Var[A+B+C] = Var [A] + Var [B] + Var [C]

Question

I know that this relationship is correct when A,B and C are pairwise independent. However, I'm finding it difficult to calculate it. Let's say that: Var [A] = Var [B] = Var [C] = 0.25. How can I prove that Var[A+B+C] = 0.75?

I started to work with the formula: Var[A+B+C] =E[(A+B+C)^2]−(E[A+B+C])^2 but I got stuck.

Thanks!

Answer 1

Expand your formula, using the fact that for pairwise independent random variables $X$ and $Y$, we have $E(XY) = E(X)E(Y)$: \begin{align*} \text{Var}(A + B + C) & = E((A + B + C)^2) - (E(A + B + C))^2\\ & = E(A^2) + E(B^2) + E(C^2) + 2E(AB) + 2E(AC) + 2E(BC)\\ & \quad - E(A)^2 - E(B)^2 -E(C)^2 - 2E(A)E(B) + 2E(A)E(C) + 2E(B)E(C)\\ & = E(A^2) + E(B^2) + E(C^2) + 2E(A)E(B) + 2E(A)E(C) + 2E(B)E(C)\\ & \quad - E(A)^2 - E(B)^2 -E(C)^2 - 2E(A)E(B) + 2E(A)E(C) + 2E(B)E(C)\\ & = (E(A^2) - E(A)^2) + (E(B^2) - E(B)^2) + (E(C^2) - E(C)^2)\\ & = \text{Var}(A) + \text{Var}(B) + \text{Var}(C). \end{align*}

Answer 2

First notice (using the fact that $A$, $B$ and $C$ are pairwise independent) that $C$ is independent to $A+B$ :

\begin{align*} E((A+B)C) &= E(AC+BC) \\ &= E(AC)+E(BC) \\ &= E(A)E(C)+E(B)E(C) \\ &= (E(A)+E(B))E(C)\\ &= E(A+B)E(C). \end{align*}

Then one has

\begin{align*} \operatorname{Var}(A+B+C) &= \operatorname{Var}(A+B)+\operatorname{Var}(C) \\ &=\operatorname{Var}(A)+\operatorname{Var}(B)+\operatorname{Var}(C). \end{align*}