Stochastic

# 6 exponential martingale let y1 y2 be independent

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Unformatted text preview: particular when B = ⌦ we have EY = Pk j =1 P (Aj ) P (B ) E (Y |Aj ) · P (Aj ). 161 5.2. EXAMPLES, BASIC PROPERTIES Proof. Using the deﬁnition of conditional expectation, Lemma 5.3, then doing some arithmetic and using the deﬁnition again, we have E (Y |B ) = E (Y ; B )/P (B ) = = k X E (Y ; Aj )/P (B ) j =1 k k X E (Y ; Aj ) P (Aj ) X P (Aj ) · = E (Y |Aj ) · P (Aj ) P (B ) P (B ) j =1 j =1 which proves the desired result. In the discussion in this section we have concentrated on the properties of conditional expectation given a single set A. To connect with more advanced treatments, we note that given a partition A = {A1 , . . . An } of the sample space, (i.e., disjoint sets whose union in ⌦) then the conditional expectation given the partition is a random variable: E (X |A) = E (X |Ai ) on Ai In this setting, Lemma 5.4 says E [E (X |A)] = EX i.e., the random variable E (X |A) has the same expected value as X . Lemma 5.1 says that if X is constant on each part of the partition then E...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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