Biased Walks on the Integer Line

Drift Changes the Hitting Problem

A nearest-neighbor walk on the integer line moves one unit at each step. When the probabilities of moving right and left are unequal, its long-term drift affects whether it is likely to return to places behind it. The resulting hitting probabilities are not symmetric: a walk drifting right can still hit a lower level, but that probability decreases with the distance to that level.

We will calculate the probability of ever reaching a specified lower level, then use that result to find expected visit counts. The cases where one step direction has probability zero require care: formulas involving a ratio of step probabilities must not be applied when their denominator vanishes.

Walk, Hitting Times, and Visits

Let $p,q\in[0,1]$, with $p+q=1$ and $p\ne q$. Let $(S_n)_{n\in\mathbb{N}_0}$ be a walk on $\mathbb{Z}$ whose increments are independent, with each increment equal to $1$ with probability $p$ and $-1$ with probability $q$. Write $\mathbb{P}_x$ and $\mathbb{E}_x$ for probability and expectation when $S_0=x$.

Definition: For a level $a\in\mathbb{Z}$, the hitting time is $\tau_a=\inf\{n\in\mathbb{N}_0:S_n=a\}$, with $\inf\varnothing=\infty$. The number of visits to a site $y$, including a visit at time zero, is $V_y=\sum_{n=0}^{\infty}\mathbf{1}_{\{S_n=y\}}$.

We first start at $k>0$ and ask for the probability of ever hitting $0$. Translation invariance will then give the result for any lower level $a$.

A Finite-Interval Calculation

For an integer $N>k$, stop the walk when it reaches either endpoint of $\{0,1,\ldots,N\}$. Let $h_i$ be the probability, starting at $i$, of hitting $0$ before $N$. The boundary values are $h_0=1$ and $h_N=0$. A first-step decomposition gives, for $1\le i\le N-1$,

h_i=p h_{i+1}+q h_{i-1}.

When $p>0$, define $r=q/p$. For unequal probabilities, the recurrence has solutions that are linear combinations of $1$ and $r^i$. Substitution into the recurrence verifies this: the constant solution works because $p+q=1$, while $p r+q/r=q+p=1$. Applying the two boundary values yields the following finite-interval formula.

Theorem: If $p>0$, $p\ne q$, and $0\le k\le N$, then $$ h_k=\frac{r^k-r^N}{1-r^N},\qquad r=\frac{q}{p}, $$ where the formula at $k=0,N$ agrees with the boundary values.

Proof: For $0<k<N$, the function $g_i=(r^i-r^N)/(1-r^N)$ is well-defined because $r\ne1$. Its boundary values are $g_0=1$ and $g_N=0$. Each of the constant and $r^i$ terms satisfies the first-step recurrence, so $g_i=p g_{i+1}+q g_{i-1}$ for interior $i$. The finite-interval hitting probabilities satisfy the same recurrence and boundary conditions. Uniqueness follows from the maximum principle on this finite set: the difference of two solutions is harmonic with zero boundary values, and a positive maximum or negative minimum at an interior point must propagate to a boundary. Thus the difference is zero everywhere, and $h_i=g_i$. If $q=0$, then $r=0$; for interior $i\ge1$, $r^i=0$, so the formula gives $h_i=0$, as required for a walk that moves only right. End of proof. $\square$

Probability of Ever Hitting a Lower Level

For $p>0$, let $N$ increase in the finite-interval formula. The events “hit $0$ before $N$” increase with $N$, and their union is exactly the event of hitting $0$ at a finite time: any path that hits $0$ in finite time has a finite maximum before that hit. If $p>q$, then $0\le r<1$, so $r^N$ tends to zero. If $p<q$, then $r>1$; dividing numerator and denominator by $r^N$ shows that the ratio tends to $1$.

Theorem: For $k>0$, the probability of ever hitting $0$ is $$ \mathbb{P}_k(\tau_0<\infty)= \begin{cases} (q/p)^k,&p>q,\\ 1,&p<q. \end{cases} $$ The first case includes $q=0$, for which the probability is zero. If $p=0$ (so $q=1$), the walk moves deterministically left and $\mathbb{P}_k(\tau_0<\infty)=1$.

Proof: When $p>0$, the increasing-event limit of the finite-interval probabilities is $$ \lim_{N\to\infty}\frac{r^k-r^N}{1-r^N}. $$ For $p>q$, this limit is $r^k$. For $p<q$, division by $r^N$ gives $(r^{k-N}-1)/(r^{-N}-1)$, whose limit is $1$. This establishes the formula for every $p>0$ with $p\ne q$, including $p=1,q=0$. If $p=0,q=1$, every step is left, so starting at $k>0$ the walk reaches zero after exactly $k$ steps. For a general lower level $a<x$, translate the walk by $-a$ and put $k=x-a$. End of proof. $\square$

Thus the lower-level hitting probability from $x>a$ is $1$ when the walk drifts left, and $(q/p)^{x-a}$ when it drifts right. The symmetric case is not part of the unequal-probability theorem; it has different behavior and was treated with the symmetric walk.

Worked Example: A Right-Drifting Walk and a Lower Level

Let $p=3/5$, $q=2/5$, and start at $x=4$. The walk drifts right, and $q/p=2/3$. Therefore, the probability of ever hitting $0$ is

\mathbb{P}_4(\tau_0<\infty)=\left(\frac23\right)^4=\frac{16}{81}.

For a finite check, take the upper boundary $N=5$. Starting from $k=2$, the probability of hitting zero before five is $$ \frac{(2/3)^2-(2/3)^5}{1-(2/3)^5} =\frac{4/9-32/243}{1-32/243} =\frac{76/243}{211/243} =\frac{76}{211}. $$ This finite-interval value is smaller than the probability $(2/3)^2=4/9$ of ever hitting zero, since paths that reach five first are excluded.

Worked Example: The Deterministic Boundary Case

Take $p=0$, $q=1$, and start at $x=3$, with lower target $a=0$. The ratio $q/p$ is undefined, so the ratio formula cannot be used. Instead, the walk is deterministic: $S_n=3-n$. Hence $S_3=0$, and the probability of hitting zero is $1$. More generally, from any $x>a$, it reaches $a$ after exactly $x-a$ steps.

Expected Visits in the Transient Direction

When $p\ne q$, the walk is transient: the expected number of visits to any fixed site is finite. We obtain the visit count by separating two questions. First, what is the probability of ever reaching the site? Second, given a visit, how many visits should we expect in total, counting the first one?

At a site $y$, after leaving it, a right step occurs with probability $p$ and a left step with probability $q$. If $p>q$, a left departure returns to $y$ almost surely, whereas after a right departure the probability of returning is $q/p$, by the lower-level hitting formula. Thus the probability of returning after a visit is $p(q/p)+q=2q$. If $p<q$, a right departure returns almost surely, and after a left departure the return probability is $p/q$, giving $p+q(p/q)=2p$.

Theorem: For $p\ne q$, the expected number of visits to $y$, starting at $y$, is $$ \mathbb{E}_y[V_y]=\frac{1}{|p-q|}. $$ More generally, $$ \mathbb{E}_x[V_y]= \begin{cases} \dfrac{1}{p-q},&p>q,\ x\le y,\\[6pt] \dfrac{(q/p)^{x-y}}{p-q},&p>q,\ x>y,\\[6pt] \dfrac{1}{q-p},&p<q,\ x\ge y,\\[6pt] \dfrac{(p/q)^{y-x}}{q-p},&p<q,\ x<y. \end{cases} $$

Proof: Let $\rho$ be the probability of returning to $y$ after leaving it. The first-return calculations above give $\rho=2q$ if $p>q$, and $\rho=2p$ if $p<q$. By the Markov property at each return, the probability of at least $j$ visits starting at $y$ is $\rho^{j-1}$ for $j\in\mathbb{N}$. Therefore, using the tail-sum formula for a nonnegative integer-valued random variable, $$ \mathbb{E}_y[V_y]=\sum_{j=1}^{\infty}\mathbb{P}_y(V_y\ge j) =\sum_{j=1}^{\infty}\rho^{j-1} =\frac{1}{1-\rho}. $$ For $p>q$, $1-\rho=1-2q=p-q$; for $p<q$, $1-\rho=1-2p=q-p$. These quantities are positive, so the geometric series converges. Now start from $x$. If $H_y=\inf\{n\in\mathbb{N}_0:S_n=y\}$, then $V_y=0$ on $\{H_y=\infty\}$. On $\{H_y<\infty\}$, the Markov property at $H_y$ makes the expected number of visits from that time onward equal to $\mathbb{E}_y[V_y]$. Thus $$ \mathbb{E}_x[V_y]=\mathbb{P}_x(H_y<\infty)\mathbb{E}_y[V_y]. $$ For $p>q$, a site to the left of or equal to $y$ is hit almost surely, while from $x>y$ the hitting probability is $(q/p)^{x-y}$. For $p<q$, a site to the right of or equal to $y$ is hit almost surely, while from $x<y$ it is $(p/q)^{y-x}$; this follows by reflecting the walk and applying the lower-level formula with the step probabilities exchanged. Substitution gives all four cases. End of proof. $\square$

Worked Example: Visit Counts for a Right Drift

Let $p=3/5$, $q=2/5$, and count visits to $y=0$. The expected number of visits when starting at zero is $1/(p-q)=5$. Starting at $x=2$, the probability of ever reaching zero is $(2/3)^2=4/9$. Therefore, the expected total number of visits to zero is $$ \mathbb{E}_2[V_0]=\frac49\cdot 5=\frac{20}{9}. $$ Starting instead at $x=-1$, the walk drifts toward zero and hits it almost surely, so $\mathbb{E}_{-1}[V_0]=5$.

Worked Example: Visit Counts for a Left Drift

Let $p=1/3$, $q=2/3$, and count visits to $y=1$. Since $p<q$, the walk drifts left. Starting at $x=4$, it hits the lower site $1$ almost surely. The return probability after a visit is $2p=2/3$, so the expected visits starting at the site are $1/(1-2/3)=3$. Consequently, $$ \mathbb{E}_4[V_1]=1\cdot3=3. $$ If instead the walk starts at $x=-1$, it must move against its drift to reach $1$. The probability of doing so is $(p/q)^2=(1/2)^2=1/4$, giving $\mathbb{E}_{-1}[V_1]=(1/4)\cdot3=3/4$.

Interpreting the Formulas

The expected visit count is a Green function: it records the expected number of visits to a specified site from a given starting point. Here it factors into the probability of reaching that site and the expected number of visits once there. This factorization is useful because the two pieces have distinct meanings and are calculated separately.

A common error is to infer that “drifting away” means a site cannot be reached. For a right-drifting walk, a lower site remains reachable with positive probability whenever $q>0$; it is simply reached with probability $(q/p)^{x-y}$ from $x>y$. Another error is to use a ratio formula when its denominator is zero. For $p=0$, the path is deterministic and must be analyzed directly. Likewise, the expected number of visits is finite but can be large when the drift is weak, because $1/|p-q|$ grows as the probabilities approach equality.

Check Your Understanding

Use the definitions and formulas to answer the following questions.

For $p=4/7$, $q=3/7$, and $x=3$, what is the probability of ever hitting $0$?
Why is it invalid to substitute $p=0$ into the ratio $q/p$, and what is the hitting probability in that case?
For $p>q$, explain why the probability of returning to $y$ after leaving it is $2q$.
For a left-drifting walk, what is the expected number of visits to a site when starting at that site?
For $p=1/3$, $q=2/3$, what is the probability of hitting $y=0$ from $x=-2$, and what is the expected number of visits to zero?