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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 04/07/2014.

Ask a homework question - tutors are online