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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 < ≤ 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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