Lesson 3

Mathematical expectation is a number (if it is finite), which in some sense predicts a random variable. What number \(e \in \mathbb{R}\) could be taken for this? Clearly, we need to give some point around which the value clusters. Depending on what “measure of clustering” (measure of dispersion) we take, different results will be obtained. The standard such measure is mean square deviation: \[\operatorname{E}\left[(\alpha - e)^{2}\right] \ .\] It is minimized precisely at \(e=\operatorname{E}[\alpha]\): \[\begin{align*}\operatorname{E}\left[(\alpha - e)^{2}\right]&=\operatorname{E}\left[(\alpha - \operatorname{E}[\alpha]+\operatorname{E}[\alpha]-e)^{2}\right]=&\text{(linearity)}\\&=\operatorname{E}\left[(\alpha-\operatorname{E}[\alpha])^{2}\right]+\operatorname{E}\left[(\operatorname{E}[\alpha]-e)^{2}\right] + \operatorname{E}\left[2(\alpha-\operatorname{E}[\alpha])(\operatorname{E}[\alpha]-e)\right]=\\&=\operatorname{E}\left[(\alpha-\operatorname{E}[\alpha])^{2}\right]+(\operatorname{E}[\alpha]-e)^{2} \ge\\&\ge\operatorname{E}\left[(\alpha-\operatorname{E}[\alpha])^{2}\right] \ .\end{align*}\] In the last equality, we used the fact that \(e\) and \(\operatorname{E}[\alpha]\) are constants and that \(\operatorname{E}[\alpha-\operatorname{E}[\alpha]]=0\).