Recitation 22 The only quantity here that we don' t already know is Ex(RIE), which is the expected die roll, given that the result is odd. Solving this equation for this unknown, we conclude that Ex(rie=3 To prove the Total Expectation Theorem we begin with a lemma Lemma. Let R be a random variable, e be an event with positive probability, and Ie be the indicator variable for e. then Ex(R·IE) Pr(e Proof. Note that for any outcome, s, in the sample space, Pr({s}∩E)= ∫oi(s)=0 Pr(s)if Ie(s)=1, d Pr(s)nE)=Ie(s Pr(s) Ex(R|E)=∑R(s)P({s}E (ef of Ex(E)) ∑R(s) Pr({s}∩E ( Def of pr(·|E) R(s IE(s)·Pr(s) Pr(e) (by(2) ∑ses(F(s)·Is(s)·Pr(s) Pr(e) Ex(R·IE ( Def of ex(R·IE) Now we prove the Total Expectation Theorem: Proof. Since the Eis partition the sample space, R R·I� � � � � � Recitation 22 2 The only quantity here that we don’t already know is Ex R | E , which is the expected die roll, given that the result is odd. Solving this equation for this unknown, we conclude that Ex R | E = 3. To prove the Total Expectation Theorem, we begin with a Lemma. Lemma. Let R be a random variable, E be an event with positive probability, and IE be the indicator variable for E. Then Ex (R IE) Ex (R | E) = · (1) Pr (E) Proof. Note that for any outcome, s, in the sample space, 0 if IE(s) = 0, Pr ({s} ∩ E) = Pr (s) if IE(s) = 1, and so Pr ({s} ∩ E) = IE(s) · Pr (s). (2) Now, � Ex (R | E) = s∈S R(s) · Pr ({s} | E) = � s∈S R(s) · Pr ({s} ∩ E) Pr (E) = � s∈S R(s) · IE(s) · Pr (s) Pr (E) � (Def of Ex (· | E)) (Def of Pr (· | E)) (by (2)) = s∈S(R(s) · IE(s)) · Pr (s) Pr (E) = Ex (R · IE) Pr (E) (Def of Ex (R · IE)) Now we prove the Total Expectation Theorem: Proof. Since the Ei’s partition the sample space, R = R IEi · (3) i