Expected Absorption Time for an Asymmetric Random Walk
Consider an asymmetric random walk on $\mathbb{Z}_{\geq 0}$ starting at position 1. At each step, you move right (to $i+1$) with probability $p$ and left (to $i-1$) with probability $q = 1 - p$. Position 0 is an absorbing barrier -- the walk terminates upon reaching it.
- For what values of $p$ is absorption certain?
- Compute the expected time to absorption $E_1$ as a function of $p$ and $q$, and identify when it is finite.
- In the symmetric case $p = q = 1/2$, what happens?
Hints
- Start with the first-step decomposition: from position $i$, you either step left (prob $q$) or right (prob $p$), so $E_i = 1 + pE_{i+1} + qE_{i-1}$.
- Guess a linear solution $E_i = ci$ and substitute into the recurrence to solve for $c$.
- The solution $E_i = i/(q - p)$ is valid when $q > p$. For $p = q = 1/2$, the recurrence has no finite solution -- this is the null recurrent case.
Worked Solution
How to Think About It: Picture yourself at position 1 with a coin that lands heads (step right) with probability $p$. If $p < 1/2$, you are drifting toward the barrier -- absorption is certain and should not take too long. If $p > 1/2$, you are drifting away, so there is a real chance you escape to infinity and never return. The symmetric case is the knife-edge: you will return to 0 with probability 1, but the expected time to do so is infinite (null recurrence). Before doing any algebra, you should be able to say: "drift inward means finite expected time, drift outward means absorption is not even certain, and symmetric means certain absorption but infinite expected time."
Quick Estimate: For $q > p$, the expected absorption time from position 1 should be proportional to
Approach: Set up the first-step recurrence for $E_i$ and guess a linear solution.
Formal Solution:
*Absorption probability.* Let $\rho_i$ be the probability of ever reaching 0 from position $i$. From position $i \geq 1$:
$\rho_i = p\,\rho_{i+1} + q\,\rho_{i-1}, \quad \rho_0 = 1$
The general solution is $\rho_i = A + B(q/p)^i$. With $\rho_0 = 1$ and requiring $\rho_i \leq 1$:
- If $q \geq p$: $\rho_i = 1$ for all $i$ (absorption is certain).
- If $q < p$: $\rho_i = (q/p)^i$ (absorption probability less than 1).
From position 1: absorption is certain when $p \leq q$ (i.e., $p \leq 1/2$).
*Expected absorption time.* Let $E_i$ be the expected time to reach 0 from position $i$. For $i \geq 1$:
$E_i = 1 + p\,E_{i+1} + q\,E_{i-1}, \quad E_0 = 0$
Guess $E_i = ci$ for some constant $c$. Substituting:
$ci = 1 + p\,c(i+1) + q\,c(i-1) = 1 + ci(p+q) + c(p - q) = 1 + ci + c(p - q)$
This simplifies to $0 = 1 + c(p - q)$, giving $c = 1/(q - p)$. So:
$E_i = \frac{i}{q - p}$
This is valid when $q > p$ (i.e., $p < 1/2$), ensuring $E_i > 0$ and finite.
From position 1:
$E_1 = \frac{1}{q - p}$
*Verification.* Check: $E_1 = 1 + p E_2 + q E_0 = 1 + p \cdot \frac{2}{q-p} + 0 = 1 + \frac{2p}{q-p} = \frac{q-p+2p}{q-p} = \frac{q+p}{q-p} = \frac{1}{q-p}$. Checks out.
*Case $p = q = 1/2$ (symmetric).* Absorption is certain ($\rho_1 = 1$), but $E_1 = 1/(q - p) \to \infty$. The walk is null recurrent: it returns to 0 with probability 1, but the expected return time is infinite.
*Case $p > q$ (rightward drift).* Absorption occurs with probability $q/p < 1$. The unconditional expected absorption time is infinite ($E_1 = \infty$). The conditional expected time given absorption is
Answer: Absorption is certain if and only if $p \leq 1/2$. When $p < 1/2$, the expected absorption time from position 1 is $E_1 = 1/(q - p) = 1/(1 - 2p)$, which is finite. When $p = 1/2$, absorption is certain but $E_1 = \infty$. When $p > 1/2$, absorption occurs with probability $q/p < 1$.
Intuition
This problem illustrates the three regimes of a random walk with an absorbing barrier: transient ($p > 1/2$, particle escapes to infinity with positive probability), null recurrent ($p = 1/2$, particle returns surely but in infinite expected time), and positive recurrent ($p < 1/2$, particle returns in finite expected time). The formula $E_1 = 1/(q - p)$ has a clean interpretation: the expected absorption time equals the reciprocal of the net drift speed toward the barrier.
In quantitative finance, this framework models the time until a portfolio or trading strategy hits a stop-loss. If your strategy has a slight negative drift (losing money on average), the time to ruin is