8 the general formula for the mean of a function g x

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Unformatted text preview: re six choices for the first number of an outcome in {Y = 3}, then five choices for the second, and then four choices for the third. So |{Y = 3}| = 6 · 5 · 4 = 120. To double check our work, we note that 6+90+120=216, as 6 1 90 120 expected. So, pY (1) = 216 = 36 , pY (2) = 216 = 15 , and pY (3) = 216 = 20 . The mean of Y is 36 36 NOT simply 1+2+3 because the three possible values are not equally likely. The correct value is 3 E [Y ] = 1+2·15+3·20 = 91 ≈ 2.527. 36 36 24 CHAPTER 2. DISCRETE-TYPE RANDOM VARIABLES Example 2.2.4 Suppose X is a random variable taking values in {−2, −1, 0, 1, 2, 3, 4, 5}, each with 1 probability 8 . Let Y = X 2 . Find E [Y ]. Solution. The pmf of X is given by 1 8 −2 ≤ u ≤ 5 0 else. pX (u) = The definition of E [Y ] involves the pmf of Y, so let us find the pmf of Y. We first think about the possible values of Y (which are 0, 1, 4, 9, 16, and 25), and then for each possible value u, find pY (u) = P {Y = u}. For example, for u = 1, pY (1) = P {Y = 1} = P {X = −1 or X = 1} = 2/8. The complete lis...
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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 Institute of Technology.

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