Lesson 8

When multiplying by a constant, the variance is multiplied by the square of that constant: \[\operatorname{Var}[c\alpha]=c^{2}\operatorname{Var}[\alpha]\] Indeed, \(\operatorname{E}[(c\alpha)^{2}]=c^{2}\operatorname{E}[\alpha^{2}]\) and \(\operatorname{E}[(c\alpha)]^{2}=(c\operatorname{E}[\alpha])^{2}=c^{2}\operatorname{E}[\alpha]^{2}\).

In the general case, variance is not linear, i.e., \(\operatorname{Var}[\alpha+\beta] \neq \operatorname{Var}[\alpha]+\operatorname{Var}[\beta]\). It is easy to see this by taking \(\alpha=\beta\): then, \[\operatorname{Var}[\alpha+\alpha]=\operatorname{Var}[2\alpha]=4\operatorname{Var}[\alpha].\] However, for independent random variables, the variance of their sum equals the sum of their variances.

Theorem. For independent random variables \(\alpha\) and \(\beta\), the following holds: \[\operatorname{Var}[\alpha+\beta]=\operatorname{Var}[\alpha]+\operatorname{Var}[\beta] \ .\]

Proof. \[\begin{align*}\operatorname{Var}[\alpha+\beta]&=\operatorname{E}[(\alpha+\beta)^{2}]-\operatorname{E}[\alpha+\beta]^{2}=\\&=(\operatorname{E}[\alpha^{2}]+\operatorname{E}[\beta^{2}]+2\operatorname{E}[\alpha\beta])- (\operatorname{E}[\alpha]^{2}+\operatorname{E}[\beta]^{2}+2\operatorname{E}[\alpha]\operatorname{E}[\beta])=\\&=\operatorname{Var}[\alpha]+\operatorname{Var}[\beta] \ .\end{align*}\]◼