EZ)=∑E(Z|A)P( Using the total expectation theorem, we obtain a+e Ea G(a+e=G(a) a+ (a+E)2 2(a+e)2+o(e2) G(a) o(e2) a+ a+E where o(e )is a collection of terms of order e or higher. If E is small, we can ignore o(e ) Hence we have G(a+)≈G(a From the formula of the sum of an infinite geometric series, we know no ing higher order terms of E, we get E This gives the following approximation 2E ≈E(1--)=E--≈E Therefore, we can rewrite G(a+a)as G(a+e)≈G(a Rearranging terms, we have G(a+a-G(a) 2G(a) G(a+e)-G(a)E(Z) =
n i=1 E(Z | Ai)P(Ai) . Using the total expectation theorem, we obtain G(a + ε) = G(a) a a + ε 2 + a + ε 2 εa (a + ε)2 + a + ε 2 εa (a + ε)2 + o(ε2) = G(a) a a + ε 2 + εa a + ε + o(ε2) , where o(ε2) is a collection of terms of order ε2 or higher. If ε is small, we can ignore o(ε2). Hence we have G(a + ε) ≈ G(a) a a + ε 2 + εa a + ε . From the formula of the sum of an infinite geometric series, we know a a + ε = 1 1 + ε a = 1 − ε a + ε a 2 − ε a 3 + ··· . Ignoring higher order terms of ε, we get a a + ε ≈ 1 − ε a . This gives the following approximations: a a + ε 2 ≈  1 − ε a 2 = 1 − 2ε a + ε2 a2 ≈ 1 − 2ε a , εa a + ε ≈ ε  1 − ε a  = ε − ε2 a ≈ ε . Therefore, we can rewrite G(a + ε) as G(a + ε) ≈ G(a) 1 − 2ε a + ε . Rearranging terms, we have G(a + ε) − G(a) = ε −2G(a) a + 1 . ⇒ G(a + ε) − G(a) ε = −2G(a) a + 1 . 2