851_lectures17_24

# In this case we dene ex to be ex ex ex denition

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Unformatted text preview: is, and so that m ￿ k=1 ak P {Ak,n,￿ } → (1 − ￿) m ￿ k=1 ak P {Ak } = (1 − ￿)E(Y ) ≤ a. E (Yn,￿ ) ≤ E(Xn ) E (Yn,￿ ) ↑ (1 − ￿)E(Y ) and E(Xn ) ↑ a (1 − ￿)E(Y ) ≤ a (which follows since everything is increasing). Since 0 &lt; ￿ ≤ 1 is arbitrary, E(Y ) ≤ a which, as noted earlier in the proof, is suﬃcient to conclude that E(X ) ≤ a. Combined with our earlier result that a ≤ E(X ) we conclude E(X ) = a and the proof is complete. 18–3 Step 3: General Random Variables Now suppose that X is any random variable. Write X + = max{X, 0} and X − = − min{X, 0} for the positive part and the negative part of X , respectively. Note that X + ≥ 0 and X − ≥ 0 so that the positive part and negative part of X are both positive random variables and X = X + − X − . Deﬁnition. A random variable X is called integrable (or has ﬁnite expectation ) if both E(X + ) and E(X − ) are ﬁnite. In this case we deﬁne E(X ) to be E(X ) = E(X + ) − E(X − ). Deﬁn...
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## This document was uploaded on 04/07/2014.

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