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# E_diskrjak - J Virtamo 38.3143 Queueing Theory Discrete...

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J. Virtamo 38.3143 Queueing Theory / Discrete Distributions 1 DISCRETE DISTRIBUTIONS Generating function (z-transform) Definition Let X be a discrete r.v., which takes non-negative integer values, X ∈ { 0 , 1 , 2 , . . . } . Denote the point probabilities by p i p i = P { X = i } The generating function of X denoted by G ( z ) (or G X ( z ); also X ( z ) or ˆ X ( z )) is defined by G ( z ) = summationdisplay i =0 p i z i = E[ z X ] Rationale: A handy way to record all the values { p 0 , p 1 , . . . } ; z is a ‘bookkeeping variable’ Often G ( z ) can be explicitly calculated (a simple analytical expression) When G ( z ) is given, one can conversely deduce the values { p 0 , p 1 , . . . } Some operations on distributions correspond to much simpler operations on the generating functions Often simplifies the solution of recursive equations

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J. Virtamo 38.3143 Queueing Theory / Discrete Distributions 2 Inverse transformation The problem is to infer the probabilities p i , when G ( z ) is given. Three methods 1. Develop G ( z ) in a power series, from which the p i can be identified as the coefficients of the z i . The coefficients can also be calculated by derivation p i = 1 i ! d i G ( z ) dz i vextendsingle vextendsingle vextendsingle vextendsingle z =0 = 1 i ! G ( i ) (0) 2. By inspection: decompose G ( z ) in parts the inverse transforms of which are known; e.g. the partial fractions 3. By a (path) integral on the complex plane p i = 1 2 πi contintegraldisplay G ( z ) z i +1 dz path encircling the origin (must be chosen so that the poles of G ( z ) are outside the path)
J. Virtamo 38.3143 Queueing Theory / Discrete Distributions 3 Example 1 G ( z ) = 1 1 - z 2 = 1 + z 2 + z 4 + · · · p i = 1 for i even 0 for i odd Example 2 G ( z ) = 2 (1 - z )(2 - z ) = 2 1 - z - 2 2 - z = 2 1 - z - 1 1 - z/ 2 Since A 1 - az corresponds to sequence A · a i we deduce p i = 2 · (1) i - 1 · ( 1 2 ) i = 2 - ( 1 2 ) i

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J. Virtamo 38.3143 Queueing Theory / Discrete Distributions 4 Calculating the moments of the distribution with the aid of G ( z ) Since the p i represent a probability distribution their sum equals 1 and G (1) = G (0) (1) = summationdisplay i =1 p i · 1 i = 1 By derivation one sees G (1) ( z ) = d dz E[ z X ] = E[ Xz X - 1 ] G (1) (1) = E[ X ] By continuing in the same way one gets G ( i ) (1) = E[ X ( X - 1) · · · ( X - i + 1)] = F i where F i is the i th factorial moment .
J. Virtamo 38.3143 Queueing Theory / Discrete Distributions 5 The relation between factorial moments and ordinary moments (with respect to the origin) The factorial moments F i = E[ X ( X - 1) · · · ( X - i + 1)] and ordinary moments (with resect to the origin) M i = E[ X i ] are related by the linear equations: F 1 = M 1 F 2 = M 2 - M 1 F 3 = M 3 - 3 M 2 + 2 M 1 . . . M 1 = F 1 M 2 = F 2 + F 1 M 3 = F 3 + 3 F 2 + F 1 . . . For instance, F 2 = G (2) (1) = E[ X ( X - 1)] = E[ X 2 ] - E[ X ] M 2 = E[ X 2 ] = F 2 + F 1 = G (2) (1) + G (1) (1) V[ X ] = M 2 - M 2 1 = G (2) (1) + G (1) (1) - ( G (1) (1)) 2 = G (2) (1) + G (1) (1)(1 - G (1) (1))

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J. Virtamo 38.3143 Queueing Theory / Discrete Distributions 6 Direct calculation of the moments
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