Let $S_n$ be the usual Simple Symmetric Random Walk on $\mathbb Z$. That is, $X_n \sim \mbox{Ber}_{\pm 1}(\frac 12)$ are iid with $X_0 = 0$ and $S_n = X_0+...+X_n$. (Where $\mbox{Ber}_{\pm 1}(p)$ for $p \in [0,1]$ is $+1$ with probability $p$ and $-1$ otherwise).
This has its own hitting probabilities for various sets and other growth functionals.
Suppose at one $z \in \mathbb Z$, I perturb the probability of the random walk and create a new stochastic process. That is, fix a $p \in [\frac 12,1]$ and define $S'_n$ recursively as follows : $S'_0 = 0$, and
if $S'_{n-1}\neq z$ then $S'_n = S'_{n-1} + \mbox{Ber}_{\pm 1}$.
On the other hand, if $S'_{n-1} = z$ then $S'_n = S'_{n-1} + \mbox{Ber}_{\pm 1}(p)$.
Since $p \geq \frac 12$, the stochastic process has a right drift at the point $z$ since if it reaches $z$ then it is more likely to go right than left.
This is like a one-point perturbation of the stochastic process.
I believe that $S'_n$ dominates $S_n$ stochastically. That is, for every $n$, and every $t \in \mathbb Z$ we have $\mathbb P(S'_n>t) \geq \mathbb P(S_n > t)$. This should be provable from one of the theorems in Shaked and Shanthikumar, which goes through a lot of these "stochastic comparison of random variables" results, so I am not worried about that.
My question is this: can we find an explicit coupling of $S'_n$ and $S_n$, for which $S'_n \geq S_n$ almost surely, thereby implying stochastic dominance of $S'_n$ over $S_n$? Such a coupling non-explicitly exists through Strassen's theorem, but in this simple situation, I would like an explicit example.
While I have seen coupling used to prove Stochastic dominance, I have seen this only in cases where the parameter is uniformly changed (e.g. $\mbox{Bin}(p)$ dominates $\mbox{Bin}(q)$ with $p>q$), but not at one point.
EDIT : Thank you for placing the bounty, @Restless.
I propose something: let $U_1,\cdots,U_n,\cdots$ be independent uniformly distributed on $(0,1)$. For all $n\in\mathbb N^*$, let $X_n=1_{\{U_n<\frac12\}}-1_{\{U_n\ge\frac12\}}$, that is $X_n=1$ if $U_n<\frac12$ and $X_n=-1$ else, so that $X_n\sim\textrm{Ber}_{\pm1}(\frac12)$ and $X_1,\cdots,X_n,\cdots$ are iid. Let $S_0=0$ and for $n\in\mathbb N^*$ let $S_n=X_1+\cdots+X_n$, $$ X'_n=X_n1_{\{S_{n-1}\neq z\}}+\left(1_{\{U_n<p\}}-1_{\{U_n\ge p\}}\right)1_{\{S_{n-1}=z\}} $$ and $S'_n=X'_1+\cdots X'_n$. Then for all $n\in\mathbb N^*$, $X_n\le X'_n$, and of course $S_n\le S'_n$.