Isye 2027

# What are the possible values of y yes y takes values

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Unformatted text preview: t die and a 5 shows on the second, then Y = 5. That is, Y ((3, 5)) = 5. In general, Y ((i, j )) = max{i, j } for (i, j ) ∈ Ω. Determine the pmf of Y. Solution: The possible values of Y are 1, 2, 3, 4, 5, and 6. There is only one outcome in Ω such that Y = 1. Speciﬁcally, {Y = 1} = {(1, 1)}. Similarly, {Y = 2} = {(2, 2), (1, 2), (2, 1)}, {Y = 3} = {(3, 3), (1, 3), (2, 3), (3, 1), (3, 2)}, and so forth. The pmf of Y is shown in Fig. 2.2. p 11/36 9/36 7/36 5/36 3/36 1/36 Y 12 3 4 5 6 Figure 2.2: The pmf for the maximum of numbers showing for rolls of two fair dice. 2.2. THE MEAN AND VARIANCE OF A RANDOM VARIABLE 2.2 23 The mean and variance of a random variable The mean of a random variable is a weighted average of the possible values of the random variable, such that the weights are given by the pmf: Deﬁnition 2.2.1 The mean (also called expectation) of a random variable X with pmf pX is denoted by E [X ] and is deﬁned by E [X ] = i ui pX (ui ), where u1 , u2 , . . . is the list of possible values of X. Example 2....
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## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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