Unformatted text preview: ≥ 0, we know that E(X + ) ≥ 0 and E(X − ) ≥ 0. We now use the fact that if
the sum of two nonnegative numbers is ﬁnite, then each number must be ﬁnite to conclude
that E(X + ) < ∞ and E(X − ) < ∞. Thus, by deﬁnition, E(X ) = E(X + ) − E(X − ) < ∞ so
that X ∈ L1 .
Corollary. If X ∈ L1 , then E(X ) ≤ E(X ). In particular, if E(X ) = 0, then E(X ) = 0.
Proof. Since E(X + ) ≥ 0 and E(X − ) ≥ 0, we have from the triangle inequality
E(X ) = E(X + −X − ) = E(X + )−E(X − ) ≤ E(X + )+E(X − ) = E(X + )+E(X − ) = E(X ).
Since E(X ) ≥ 0, if E(X ) = 0, then 0 ≤ E(X ) ≤ E(X ) = 0 implying E(X ) = 0.
Theorem 20.3. If X, Y : (Ω, F , P) → (R, B ) are integrable random variables with X = Y
almost surely, then E(X ) = E(Y ).
Proof. Suppose that X = Y almost surely so that P {ω ∈ Ω : X (ω ) = Y (ω )} = 1. To begin,
assume that X ≥ 0 and Y ≥ 0 and let A = {ω : X (ω ) = Y (ω )} so that P {A} = 0. Write
E ( Y...
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 Fall '14
 Statistics, Probability, Probability theory, Xn, Prof. Michael Kozdron

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