J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
1
CONTINUOUS DISTRIBUTIONS Laplace transform (Laplace-Stieltjes transform) Definition The Laplace transform of a non-negative random variable X ≥ 0 with the probability density function f (x) is defined as ∗
f (s) =
Z
∞ −st e f (t)dt 0
= E[e
−sX
]
=
∞ −st e dF (t) 0
Z
also denoted as LX (s)
• Mathematically it is the Laplace transform of the pdf function. • In dealing with continuous random variables the Laplace transform has the same role as the generating function has in the case of discrete random variables. – if X is a discrete integer-valued (≥ 0) r.v., then f ∗(s) = G(e−s)
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
Laplace transform of a sum Let X and Y be independent random variables with L-transforms fX∗ (s) and fY∗ (s). ∗ fX+Y (s) = E[e−s(X+Y )]
= E[e−sX e−sY ] = E[e−sX ]E[e−sY ] = fX∗ (s)fY∗ (s) ∗ fX+Y (s) = fX∗ (s)fY∗ (s)
(independence)
2
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
Calculating moments with the aid of Laplace transform By derivation one sees d f ∗0 (s) = E[e−sX ] = E[−Xe−sX ] ds Similarly, the nth derivative is dn ∗(n) f (s) = n E[e−sX ] = E[(−X)n e−sX ] ds Evaluating these at s = 0 one gets E[X]
= −f ∗ 0 (0)
E[X 2] = +f ∗ 0 0 (0) ... E[X n] = (−1)n f ∗(n)(0)
3
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
4
Laplace transform of a random sum Consider the random sum Y = X1 + · · · + XN
where the Xi are i.i.d. with the common L-transform fX∗ (s) and N ≥ 0 is a integer-valued r.v. with the generating function GN (z). fY∗ (s) = E[e−sY ]
= E[E e
−sY
|N ]
(outer expectation with respect to variations of N)
= E[E e−s(X1+···+XN ) | N ]
(in the inner expectation N is fixed)
= E[E[e−s(X1)] · · · E[e−s(XN )]]
(independence)
= GN (fX∗ (s))
(by the definition E[z N ] = GN (z))
= E[(fX∗ (s))N ]
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
5
Laplace transform and the method of collective marks We give for the Laplace transform f ∗(s) = E[e−sX ],
X ≥ 0,
the following
Interpretation: Think of X as representing the length of an interval. Let this interval be subject to a Poissonian marking process with intensity s. Then the Laplace transform f ∗(s) is the probability that there are no marks in the interval. P{X has no marks} = E[P{X has no marks | X}]
(total probability)
= E[P{the number of events in the interval X is 0| X}] = E[e−sX ] = f ∗(s)
ì í î
intensiteetti s
X
P{there are n events in the interval X | X}
=
(sX)n −sX e n!
P{the number of events in the interval X is 0 | X} = e−sX
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
6
Method of collective marks (continued) Example: Laplace transform of a random sum Y = X1 + · · · + XN ,
where
X1 ∼ X2 ∼ · · · ∼ XN , common L-transform f ∗(s) N is a r.v. with generating function GN (z)
intensiteetti s
fY∗ (s) = P{none of the subintervals of Y is marked} =
) GN ( fX∗ (s) | {z } probability that a single subinterval has no {zmarks | } probability that none of the subintervals is marked
X1 X2
...
XN
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
Uniform distribution X ∼ U(a, b) The pdf of X is constant in the interval (a, b):
f (x) =
1 b−a
a t} P{X > t + x} = P{X > t} e−λ(t+x) = = e−λx = P{X > x} −λt e
P{X > t + x | X > t} =
P{X > t + x | X > t} = P{X > x} • The distribution of the remaining duration of the call does not at all depend on the time the call has already lasted • Has the same Exp(λ) distribution as the total duration of the call.
exp(- λ t)
exp(- λ (t-u))
u
t
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
12
Example of the use of the memoryless property A queueing system has two servers. The service times are assumed to be exponentially distributed (with the same parameter). Upon arrival of a customer () both servers are occupied (×) but there are no other waiting customers.
´ ¨
´
The question: what is the probability that the customer () will be the last to depart from the system? The next event in the system is that either of the customers (×) being served departs and the customer enters () the freed server.
¨
´
By the memoryless property, from that point on the (remaining) service times of both customers () and (×) are identically (exponentially) distributed. The situation is completely symmetric and consequently the probability that the customer () is the last one to depart is 1/2.
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
The ending probability of an exponentially distributed interval Assume that a call with Exp(λ) distributed duration has lasted the time t. What is the probability that it will end in an infinitesimal interval of length h? P{X ≤ t + h | X > t} = P{X ≤ h} (memoryless) = 1 − e−λh
= 1 − (1 − λh + 21 (λh)2 − · · ·)
= λh + o(h)
The ending probability per time unit = λ
(constant!)
13
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
14
The minimum and maximum of exponentially distributed random variables Let X1 ∼ · · · ∼ Xn ∼ Exp(λ)
(i.i.d.)
The tail distribution of the minimum is P{min(X1 , . . . , Xn) > x} = P{X1 > x} · · · P{Xn > x}
(independence)
= (e−λx )n = e−nλx
The minimum obeys the distribution Exp(nλ). n parallel processes each of which ends with intensity λ independent of the others
The cdf of the maximum is X2
P{max(X1 , . . . , Xn) ≤ x} = (1 − e−λx )n
X3
ì í î
ì í î
ì í î
X5 ì í î
The expectation can be deduced by inspecting the figure 1 1 1 + + ···+ E[max(X1 , . . . , Xn)] = nλ (n − 1)λ λ
X1
ì í î
The ending intensity of the minimum = nλ
~Exp(nl) ~Exp((n-1)l) ~Exp((n-2)l)
~Exp(l)
X4
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
Erlang distribution X ∼ Erlang(n, λ)
15
Also denoted Erlang-n(λ).
X is the sum of n independent random variables with the distribution Exp(λ) X = X1 + · · · + Xn
Xi ∼ Exp(λ)
(i.i.d.)
The Laplace transform is f ∗(s) = (
λ n ) λ+s
By inverse transform (or by recursively convoluting the density function) one obtains the pdf of the sum X (λx)n−1 −λx f (x) = λe (n − 1)!
x≥0
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
16
Erlang distribution (continued): gamma distribution The formula for the pdf of the Erlang distribution can be generalized, from the integer parameter n, to arbitrary real numbers by replacing the factorial (n − 1)! by the gamma function Γ(n): (λx)p−1 −λx λe f (x) = Γ(p)
Gamma(p, λ) distribution
Gamma function Γ(p) is defined by Γ(p) = 0.5 0.4
By partial integration it is easy to see that when p is an integer then, indeed, Γ(p) = (p − 1)!
∞ −u p−1 e u du 0
Z
The expectation and variance are n times those of the Exp(λ) distribution:
n=1 n=2
0.3
n=3
n=4
0.2
E[X] =
n=5
0.1 0
2
4
6
8
10
n λ
V[X] =
n λ2
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
17
Erlang distribution (continued) ~Erlang(2,l)
ì í î
Example. The system consists of two servers. Customers arrive with Exp(λ) distributed interarrival times. Customers are alternately sent to servers 1 and 2.
~Exp(l)
The interarrival time distribution of customers arriving at a given server is Erlang(2, λ). Proposition. Let Nt , the number of events in an interval of length t, obey the Poisson distribution: Nt ∼ Poisson(λt)
Then the time Tn from an arbitrary event to the nth event thereafter obeys the distribution Erlang(n, λ). 1
2
3
n
... t
ì í î
0 Nt tapahtumaa
Tn
Proof. FTn (t) = P{Tn ≤ t} = P{Nt ≥ n} (λt)i −λt = P{Nt = i} = e i=n i=n i! ∞ X
fTn =
d F (t) dt Tn
∞ X
∞ (λt)i iλ (λt)i−1 −λt X = e − λe−λt i! i=n i=n i! ∞ X
∞ (λt)i (λt)i−1 −λt X = λe − λe−λt i=n (i − 1)! i=n i! ∞ X
(λt)n−1 −λt λe = (n − 1)!
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
18
Normal distribution X ∼ N(µ, σ 2) The pdf of a normally distributed random variable X with parameters µ ja σ 2 is f (x) = √
Parameters µ and σ 2 are the expectation and variance of the distribution
1 1 2 2 e− 2 (x−µ) /σ 2πσ
Proposition: If X ∼ N(µ, σ 2), then Y = αX + β ∼ N(αµ + β, α2σ 2). Proof: FY (y) = P{Y ≤ y} = P{X ≤
Seuraus:
=
Z
(y−β)/α
=
Z
y
−∞
y−β } α
= FX ( y−β ) α
1 2 2 √ 1 e− 2 (x−µ) /σ dx 2πσ
z = αx + β
− 21 (z−(αµ+β))2 /(ασ)2 √ 1 e dz −∞ 2π(ασ)
Z=
X −µ ∼ N(0, 1) σ
(α = 1/σ, β = −µ/σ)
Denote the pdf of a N(0,1) random variable by Φ(x). Then FX (x) = P{X ≤ x} = P{Z ≤
x−µ } σ
= Φ( x−µ ) σ
E[X] = µ V[X] = σ 2
J. Virtamo
38.3143 Queueing Theory / Continuous Distributions
19
Multivariate Gaussian (normal) distribution Let X1, . . . , Xn be a set of Gaussian (i.e. normally distributed) random variables with expectations µ1, . . . , µn and covariance matrix
Γ=
2 σ11
2 σ1n
··· ... . . . ... 2 2 σn1 · · · σnn
σij2 = Cov[Xi , Xj ]
(σii2 = V[Xi])
Denote X = (X1 , . . . , Xn)T . The probability density function of the random vector X is 1 (x−µ)TΓ−1 (x−µ) 1 − e 2 f (x) = r n (2π) |Γ|
where |Γ| is the determinant of the covariance matrix. −1/2 By a change of variables one sees easily that the pdf of the random vector Z = Γ (X − µ) √ √ 2 2 is (2π)−n/2 exp(− 12 zT z) = 2πe−z1 /2 · · · 2πe−zn/2 .
Thus the components of the vector Z are independent N(0,1) distributed random variables. Conversely, X = µ + Γ1/2 Z by means of which one can generate values for X in simulations.