Unformatted text preview: j E [(Mn
X Mk )Mj Av ] E [(Mn Mk )Av ] = 0 x proves the desired formula.
The last result has the following useful corollary
E (Mn M0 )2 = n
X E (Mk Mk 2
1) (5.6) k=1 Proof. Expanding out the square of the sum
E n
X k=1 Mk Mk 1 !
2 = n
X k=1 +2 E (Mk
X Mk
E [(Mk 1j<kn and the second sum vanishes by Lemma 5.8. 2
1) Mk 1 )(Mj Mj 1 )] 165 5.3. GAMBLING STRATEGIES, STOPPING TIMES 5.3 Gambling Strategies, Stopping Times The ﬁrst result should be intuitive if we think of supermartingale as betting on
an unfavorable game: the expected value of our fortune will decline over time.
Theorem 5.9. If Mm is a supermartingale and m n then EMm E Mn . Proof. It is enough to show that the expected value decreases with each time
step, i.e., EMk E Mk+1 . To do this, we will again use the notation from (5.5)
Av = {Xn = xn , Xn 1 = xn 1 , . . . , X0 = x0 , M0 = m} and note that linearity in the conditioning set (Lemma 5.3) and the deﬁnition
of conditional expectation imply
X
E (Mk+1 Mk ) =
E (Mk+1 Mk ; Av )
v = X P (Av )E (Mk+1 Mk Av ) 0 v since supermartingales have E (Mk+1 Mk ...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

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