Lesson 12

One might suggest that since the expectation of the sum of random variables equals the sum of their expectations, the same holds for the product, but this is not generally true. However, it is valid for independent random variables.

Theorem. For independent random variables \(\alpha\) and \(\beta\) such that \(\operatorname{E}[|\alpha|]\) and \(\operatorname{E}[|\beta|]\) are finite, we have \[\operatorname{E}[\alpha\beta]=\operatorname{E}[\alpha]\operatorname{E}[\beta] \ .\]

Proof. \[\begin{align*}\operatorname{E}[\alpha]\operatorname{E}[\beta]&=\left(\sum_{a}a\Pr[\alpha=a]\right) \cdot \left(\sum_{b}b\Pr[\beta=b]\right) =\\&=\sum_{a,b}ab\Pr[\alpha=a]\Pr[\beta=b]=&\text{(independence)}\\&= \sum_{a,b}ab\Pr[\alpha=a, \beta=b]=\\&=\sum_{c}\sum_{a,b \colon ab=c}c\Pr[\alpha=a,\beta=b]=\\&=\sum_{c}\left(c\sum_{a, b \colon ab=c}\Pr[\alpha=a, \beta=b]\right) =\\&=\sum_{c} c \Pr[\alpha\beta=c]=\\&=\operatorname{E}[\alpha\beta] \ .\end{align*}\]◼