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Proof of equation 51 when x and y are discrete we

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Proof of Equation (5.1) whenXandYAre Discrete:We must show thatE[X]=yE[X|Y=y]P{Y=y}(5.2)
334Chapter 7Properties of ExpectationNow, the right-hand side of Equation (5.2) can be written asyE[X|Y=y]P{Y=y} =yxxP{X=x|Y=y}P{Y=y}=yxxP{X=x,Y=y}P{Y=y}P{Y=y}=yxxP{X=x,Y=y}=xxyP{X=x,Y=y}=xxP{X=x}=E[X]and the result is proved.One way to understand Equation (5.2) is to interpret it as follows: To calculateE[X], we may take a weighted average of the conditional expected value ofXgiventhatY=y, each of the termsE[X|Y=y] being weighted by the probability ofthe event on which it is conditioned. (Of what does this remind you?) This is anextremely useful result that often enables us to compute expectations easily by firstconditioning on some appropriate random variable. The following examples illustrateits use.EXAMPLE 5cA miner is trapped in a mine containing 3 doors. The first door leads to a tunnel thatwill take him to safety after 3 hours of travel. The second door leads to a tunnel thatwill return him to the mine after 5 hours of travel. The third door leads to a tunnelthat will return him to the mine after 7 hours. If we assume that the miner is at alltimes equally likely to choose any one of the doors, what is the expected length oftime until he reaches safety?
Section 7.5Conditional Expectation335E[X|Y=2]=5+E[X]. The argument behind the other equalities in Equation (5.3)is similar. Hence,E[X]=13(3+5+E[X]+7+E[X])orE[X]=15.EXAMPLE 5d Expectation of a sum of a random number of random variablesSuppose that the number of people entering a department store on a given day isa random variable with mean 50. Suppose further that the amounts of money spentby these customers are independent random variables having a common mean of $8.Finally, suppose also that the amount of money spent by a customer is also inde-pendent of the total number of customers who enter the store. What is the expectedamount of money spent in the store on a given day?

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