Reading E[X + Y] = E[X] + E[Y]
| E[ · ] | "The expected value of" whatever is inside the brackets. |
| X, Y | Two random quantities, such as die 1 and die 2. |
| X + Y | A new random quantity: their total on each roll. |
Why it always holds. An average is "add everything, then divide", and the order you add things in doesn't matter. Over 600 rolls of two dice, the grand total of all the sums equals the total of die 1's column plus the total of die 2's column. Dependence changes which values appear together on the same row, but not what each column adds up to.
Reading Var(X + Y) = Var(X) + Var(Y) + 2·Cov(X, Y)
| Cov(X, Y) | Covariance: do X and Y tend to be high together (positive), move in opposite directions (negative), or ignore each other (zero)? It is the average of (X − μₓ)(Y − μᵧ). |
| independent | Knowing one tells you nothing about the other. Then Cov = 0 and the variances simply add. |
| Setup | Relationship | Variance | σ |
|---|
| Two separate dice, X + Y | Independent, Cov = 0 | 2.92 + 2.92 = 5.83 | 2.42 |
| One die counted twice, X + X | Perfectly together | 4 × 2.92 = 11.67 | 3.42 |
| X + (7 − X) | Perfectly opposite | 0 | 0 |
The last row is a hedge: two risky positions whose errors cancel exactly. All three setups have EV 7, so only the spread differs.
Where √n comes from
Add n independent copies, each with variance σ₁². The variances add to n·σ₁², so σ = √(n·σ₁²) = σ₁·√n. Ten dice: 1.71 × √10 ≈ 5.40, against an EV of 35. In finance, this is why diversification works: holding many assets that aren't perfectly correlated grows expected return faster than it grows risk.