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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online