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Unformatted text preview: 7.4 The zTransform 251 Example 7.6 Is the following function a valid ztransform of a PMF? If so, what is the PMF? G X ( z ) = 1 a 1 az < a < 1 Solution First, we note that G X ( 1 ) = 1, which means that the function G X ( z ) is potentially a valid ztransform of a PMF. Next, we test the coefficients of z in the function. Now, G X ( z ) = ( 1 a ) ∞ summationdisplay k = ( az ) k = ( 1 a ) braceleftbig 1 + az + a 2 z 2 + a 3 z 3 + ··· + a k z k + ··· bracerightbig Since 0 < a < 1, we see that all the coefficients of z are nonnegative quantities that are no greater than 1. Therefore, the function is a valid ztransform of a PMF, and the PMF is the following: p X ( x ) = ( 1 a ) a x x = , 1 , 2 ,... trianglesolid Example 7.7 Explain why the function F ( z ) = z 2 + z 1 is or is not a valid ztransform of the PMF of a random variable. Solution One of the tests for a function of z to be a valid ztransform of a PMF is that it must be equal to 1 when evaluated at z = 1. As can be seen, F ( 1 ) = 1, so the function has passed the first test. The second test is that the coefficients of z must be nonnegative since, for example, the coefficient of z k is the probability that the random variable takes the value k . In the function above, the constant term, which represents the probability that the supposed random variable takes the value 0, is negative 1. This means that the function cannot be a valid ztransform of a PMF. trianglesolid Example 7.8 The ztransform of the PMF of a discrete random variable K is given by G K ( z ) = A bracketleftbigg 10 + 8 z 2 ( 2 z ) bracketrightbigg (a) What is the expected value of K ? (b) Find p K ( 1 ) , the probability that K has the value 1. 252 Chapter 7 Transform Methods Solution Before answering both questions, we need to obtain the numerical value of A . For G K ( z ) to be a valid ztransform, it must satisfy the condition G K ( 1 ) = 1. Thus, we have that G K ( 1 ) = A bracketleftbigg 10 + 8 ( 2 1 ) bracketrightbigg = 18 A = 1 This means that A = 1 / 18. (a) The expected value of K is E [ K ] = d dz G K ( z )  z = 1 = A braceleftbigg ( 2 z ) 16 z ( 10 + 8 z 2 )( 1 ) ( 2 z ) 2 vextendsingle vextendsingle vextendsingle vextendsingle z = 1 bracerightbigg = 34 18 = 1 . 9 (b) To obtain the PMF of K , we observe that G K ( z ) = A bracketleftbigg 10 + 8 z 2 ( 2 z ) bracketrightbigg = A 2 bracketleftbigg 10 + 8 z 2 1 z 2 bracketrightbigg = A ( 10 + 8 z 2 ) 2 ∞ summationdisplay m = parenleftbigg z 2 parenrightbigg m = A ( 10 + 8 z 2 ) 2 braceleftbigg 1 + z 2 + z 2 4 + z 3 8 + z 4 16 + ··· bracerightbigg = A 2 braceleftbigg 10 + z bracketleftbigg 10 2 bracketrightbigg + z 2 bracketleftbigg 10 4 + 8 bracketrightbigg + z 3 bracketleftbigg 10 8 + 4 bracketrightbigg + z 4 bracketleftbigg 10 16 + 2 bracketrightbigg + ··· bracerightbigg Thus, the probability that K has a value 1 is the coefficient of z in G K ( z ) , which is p K ( 1 ) = A 2 × 10 2 = 5 36 trianglesolid 7.4.1 MomentGenerating Property of the zTransform...
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This note was uploaded on 01/05/2010 for the course STAT 350 taught by Professor Carlton during the Fall '07 term at Cal Poly.
 Fall '07
 Carlton

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