_1\{X_m=0\})^{L+1}$$

$^1$ – by the strong Markov property;

$^2$ – $P_x(X_k=0)=P_1(X_k=0)^x$;

$^3$ – $X_T\ge L+1$ and $P\{X_k=0\}$ is non-decreasing in $k$.

Consequently,

$$P_1\{\max_{0\le n\le m}X_n>L|X_m=0\}=\frac{P_1\{\max_{0\le n\le m}X_n>L,X_m=0\}}{P_1\{X_m=0\}}\le (P_1\{X_m=0\})^L$$