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 $p_n + \epsilon &lt; 1$ 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 $(1/3)t + 2 \epsilon t$.
Write $Z(t) = Y(t) - X(t)$. Then
\[
Y(t)-(1/3 + 2\epsilon)t = \Big(X(t) - (1/3)t\Big) + \Big(E[Z(t)]-2\epsilon t\Big) + \Big(Z(t) - E[Z(t)]\Big).
\]
The concentration of the first term is given by your assumption.
For the third term, $Z(t)$ is the sum of $t$ iid bounded random variables,
so $P(|Z(t) -E[Z(t)] | \ge x \sqrt{t} ) \le 2 e^{-C x^2}$ for some constant $C&gt;0$ by Hoeffding's inequality; this takes care of the third term above.

For the middle term, $E[Z(t)]-2\epsilon t=0$ by noticing that $Z(t)-2\epsilon t$ is a Martingale with zero mean at time 0.

By the union bound, with probability at least $1-2e^{-C x^2} - e^{-c^2}$ we get
\[
|Y(t)-(1/3 + 2\epsilon)t| \le (c+x)\sqrt t.
\]
NOTE: This is totally wrong - see comments.

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 $\frac c2\sqrt t$.