Recipe

Find the mean, variance, and SD of a sum of independent random variables

Use when: you're asked for E(Y), \operatorname{Var}(Y), or the standard deviation of a sum Y=X_1+X_2+\cdots of random variables

Goal: the sum's own mean, variance, and, if needed, standard deviation, built directly from each piece's own mean and variance

Taught in Sums of independent random variables

1. 1

   Check whether the problem states, or otherwise justifies, that the variables being summed are independent of each other. The mean rule in Step 2 holds either way, but the variance rule in Step 3 only holds when independence is genuine.

   • the variables are independent — continue to Step 3 and add the variances directly.
   • the variables are not independent (or independence isn't given) — Step 3's rule doesn't apply as stated; an extra covariance term (chapter 9's covariance section) has to be included instead of a plain sum.

   Why: independence is exactly what makes the cross terms in Var(X1+X2+...) vanish, so it has to be checked before trusting Step 3.

2. 2

   Regardless of independence, E(Y) = E(X_1) + E(X_2) + \cdots; add each piece's own mean.

   Example

   X_1, X_2, X_3 have means 50, 30, and 20. E(Y) = 50+30+20 = 100.

   Why: linearity of expectation holds unconditionally, so this step never needs the Step 1 independence check at all.

3. 3

   With independence confirmed in Step 1, \operatorname{Var}(Y) = \operatorname{Var}(X_1) + \operatorname{Var}(X_2) + \cdots; add each piece's own variance, never its standard deviation.

   Example

   The same X_1, X_2, X_3 have variances 16, 9, and 4. \operatorname{Var}(Y) = 16+9+4 = 29.

   Why: independence is precisely the condition that cancels the covariance terms that would otherwise appear when expanding \operatorname{Var}(X_1+X_2+\cdots) term by term.

4. 4

   If the question asks for the sum's own standard deviation, take the square root of the SUMMED VARIANCE from Step 3. Never add the individual standard deviations first.

   Example

   \operatorname{SD}(Y) = \sqrt{29} \approx 5.385, not \sqrt{16}+\sqrt{9}+\sqrt{4} = 4+3+2 = 9, which is wrong by a wide margin.

   Why: only variance is additive under independence; standard deviation is a square root, and a square root of a sum is never the sum of the square roots.