Why is this true with random walks?

Question

Let $S_n=\sum_n{X_n}$ a random walk with $P[X=1]=p=1-P[X=-1]$. Prove that for any $k \in Z$ $P[\cup_{n \geq1} \{S_n\geq k\}]= (P[\cup_{n \geq1} \{S_n\geq 1\}])^k$

I do not understand why is this true, should I use the reflection principle to prove this?

Thanks for your support.

Answer 1

I don't know of a way to use the reflection principle to prove this. It might be easier to think this way: for any walk to go from $0$ to $k$ it must, in turn, go from $0$ to $1$, then from $1$ to $2$, ..., then from $k-1$ to $k$.

Conversely, any sequence of walks from $0$ to $1$ can be joined end-to-end to form a single path from $0$ to $k$.

So,

\begin{align} P(\cup_{n\geq 1} \{S_n\geq k\}) &= P(\text{Walk from $0$ to $1$}\;\cap\;\text{Walk from $1$ to $2$}\;\cap\;\cdots\cap\text{Walk from $k-1$ to $k$}) \\ &= P(\text{Walk from $0$ to $1$})\;P(\text{Walk from $1$ to $2$})\cdots P(\text{Walk from $k-1$ to $k$}) \\ & \qquad\qquad\qquad\qquad\qquad\qquad\text{by independence of the separate sub-walks} \\ &= P(\text{Walk from $0$ to $1$})\;P(\text{Walk from $0$ to $1$})\cdots P(\text{Walk from $0$ to $1$}) \\ &= (P[\cup_{n \geq1} \{S_n\geq 1\}])^k. \end{align}