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Unformatted text preview: EX 3 = ∞ X x =1 x 3 (1p ) x1 p = p + ∞ X x =2 x 3 (1p ) x1 p = p + (1p ) ∞ X x =2 (( x1) + 1) 3 (1p ) ( x1)1 p so that making the change of variable z = x1 leads to = p + (1p ) ∞ X z =1 ( z + 1) 3 (1p ) z1 p = p + (1p ) E µ ( X + 1) 3 ¶ = p + (1p ) ( 1 + 3 EX + 3 EX 2 + EX 3 ) same as before....
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This note was uploaded on 12/15/2011 for the course STAT 5326 taught by Professor Frade during the Fall '10 term at FSU.
 Fall '10
 Frade

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