Chapt 7c

# Chapt 7c - 7.6 Chapter Summary 261 Since the number of...

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Unformatted text preview: 7.6 Chapter Summary 261 Since the number of items purchased by each customer is independent of the number purchased by other customers, we have that G Y | K ( z ) = [ G N ( z ) ] k Thus, G Y ( z ) = ∞ summationdisplay k = p K ( k ) G Y | K ( z ) = ∞ summationdisplay k = p K ( k ) [ G N ( z ) ] k = G K ( G N ( z )) Since G K ( z ) = e λ( z- 1 ) and G N ( z ) = e μ ( z- 1 ) , we have that G Y ( z ) = e λ( e μ ( z- 1 )- 1 ) trianglesolid 7.6 Chapter Summary This chapter discussed three transform methods that are frequently used in the analysis of probabilistic problems. These are the characteristic function, the s-transform, and the z-transform. Both the s-transform and the z-transform are used for random variables that take only nonnegative values, which include many random variables that are used to model practical systems. The moment- generating properties of the different transforms have also been demonstrated. Table 7.1 is a summary of the different transforms of some of the well-known PMFs and PDFs. 7.7 Problems Section 7.2: Characteristic Functions 7.1 Find the characteristic function of the random variable X with the following PDF: f X ( x ) = braceleftBigg 1 b- a a < x < b otherwise 7.2 Find the characteristic function of the random variable Y with the following PDF: f Y ( y ) = braceleftbigg 3 e- 3 y y ≥ otherwise 262 Chapter 7 Transform Methods Table 7.1 Summary of the Transforms of Well-Known PMFs and PDFs Characteristic PMF PDF z-Transform s-Transform Function Bernoulli p X ( x ) = braceleftbigg 1- p x = p x = 1- 1- p + zp Binomial p X ( n ) ( x ) = ( n x ) p x ( 1- p ) n- x where x = , 1 , 2 ,..., n- ( 1- p + zp ) n Geometric p X ( x ) = p...
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Chapt 7c - 7.6 Chapter Summary 261 Since the number of...

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