Annus MirabilisEinstein in 1905
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Foundation lesson

Probability and independence

Squaring a sum of two steps gives each step squared plus twice their product. If the steps are independent and each averages zero, the product averages zero too, and only the squares are left to add up.

Written for this edition, not translated from Einstein.

When you square a sum of random steps, why do the mixed terms average to zero?

Flip a coin twice and take one step for each flip: a metre to the right for heads, a metre to the left for tails. How far from the start do you end up, on average?

  1. The four outcomes, right-right, right-left, left-right and left-left, are equally likely: one chance in four each.
  2. They leave you 2 m to the right, back at the start, back at the start, or 2 m to the left.
  3. Square those distances so that left and right count alike: 4, 0, 0 and 4. Their average is 2, one square metre for each step.

Combining the two steps added nothing beyond one square metre each. In symbols, with Δ₁ and Δ₂ for the two steps, squaring the sum gives each step squared and a mixed term, twice their product:

(Δ1+Δ2)2=Δ12+2 Δ1 Δ2+Δ22\left(\Delta_{1} + \Delta_{2}\right)^{2} = \Delta_{1}^{2} + 2\,\Delta_{1}\,\Delta_{2} + \Delta_{2}^{2}

Delta one plus delta two, squared, equals delta one squared plus two delta one delta two plus delta two squared.

In the coin walk the product of the two steps is +1 for right-right and left-left and −1 for the other two, so on average it is zero and the mixed term adds nothing. Two conditions make that happen. Independence: learning the first step does not change the chances for the second, so the chance of a pair is the product of the two chances. Centring: each step averages zero, left as likely as right. Then the average of the product is the product of the averages, zero times zero:

⟨Δ1Δ2⟩=⟨Δ1⟩⟨Δ2⟩=0\langle \Delta_1 \Delta_2 \rangle = \langle \Delta_1 \rangle \langle \Delta_2 \rangle = 0

The average of delta one times delta two equals the average of delta one times the average of delta two, which is zero.

Both conditions matter. If each step drifts, averaging +1, the product of the averages is 1, not 0. If the second step tends to copy the first, the product is positive more often than negative. The Brownian paper's §4 assumes that a particle's motions in successive intervals are independent, as long as the intervals are not chosen too small.

Worked exampleWhat changes when the second step copies the first

  1. Now let the second flip always copy the first. Only right-right and left-left can happen, one chance in two each.
  2. The product of the two steps is +1 every time, so the mixed term no longer averages zero.
  3. You always end 2 m from the start, so the squared distance is 4 every time: the average is 4, not 2. Independence removed that extra 2.

Where this lesson stops

Before dropping a mixed term, ask two things: does learning one step change the chances for the next, and does each step average zero?

If you came here from a passage, Back returns you to the exact place you left.

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