Suppose we have a simple symmetric random walk starting at $1$, and define $$T_{\text{hit zero}} = \min\{n\geq 1 : S_n = 0\}.$$ I was trying to argue that $$\star =\mathbb{E}_1\big[S_n\textbf{1}_{\{T_{\text{hit zero}} > n\}}\big] \not\to 0$$ as $n\to\infty$.
I can think of it with using the FKG inequality, as the expectation can be lower bounded by $$ \star \geq \sqrt{n} \; \mathbb{P}_1( \{S_n \geq \sqrt n\}\cap \{T_{\text{hit zero}}>n\}).$$ Both events are increasing (if we flip any of the steps from a $-1$ to $1$, they are more likely to happen), and so we get $$\star\geq \sqrt n \; \mathbb{P}_1\{S_n \geq \sqrt n\} \mathbb{P}_1\{T_{\text{hit zero}}>n\} \geq \sqrt n C_1 \frac{C_2}{\sqrt n} > \epsilon.$$
Is there an easy to to show this claim without FKG inequality?
Claim: $\star_n =1$ for all nonnegative integers $n$.
Proof: I assume we walk over the integers, $S_n$ is the integer location at time $n \in \{0, 1, 2, ...\}$, and $S_0=1$. Then $$ S_n = 1 + \sum_{i=1}^n X_i \quad \forall n\in \{1, 2,3, ...\}$$ where $\{X_i\}$ are i.i.d. with $P[X_i=1]=P[X_i=-1]=1/2$ for all $i$. Thus $$E[S_n] = 1 \quad \forall n \in \{1, 2, 3, ....\}$$ For each $k \in \{1, 2, 3, ...\}$ define $H_k$ as the event that we first hit zero at time $k$. Define $G_k$ as the event that we have not yet hit 0 up to and including time $k$. Then for any $n\in \{1, 2, 3, ...\}$ we obtain \begin{align} 1 &= E[S_n] \\ &= E[S_n|G_n]P[G_n]+\sum_{k=1}^{n}E[S_n|H_k]P[H_k] \\ &=E[S_n|G_n]P[G_n]\\ &= E[S_n 1_{G_n}]\\ &= \star_n \end{align} where we have used the fact that $E[S_n|H_k]=0$ for $n\geq k$. Thus, $\star_n =1$ for all positive integers $n$. Finally, it is clear that $\star_0=1$. $\Box$