Prove identity for symmetric random walk

Question

Let $S=(S(n))_{n\geq 0}$ be a symmetric random walk starting at zero. Prove that for all $m \geq 1$ $$ 2\mathbb{P}(S(1)>0, \ldots, S(2m)>0)=\mathbb{P}(S(1)\geq 0, \ldots, S(2m)\geq 0).$$

There is a theorem that says that if $S(0)=0$ then $$\mathbb{P}(S(1)\neq 0, \ldots, S(2m-1)\neq 0, S(2m)=b)=\frac{|b|}{2m}\mathbb{P}(S(n)=b).$$

Using this theorem, I tried $$\mathbb{P}(S(1)>0, \ldots, S(2m)>0)=\sum_{k=1}^\infty \mathbb{P}(S(1)>0, \ldots, S(2m-1)>0, S(2m)=k) =\sum_{k=1}^\infty \frac{k}{2m}\mathbb{P}(S(2m)=k) =\sum_{k=1}^\infty \frac{k}{2m} \binom{2m}{1/2(2m+k)}(1/2)^{2m}.$$

I am stuck here. Perhaps using this theorem was a bad idea?

Answer 1

To prove this identity, one simple observation is enough.

image source

Let ${\cal P}_{2n}$ and ${\cal N}_{2n}$ be the sets of paths satisfying $\{S(1)>0, \ldots, S(2n)>0\}$ and $\{S(1)\geq 0, \ldots, S(2n)\geq 0\}$ respectively for all $n \in \Bbb{N}$.

Lemma: There exists a bijection between ${\cal P}_{2n}$ and ${\cal N}_{2n-1}$ for all $n \in \Bbb{N}$.

Proof: Each $\color{blue}{S} \in {\cal P}_{2n}$ satisfies $\color{blue}{S} \ge 1$. Shift the origin to $(1,1)$ to create a path $\color{red}{S'} \in {\cal N}_{2n-1}$. For each path in ${\cal N}_{2n-1}$, an inverse process gives a path in ${\cal P}_{2n}$.

$$2\Bbb{P}({\cal P}_{2m}) = 2\;\frac{|{\cal P}_{2m}|}{2^{2m}} = \frac{|{\cal N}_{2m-1}|}{2^{2m-1}} = \Bbb{P}({\cal N}_{2m-1})$$