Lesson 5

The expectation \(\operatorname{E}[\alpha]\) can also be found in another way. We partition the probability space \(U\) into two events: \(A\) and \(\overline{A}= U \setminus A\), where \(A\) represents “the first trial was a success.” By the law of total expectation, we know that \[\operatorname{E}[\alpha]=\operatorname{E}[\alpha \mid A] \Pr[A]+\operatorname{E}[\alpha \mid \overline A] \Pr[\overline A] \ .\] We know that \(\Pr[A]=p\), \(\Pr[\overline A]=1-p\), and that \(\operatorname{E}[\alpha \mid A]=1\). A simple but key observation about the remaining term: \[\operatorname{E}[\alpha \mid \overline A] =1+\operatorname{E}[\alpha] \ .\] Indeed, since success did not happen on the first step, we can simply ignore this step—what remains is exactly the same experiment. Substituting everything into the original equation, we obtain \[\operatorname{E}[\alpha] = \frac{1}{p} \ .\]