Random walks: why you end up about √n away

Flip a coin. Heads, step right; tails, step left. After n flips you are typically about √n steps from where you started — not n, and not zero.

Why it matters

This is the rule behind diffusion. Perfume crossing a room, heat spreading through a bar, a stock price drifting, a molecule jostled by its neighbours: each is a sum of many small random pushes, and each spreads as the square root of time. Doubling the time does not double the spread. To spread twice as far takes four times as long. Every estimate of how long mixing takes rests on this one fact.

The mechanism

Call each step s_k, equal to +1 or −1 with equal chance. Your position after n steps is the sum X = s₁ + s₂ + … + s_n.

The average of X is zero: every step is as likely to go left as right, so the pluses and minuses cancel on average. That tells you nothing about how far you end up, because "average position zero" is consistent with being far away on either side.

So look at X² instead, which is always positive. Multiply the sum by itself. You get n terms of the form s_k·s_k, each equal to 1 because (+1)² = (−1)² = 1. And you get cross terms s_j·s_k with j ≠ k. Each cross term is +1 or −1 with equal chance, because the two steps are independent, so on average each is zero.

The average of X² is therefore exactly n. The typical size of X is the square root of that: √n. Each step adds a fixed amount to the squared distance, so distance itself grows like a square root.

The distance is typical, not guaranteed. Some walks end at zero and some end far out, but the spread of endpoints has width √n, and as n grows the shape of that spread settles into a bell curve.

In one breath

Random steps cancel on average but not in square. Squaring the sum gives n guaranteed ones plus cross terms that average out, so the mean squared distance is n and the typical distance is √n. That is why anything driven by many small random pushes spreads as the square root of time.