# What happens to a random walk when we increase the probabilities of going right?

Consider a random walk on the integers where the probability of transitioning from $n$ to $n+1$ is $p_n$ (and of course, the probability of transitioning from $n$ to $n-1$ is $1-p_n$); we assume all $p_n$ are strictly less than $1$. Suppose we know that this random walk is pretty well concentrated; for example, let us assume that we know that $$P(|X(t) - (1/3)t| \geq c \sqrt{t}) \leq e^{-c^2}$$ where $X(t)$ is the state of the walk after $t$ steps.

Now suppose we increase every $p_n$ by $\epsilon$ (and correspondingly decrease the probability of transitioning from $n$ to $n-1$ by $\epsilon$), where $\epsilon$ is some number such that $p_n + \epsilon < 1$ for all $n$. Let $Y(t)$ be the state of the new chain after $t$ steps. My question is: does a similar concentration result hold for $Y(t)$?

It seems very intuitive that $Y(t)$ should concentrate around $(1/3)t + 2 \epsilon t$.

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Let $E(t)$ be the one-way random walk that adds one to its current state with probability $\epsilon$ and remains in its current state with probability $1-\epsilon$. Is it not true that $Y(t) = X(t) + E(t)$ exactly? - that is, the distributions of the random variables $Y(t)$ and $X(t) + E(t)$ are exactly the same?
Assuming I'm right that they are the same, then a concentration result for $Y$ follows from the hypothesized concentration result for $X$ and the standard concentration that can be calculated for $E(t)$: if $Y(t) - (\frac13+\epsilon)t$ exceeds $c\sqrt t$, then one of $X(t) - \frac13t$ or $E(t) - \epsilon t$ must exceed $\frac c2\sqrt t$.
There is no reason for the distribution of $Y(t)$ to be the same as the distribution of $X(t)+E(t)$, as far as I can tell. What makes this "obvious" problem difficult for me is the dependency created - depending on whether $E(t)=0$ or $E(t)=1$ the random walk will take a different probability to go right. – yves Feb 23 '13 at 2:06
Sigh. You're totally right. $X(t)$ and $Y(t)$ are both supported on integers with the same parity as $t$, while neither $E(t)$ nor $X(t)+E(t)$ have this property. – Greg Martin Feb 23 '13 at 4:47