2

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.

Sarvesh Ravichandran Iyer
  • 77,817
  • Which theorem of Strassen are you referring to? The one I know is related to convex order and not this stochastic order. Moreover $X$ is dominated by $S$ in the stochastic order iff $F_X^{-1}(U)\le F_S^{-1}(U)$ everywhere, where $F^{-1}$ denotes the quantile function and $U$ is uniformly distributed on $(0,1)$. Of course the latter makes sense only when $X$ and $S$ take values in $\mathbb R$. – Will Feb 01 '21 at 14:04
  • I'll edit it, @Will. – Sarvesh Ravichandran Iyer Feb 01 '21 at 14:08
  • @Will, the Strassen theorem I know states this : if $X,Y$ are real-valued random variables such that $P(X <t) \leq P(Y < t)$ for all $t \in \mathbb R$, then there exists a probability space, and real valued random variables $X',Y'$ out of that probability space, such that $X=X'$ and $Y=Y'$ in distribution, and $X' \leq Y'$ almost surely. Now, each $S_n$ dominates $X_n$ under that definition, according to results from SS. I want two (real/integer valued) stochastic processes $S'$ and $X'$ such that $S' = S$, $X'=X$ in distribution, and $S' \geq X'$ a.s. – Sarvesh Ravichandran Iyer Feb 01 '21 at 14:23
  • 1
    Okay I didn't know this was called Strassen's theorem as well. There is not much mystery behind this. As I said, you can take $X'=F_X^{-1}(U)$ and $Y'=F_Y^{-1}(U)$, where $U$ is uniformly distributed on $(0,1)$. Then $X'=X$ and $Y'=Y$ in distribution (that's the inverse transform sampling), and $X\le_{st} Y$ iff $X'\le Y'$, where $\le_{st}$ denotes the stochastic order, and $\le$ denotes the usual partial order on $\mathbb R$. – Will Feb 01 '21 at 15:53
  • @Will You are correct, I see what you are saying. I was just hoping there was a construction so that $S',X'$ themselves could be grid walks, or just more meaningful random variables than the ones you describe. For example, if $X_n,S_n$ had parameters $p \neq q$ at every point, then you could construct $X_n',S_n'$ as random walks themselves. But I get your point, maybe I'm not being very precise over this. – Sarvesh Ravichandran Iyer Feb 01 '21 at 15:59
  • 1
    Are you sure that $S_n$ dominates $X_n$ stochastically? For instance $\mathbb P(S_2=-2)>0$ whereas $\mathbb P(X_2=-2)=0$. – Will Feb 01 '21 at 17:29
  • @Will Oh, I made an error in notation! Thanks for pointing it out. You can see the edits now : the point is, I want the random variables coming from Strassen's theorem to also look like random walks themselves, or to have more meaning than just being a Strassen prover. By the way, I sincerely appreciate you taking interest in my question, thank you very much for that. – Sarvesh Ravichandran Iyer Feb 01 '21 at 18:09
  • I see, that makes much more sense now – Will Feb 01 '21 at 18:25
  • You are welcome @Will. Indeed, it was my intention all along , but I did not spot that terrible error. I think the content is correct now. You are free to answer. However, I want to remind you I'm actually not the one with the bounty. This question of mine was unanswered, and then I randomly found out today that it was bountied! So another (very nice) user is interested in this answer as well. – Sarvesh Ravichandran Iyer Feb 01 '21 at 18:26

1 Answers1

3

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 $S_n\le S'_n$.

Will
  • 8,358