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In Self-Test Problem 7.1, we showed how to use the value of a uniform (0,1) random variable (commonly called a random number) to obtain the value of a random variable whose mean is equal to the expected number of distinct names on a list. However, its use required that one choose a random position and then determine the number of times that the name in that position appears on the list. Another approach, which can be more ef铿乧ient when there is a large amount of replication of names, is as follows: As before, start by choosing the random variable X as in Problem 7.1. Now identify the name in position X, and then go through the list, starting at the beginning, until that name appears. Let I equal 0if you encounter that name before getting to position X, and let l equal1ifyour 铿乺st encounter with the name is at position X. Show that E[mI]=d.

Hint: Compute E[I] by using conditional expectation.

Short Answer

Expert verified

The position is equality likely Therefore, to be chosen as,

E[mI]=E[mn(X)]=d

Step by step solution

01

Given Information

Show thatE[ml]=d.

02

Explanation

Name chosen as n(x),

position is equality likely Therefore, to be chosen as

E[In(X)]=P{I=1n(X)}=1N(X)

E[I]=E[In(X)]

Therefore,

E[mI]=E[mn(X)]=d

03

Final Answer

E[mI]=E[mn(X)]=d

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