**Unformatted text preview: **) = E( Y 1A + Y 1Ac ) = E( Y 1A ) + E( Y 1Ac ) = E( Y 1A ) + E( X 1Ac ) . ( ∗) We know that there exist sequences Xn and Yn of non-negative simple random variables such
that (i) Xn ↑ X and E(Xn ) ↑ E(X ), and (ii) Yn ↑ Y and E(Yn ) ↑ E(Y ). Thus, Xn 1A ↑ X 1A
and E(Xn 1A ) ↑ E(X 1A ) and similarly Yn 1A ↑ Y 1A and E(Yn 1A ) ↑ E(Y 1A ). For each n, the
random variable Xn takes on ﬁnitely many values and is therefore bounded by K , say, where
K may depend on n. Thus, since Xn ≤ K , we obtain Xn 1A ≤ K and so
0 ≤ E(Xn 1A ) ≤ E(K 1A ) = K P {A} = 0.
This implies that E(Xn 1A ) = 0 and so by uniqueness of limits, E(X 1A ) = 0. Similarly,
E(Y 1A ) = 0. Therefore, by (∗), we obtain
E( Y ) = E( Y 1A ) + E( X 1Ac ) = 0 + E( X 1Ac ) = E( X 1A ) + E( X 1Ac ) = E( X ) .
In general, note that X = Y almost surely implies that X + = Y + almost surely and
X − = Y − almost surely. 20–2 Statistics 851 (Fall 2013)
Prof. Michael Kozdron October 25, 2013 Lecture #21: Proofs of the Main Expectation Theorems
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We will continue pro...

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