Let $\left\{ S_{n}\right\} _{n\geq0}$ be the (asymmetric) simple random walk such that $S_{n+1}=S_{n}+\xi_{n+1}$ where $\xi_{n}\in\{-1,1\}$ and $\left\{ \xi_{n}\right\} _{n\geq0}$ are iid. Moreover, $\mathbb{P}(\xi_{n}=1)=p$ and $\mathbb{P}(\xi_{n}=-1)=1-p$ and ${\displaystyle p>\frac{1}{2}}.$ For $x\in\mathbb{Z}$ we define $T_{x}=\inf\left\{ n:S_{n}=x\right\} $ and $\phi(x)=\left(\frac{1-p}{p}\right)^{x}.$
Then for a<0, Show that $E\left[\phi(S_{N})\bigg|T>N\right]\leq\left(\frac{1-p}{p}\right)^{b}+\left(\frac{p}{1-p}\right)^{a}$where $T=T_{a}\wedge T_{b}$.
My thoughts:
${\displaystyle \phi(S_{N})={\displaystyle \left(\frac{1-p}{p}\right)^{S_{N}}}=\left(\frac{1-p}{p}\right)^{\sum_{i=1}^{N}\xi_{i}}=\prod_{i=1}^{N}\left(\frac{1-p}{p}\right)^{\xi_{i}}}$
Since $\left\{ \xi_{n}\right\} _{n\geq0}$ are iid, I somehow guess $E\left[\phi(S_{N})\bigg|T>N\right]=1$, but I am not sure.
Even if I am right, I can not show $\left(\frac{1-p}{p}\right)^{b}+\left(\frac{p}{1-p}\right)^{a}\geq1$.
I only know that T>N\Rightarrow a
Thank you!