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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. DISCRETETYPE 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.
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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.
 Spring '08
 Zahrn
 The Land

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