solutions2_Cexercises

# solutions2_Cexercises - EX 3 = ∞ X x =1 x 3 (1-p ) x-1 p...

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Solution to Exercise C7 Let X Geometric( p ). Recall that EX = 1 p and EX 2 = 2 - p p 2 . EX 3 = E ( X 3 | X = 1) P ( X = 1) + E ( X 3 | X > 1) P ( X > 1) = 1 · p + E ± (1 + ( X - 1)) 3 | X > 1 ² · (1 - p ) = 1 · p + E ± (1 + X ) 3 ² · (1 - p ) = 1 · p + E ( 1 + 3 X + 3 X 2 + X 3 ) · (1 - p ) = p + ( 1 + 3 EX + 3 EX 2 + EX 3 ) (1 - p ) = p + ³ 1 + 3 p + 3(2 - p ) p 2 + EX 3 ´ (1 - p ) = p 2 - 6 p + 6 p 2 + (1 - p ) EX 3 so that pEX 3 = p 2 - 6 p + 6 p 2 and EX 3 = p 2 - 6 p + 6 p 3 . Note: The pmf of the Geometric distribution is a geometric series, and if you remove the ﬁrst few terms from a geometric series, what remains is another geometric series. This observation is essentially equivalent to the Discrete Memoryless Property. The above calculation can be rewritten as
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Unformatted text preview: EX 3 = ∞ X x =1 x 3 (1-p ) x-1 p = p + ∞ X x =2 x 3 (1-p ) x-1 p = p + (1-p ) ∞ X x =2 (( x-1) + 1) 3 (1-p ) ( x-1)-1 p so that making the change of variable z = x-1 leads to = p + (1-p ) ∞ X z =1 ( z + 1) 3 (1-p ) z-1 p = p + (1-p ) E µ ( X + 1) 3 ¶ = p + (1-p ) ( 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.

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solutions2_Cexercises - EX 3 = ∞ X x =1 x 3 (1-p ) x-1 p...

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