1. Transform Analysis

2. Two Types of Transforms
Discrete random variable: z-transform
X(z) = Gp(z) = E[zX] = Σ{i = 0 to }p(i)zi,
where P{X=i} = p(i)
Note that X(1) = 1
Continuous random variable: Laplace transform
X(s) = Lf(s) = E[e–sX] {t=0 to }e–stfX(t)dt
where fX() is p.d.f. of X
Note that X(0) = 1
Note also that Xx(s) = e–sx when P{X=x} = 1

3. Basic Transform Properties
Let Z = X+Y, with X and Y two discrete independent r.v.’s
Z(z) = X(z)Y(z)
Let Z = X+Y, with X and Y two continuous independent r.v.’s with p.d.f. x(t) and y(t), t  0
Z(s) = X(s)Y(s)
Let X = A with prob. p and X = B with prob. 1-p, then
X(z) = pA(z) + (1 – p)B(z), if A & B are discrete r.v.’s
X(s) = pA(s) + (1 – p)B(s), if A & B are continuous r.v.’s

4. Other Transform Properties – (1a)
If Y is a discrete r.v. with probabilities pk, k = 0,1,2,…, and XY is a discrete/continuous r.v. that depends on Y, then
XY(z) = {k=0 to }Xk(z)pk – (discrete)
XY(s) = {k=0 to }Xk(s)pk – (continuous)
In the discrete case, we have
XY(z) = E[zXY]
= {k=0 to } E[zXY|Y=k]pk
= {k=0 to } E[zXk|Y=k]pk
= {k=0 to } Xk(z)pk
In the continuous case, we have
XY(s) = E[e–sXY]
= {k=0 to } E[e–sXY|Y=k]pk
= {k=0 to } E[e–sXy|Y=k]pk
= {k=0 to } Xy(s)pk

5. Other Transform Properties – (1b)
If Y is a continuous r.v. with p.d.f. fY(y), and XY is a discrete/continuous r.v. that depends on Y, then
XY(z) = {y=0 to }Xy(z)fY(y)dy – (discrete)
XY(s) = {y=0 to }Xy(s)fY(y)dy – (continuous)
In the discrete case, we have
XY(z) = E[zXY]
= {y=0 to } E[zXY|Y=y]fY(y)dy
= {y=0 to } E[zXy|Y=y]fY(y)dy
= {y=0 to } Xy(z)fY(y)dy
In the continuous case, we have
XY(s) = E[e–sXY]
= {y=0 to } E[e–sXY|Y=y]fY(y)dy
= {y=0 to } E[e–sXy|Y=y]fY(y)dy
= {y=0 to } Xy(s)fY(y)dy

6. Applying to # Arrivals in a Service Time
Number of Poisson arrivals A (of rate λ) during a (service) time with distribution S(t) – A is a discrete random variable and S a continuous one
AS(z) = {t=0 to }AS(z|S=t)fS(t)dt
= {t=0 to }At(z)fS(t)dt
= {t=0 to }e–λ(1–z)tfS(t)dt (see next slide)
= S(λ(1–z)) – a more direct derivation

7. z-Transform of Poisson Arrivals
Given Poisson arrivals of rate λ, what is the z-transform of the number of arrivals in t?

8. Other Transform Properties – (2)
Let Z = Y1+Y2+…+YX, where the Yi‘s are discrete/continuous i.i.d. r.v.’s and X is a discrete r.v. independent of the Yi’s
Z(z) = X(Y(z)) – z-transform of Z is z-transform of X evaluated at the z-transform of Y – (discrete)
Z(s) = X(Y(s)) – Laplace transform of Z is z-transform of X evaluated at the Laplace transform of Y – (continuous)
Based on transform of sum of i.i.d. r.v.’s is product of transform and z-transform as an expectation after conditioning
Assume Yi‘s are exponential with parameter μ, and X is Geom(p)
X(z) = zp/[1-z(1–p)]; Y(s) = μ/(μ+s)
Z(s) = μp/(μ+s–μ+μp) = μp/(s+μp)
Z is exponential with parameter μp

9. Number in System for M/G/1 – (1)
Consider the queue state at departure times
This forms an embedded DTMC
Recall that statistics seen by departures, arrivals, and at random times are identical (PASTA and level crossing argument)
πj = π0aj+ Σ{i=1 to j+1}πiaj–i+1 where aj is the probability of j arrivals during one service time
Recall that we know AS(z) = S(λ(1–z)), so we can express N(z) – z-transform of number in the system, in terms of the Laplace transform of the service time evaluated at λ(1–z)

10. Number in System for M/G/1 – (2)
We know that π0= 1 – ρ, so that we have
and we can now get all the moments of N(z)

11. Time in System for M/G/1 Let T be a job’s time in the system
The z-transform of the number of Poisson arrivals of rate λ during T is given by AT(z) = T(λ–λz)
But the number of arrivals during T is the number of jobs in the system seen by a departure, so T(λ–λz) = N(z)
Hence, we have
Similarly, since T = S + TQ, we get

12. Re-deriving the P-K Formula

13. Other Transform Properties – (3)
Given a continuous r.v. with p.d.f. b() and c.d.f. B(), then B(s) = b(s)/s
i.e., the Laplace transform of the distribution is the Laplace transform of the density function divided by s
Note also that the Laplace transform of the complementary c.d.f. is B(s) = (1 – b(s))/s