Drift-Corrected Biased Random Walks

Why Remove the Drift?

A biased random walk has a systematic tendency to move in one direction. For example, if each step is more likely to be \(+1\) than \(-1\), the walk has positive mean displacement per step. Subtracting the expected displacement accumulated so far removes this systematic trend. The result is not generally constant along a path; rather, its next increment has conditional mean zero given the information already observed. That is the feature that makes it a martingale.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((\xi_n)_{n\ge1}\) be independent random variables taking values \(+1\) and \(-1\), with \(\mathbb{P}(\xi_n=+1)=p\) and \(\mathbb{P}(\xi_n=-1)=q\), where \(p,q\ge0\) and \(p+q=1\). The walk starts at a fixed \(x_0\in\mathbb{R}\) and is \(X_0=x_0\), \(X_n=x_0+\sum_{k=1}^n\xi_k\) for \(n\ge1\). Its information filtration is \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(\xi_1,\ldots,\xi_n)\). The drift per step is \(\mathbb{E}[\xi_n]=p-q\).

Here the filtration records the steps already taken, but not the next step. Independence is what ensures that learning the past does not change the conditional probabilities of the next step. The theorem below uses that fact directly. It includes the extreme cases \(p=1,q=0\) and \(p=0,q=1\): in those cases the walk is deterministic and the corrected process is constant.

Proof. Each \(\xi_k\) is bounded, so \(X_n\) and \(M_n\), being finite sums of bounded variables and constants, are integrable. The variable \(M_n\) is a function of \(\xi_1,\ldots,\xi_n\), so it is \(\mathcal{F}_n\)-measurable. Also, \(M_{n+1}-M_n=\xi_{n+1}-(p-q)\). By independence, \(\xi_{n+1}\) is independent of \(\mathcal{F}_n\), and its probabilities of \(+1\) and \(-1\) are \(p\) and \(q\). Therefore, \(\mathbb{E}[\xi_{n+1}\mid\mathcal{F}_n]=p(1)+q(-1)=p-q\) almost surely. Taking conditional expectations in the increment identity gives \(\mathbb{E}[M_{n+1}-M_n\mid\mathcal{F}_n]=0\), hence \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely. These are all the martingale conditions. \(\square\)

The proof depends on the conditional mean of the next step, not on any claim that a realized corrected path is flat. The path can go up or down substantially. The martingale property says that, given the observed past, the average of the next corrected value over the possible next steps is the current value.

Biased-Coin Examples

These computations illustrate two different checks. Substitution into \(M_n=X_n-n(p-q)\) checks the arithmetic on the selected path. Computing the conditional mean of the next corrected increment checks the probabilistic reason the process is a martingale. A single path can verify the first, but not the second: it does not show what would have happened under the other possible next toss.

Which Linear Correction Works?

The drift correction is not an arbitrary choice of slope. For a walk with constant mean step \(p-q\), it is the unique constant linear correction that makes the result a martingale.

Proof. The increments satisfy \(Y_{n+1}-Y_n=\xi_{n+1}-c\). If \(Y_n\) is a martingale, its conditional increments have conditional mean zero, so \(0=\mathbb{E}[Y_{n+1}-Y_n\mid\mathcal{F}_n]=\mathbb{E}[\xi_{n+1}\mid\mathcal{F}_n]-c=p-q-c\) almost surely. Thus \(c=p-q\). Conversely, if \(c=p-q\), the theorem's conditional-increment calculation gives \(\mathbb{E}[Y_{n+1}-Y_n\mid\mathcal{F}_n]=0\) for each \(n\). The variables are integrable and adapted, since they differ from \(X_n\) by deterministic constants. Therefore \(Y_n\) is a martingale. This proves both directions. \(\square\)

Independent Steps with Different Drifts

The same conditional-mean reasoning applies when the step probabilities change with time. Suppose the \(\xi_n\) remain independent and take values \(+1\) and \(-1\), but \(\mathbb{P}(\xi_n=+1)=p_n\) and \(\mathbb{P}(\xi_n=-1)=q_n=1-p_n\). Define the deterministic mean \(d_n=p_n-q_n\) of step \(n\). Then the accumulated drift through time \(n\) is \(\sum_{k=1}^n d_k\), rather than \(n\) times a single constant.

Proof. The process is adapted and integrable because each \(X_n\) is a finite sum of bounded steps and each \(d_k\) is deterministic. Its increment is \(M_{n+1}-M_n=\xi_{n+1}-d_{n+1}\). Independence of the steps implies that \(\xi_{n+1}\) is independent of \(\mathcal{F}_n\). Therefore, \(\mathbb{E}[\xi_{n+1}\mid\mathcal{F}_n]=p_{n+1}-q_{n+1}=d_{n+1}\) almost surely. It follows that the conditional expectation of \(M_{n+1}-M_n\) given \(\mathcal{F}_n\) is zero. Hence \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely for every \(n\), proving the proposition. \(\square\)

Interpretation and Common Pitfalls

Drift correction isolates the unpredictable part of the walk's change. At each time, the raw increment has a conditional mean determined by the step distribution; subtracting that mean makes the corrected increment conditionally mean zero. With constant step probabilities, that accumulated subtraction is \(n(p-q)\). With changing probabilities, it is the sum of the individual step means. The path itself need not have zero displacement, and its corrected value need not equal zero.

Common mistake. Do not replace the signed mean \(p-q\) by \(p\), or test the martingale condition using only the values along one observed path. For a martingale, the relevant test is \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\), not \(M_{n+1}=M_n\) on every outcome. The wrong-drift example fails because its first conditional expectation is \(-\tfrac14\), while its current value is \(0\). Conversely, an individual corrected increment may be nonzero—as in the coin-toss paths—while its conditional mean remains zero.