Updated 12 September 2026

Random walk

A particle starts at position 0 and at each integer time step moves one unit right (with probability p) or one unit left (with probability 1 − p). Where does it end up after t steps?

The √t law

For an unbiased walk (p = 0.5), the expected position is always 0 — steps cancel on average. But the spread grows:

For the fair case, the dashed curves show ±√t around zero. They are pointwise one-SD references, not barriers that a path must stay within. At 100 steps the SD is 10; at 10,000 steps it is 100. Spread grows as the square root of time, not linearly.

Bias

With p ≠ 0.5, each step has a net drift of 2p − 1 per step:

The amber center line shows expected position, with outer curves at ±√(4p(1−p)t) around it. A bias changes both the center and variance. At p=0.9 and t=100, the expected position is 80 and SD is 6. Plotting an unbiased ±√t envelope around zero would describe the wrong experiment.

For any fixed nonzero drift, its magnitude grows linearly in time while the centered SD grows as √t. This is an asymptotic comparison of scales, not a promise that every finite walk will follow the drift.

Diffusion and Brownian motion

The random walk is the discrete ancestor of Brownian motion. For the fair walk, under diffusive scaling that divides position by √n while accelerating time by n, the process converges to a continuous Gaussian process where position after time t is distributed as Normal(0, t). The same √t scaling survives in the limit — it is a fundamental feature of diffusion, not an artifact of the discrete model.

The Central Limit Theorem explains why: the position after t steps is a sum of t independent ±1 variables, and centering and scaling that finite-variance sum gives a normal limit. The exact finite-time position remains on a lattice.

Count right steps to find the endpoint law

Let H be the number of right steps. There are T−H left steps, so S_T=H−(T−H)=2H−T, with H binomial(T,p). At T=4 and p=1/2, the chance of ending at zero is the chance of two right steps: 6/16=0.375. The chance of ending at or below zero is (1+4+6)/16=0.6875.

Ending at position one after four steps is impossible: 2H−4 is even. The point-probability panel retains this parity constraint instead of filling every real-valued location with a normal approximation. The cutoff always concerns the original signed position, even in the centered display.

Displacement is different from distance traveled

The path right, right, left, left travels four units but ends where it started. Its signed displacement and distance from the origin are both zero. At a given time, averaging signed positions can also cancel positive and negative endpoints without saying that walkers traveled little.

Each step has mean 2p−1 and second moment one, hence variance 1−(2p−1)²=4p(1−p). Adding independent steps gives the drift and variance in the table. At p=0 or 1 there is no randomness: position is exactly −T or T and the variance is zero.

References

Random Services: the simple random walk derives the endpoint distribution and increment moments. Continue to gambler’s ruin for hitting probabilities or binomial counts to inspect the right-step count.

Continue to martingales to distinguish conditional fairness from constant expectation and check bounded stopping rules.