Renewal processes

Interarrival times

• {0,T1,T2,..} is an i.i.d. sequence with a common distribution fct. F

• Si = j=1i Tj

• {Si} is a nondecreasing, positive sequence of reneval times (point)

• The distribution of Si is F(i) • F(i) = f(i) * F = F * f(i)

• f(i) is the i-fold convolution of f (f = d/dx F)

The counting process

• N(t) = maxi {Si · t}• N(t) counts the number of renewal point before

t• M(t) = E[N(t)] is the expected number of

renevals before t• M(t) = n n P(N(t)=n)

• P(N(t)=n)=P(Sn · t and Sn+1 > t)= P(Sn+1 > t) - P(Sn > t and Sn+1 > t)

The counting process

• P(N(t)=n)=P(Sn · t and Sn+1 > t)= P(Sn+1 > t) - P(Sn > t and Sn+1 > t)

• Now Sn > t => Sn+1 > t so • P(Sn > t and Sn+1 > t) = P(Sn > t)• Altogether • P(Sn · t and Sn+1 > t)

= P(Sn+1 > t) - P(Sn > t)= P(Sn · t) - P(Sn+1 · t) = F(n)(t) – F(n+1)(t)

The counting process• M(t) = n n P(N(t)=n)

=n n P(Sn · t and Sn+1 > t) = n n (F(n)(t) – F(n+1)(t))=F(1)(t) + n=2 n F(n)(t) – (n-1)F(n)(t)

=F(1)(t) + n=2

F(n)(t)

=F(t) + n=2 f(n) * F=F(t) + f * n=1 f(n) * F=F(t) + (f * M)(t)

The renewal density

• m(t)= d/dt M(t): is called the renewal density• M(t+h)-M(t) is the expected number of renewals

in [t,t+h]• When h is small P(N(t+h)-N(t)>1)=O(h2)

and P(N(t+h)-N(t)=1)=O(h)• Thus (M(t+h)-M(t)) ¼ P(N(t+h)-N(t)=1) ¼ h ¢ m(t)

• h ¢ m(t) approximates the probability of a renewal

within [t,t+h]

The renewal density

• m(t)= d/dt M(t)• M(t)=F(t) + (f * M)(t)• m(t) = f(t) + d/dt (F * M)(t)• = f(t) + d/dt(s0

t M(t-u) f(u) du)

• = f(t) + s0t m(t-u) f(u) du

• = f(t) + (f * m)(t)

Recurrence times

• Backward recurrence time (age):A(t) = t – SN(t)

• Forward recurrence time (excess):Y(t) = SN(t)+1 –t

• FA,t(a) = P(A(t) · a)

• FY,t(y) = P(Y(t) · y)

Distribution of ageF A,t(a) = P(A(t) · a) = P(t-S N(t) · a)• We condition on the first renewal, i.e. P(A(t) · a)

= s01 P(A(t) · a | S1=s) f(s) ds

= s0t P(A(t) · a | S1=s) f(s) ds

+ st1 P(A(t) · a | S1=s) f(s) ds

= s0t P(A(t-s) · a) f(s) ds

+ st1 P(A(t) · a | S1=s) f(s) ds

= s0t FA,t-s(a) f(s) ds

+ st1 I(t · a) ¢ f(s) ds

= s0

t FA,t-s(a) f(s) ds + I (t · a} R(t)= (F A,.(a) * f)(t) + I(t · a) R(t)

Distribution of excess• FY,t(y) = P(Y(t) · y)• We condition on the first renewal, i.e. P(Y(t) · y)

= s01 P(Y(t) · y | S1=s) f(s) ds

= s0t P(Y(t) · y | S1=s) f(s) ds

+ st1 P(Y(t) · y | S1=s) f(s) ds

= s0t P(Y(t-s) · y) f(s) ds

+ st1 P(Y(t) · y | S1=s) f(s) ds

= s0t FY,t-s(y) f(s) ds

+ st1 I(s-t · y) ¢ f(s) ds

= (FY,.(y) * f)(t) + st

1 I(s · (t+y)) ¢ f(s) ds= (FY,.(y) * f)(t) + F(t+y)-F(t)

General solutions

• Generally: Z = Q + Z * f• Laplace transform

Z(s) = Q(s) + Z(s)f(s)Z(s) (1-f(s)) = Q(s)Z(s) = Q(s) / (1-f(s))

Alternative solution• m(t) = f(t) + (f * m)(t)• Laplace transform

m(s) = f(s) + f(s) m(s)m(s) (1-f(s))=f(s)1-f(s)=f(s)/m(s)

• Z(s) (1-f(s)) = Q(s) Z(s) f(s) = Q(s) m(s)

• Z(s) = Q(s) + Z(s) f(s) = Q(s) + Q(s) m(s)

• Z(t) = Q(t) + (Q * m)(t)

Example (Poisson)

• Poisson process: F(t)=1-exp(-¸ t)f(t) = ¸ exp(-¸ t)R(t) = exp(-¸ t)

• m = f + m * f m(s)=f(s)/(1-f(s)) • f(s) = ¸ s exp(-st) exp(-¸ t) dt = ¸ /(s+¸)• m(s) = ¸ /(s+¸)/(1- ¸ /(s+¸)) = ¸ /(s+¸- ¸ )) = ¸ / s• m(t) = ¸ !!!

Limiting renewal densityin general

• m(t) = f(t) + (f * m)(t) • m(s) = f(s) + f(s) m(s) • m(s)=f(s)/(1-f(s))• limt -> 1 m(t) = lims -> 0 s m(s) =

• lims -> 0 s f(s)/(1-f(s)) = (l’Hospital) lims -> 0 d/ds (s f(s)) / lims -> 0 d/ds (1-f(s))

= f(0)/ ((d/ds -f(s))|s=0) = 1/E(Ti) !!!

Example (Poisson)• m(t) = ¸• FA,t(a) = (FA,.(a) * f)(t) + I(t · a) R(t) • FY,t(y)= (FY,.(y) * f)(t) + F(t+y)-F(t)• Z = Q + Z * f Z(s) = Q(s) / (1-f(s))

or Z(t) = Q(t) + (Q * m)(t)

FA,t(a) = I(t · a) R(t) + ¸ s0t I(s · a) R(s) ds

= I(t · a) R(t) + ¸ s0min(t,a) R(s) ds

= I(t · a) exp(-¸ t)+ (1-exp(-min(t,a)))

FY,t(y) = (FY,.(y) * f)(t) + F(t+y)-F(t) = F(t+y)-F(t) + ¸ s0

t F(s+y)-F(s) ds (husk -¸)= exp(-¸ t) - exp(-¸ (t+y)) - exp(-¸ t) + exp(-¸ (t+y)) + (1- exp(-¸ y) ) = 1-exp(-¸ y) = F(y) !!!

Alternating renewal process

• Used to model random on/off processes• Network traffic• Power consumption

Sn-1 Sn

Tn = Zn + Yn

Zn Yn

ON OFF

Alternating renewal process• I(t) = I(SN(t) < t · SN(t)+ZN(t))• I(t) indicates whether t belongs to an on-period.• P(ON at t) = P(I(t)=1)=O(t)• We condition on the first renewal

O(t) = P(I(t)=1) = s01 P(I(t)=1 | S1=s) f(s) ds

= s0t P(I(t)=1 | S1=s) f(s) ds

+ st1 P(I(t)=1 | S1=s) f(s) ds

= s0t P(I(t-s)=1) f(s) ds

+ st1 P(t · Z1 | S1=s) f(s) ds

= (O * f)(t) + st1 P(Z1 ¸ t| S1=s) f(s) ds

Alternating renewal process

O(t) = (O * f)(t) + st1 P(Z1 ¸ t| S1=s) f(s) ds

= (O * f)(t) + s01 P(Z1 ¸ t| S1=s) f(s) ds

= (O * f)(t) + P(Z1 ¸ t)= (O * f)(t) + 1-FZ(t)O(s)=1-FZ(s) + O(s)*f(s)

Example (2 state Markov)• 2 exponential distributions

FZ(t)=1-exp(-¸ t)FY(t)=1-exp(-¹ t)f(t) = ¸ ¹ s0

t exp(-¸ (t-s)) exp(-¹ s) ds E(T)=E(Y)+E(Z)=1/¹ + 1/¸

limt -> 1=1/E(T)=1/(1/¹+1/¸)

• O(s)=1-FZ(s) + O(s)*f(s) O(s)=(1-FZ(s))/(1-f(s)) or O(s) = 1-FZ(s) + (1-FZ(s)) m(s)

• limt -> 1 O(t) = lims -> 0 s O(s) =lims -> 0 s(1-FZ(s)) + s(1-FZ(s)) m(s) =lims -> 0 (1-FZ(s)) lims -> 0 s m(s) = s RZ(t) dt / E(T)

• s RZ(t) dt = s 1 ¢ RZ(t) dt = tRZ(t) + s t ¢ fZ(t) dt -> s t ¢ fZ(t) dt = E(Z)

• limt -> 1 O(t) = E(Z)/E(T) !!!

Autocorrelation

• CII(s) = E((It-E(I))(It+s-E(I))) = E((It It+s) – E2(I)

• E(I) = limt -> 1 O(t) = E(Z)/E(T)• E(It It+s)=P(It and It+s)

• Tn = Zn + Yn

• SN(t)=SN(t)-1+TN(t)

• A(t)=t-SN(t)

Autocorrelation

• CII(s) = E((It It+s) – E2(I)• t lies in the 1st renewal period• E(I) = E(Z)/E(T)• E(It It+s)=P(It and It+s)

• P(It and It+s) = s P(t+s · Z1 | S1=x) +

P(t · Z1 and t+s ¸ S1 | S1=x) O(t+s-x) f(x) dx

AutocorrelationP(It and It+s) = s P(t+s · Z1 | S1=x) + P(t · Z1 and t+s ¸ S1 | S1=x) O(t+s-x) f(x) dx

=s P(t+s · Z1 | S1=x) + P(t · Z1 and t+s ¸ x | S1=x) O(t+s-x) f(x) dx

=s P(t+s · Z1 | S1=x) + I(t+s ¸ x) P(t · Z1| S1=x) O(t+s-x) f(x) dx

=s P(t+s · z | Z1=z, S1=x) fZ,S(z,x) dzdx + s I(t+s ¸ x) P(t · z| Z1=z, S1=x) O(t+s-x) fZ,S(z,x) dzdx

=s I(t+s · z) fS|Z(z,x) fZ(z) dzdx + s I(t+s ¸ x) I(t · z) O(t+s-x) fS|Z(z,x) fZ(z) dzdx

=s I(t+s · z) fY(x-z) fZ(z) dzdx + s I(t+s ¸ x) I(t · z) O(t+s-x) fY(x-z) fZ(z) dzdx

AutocorrelationP(It and It+s) = s I(t+s · z) fY(x-z) fZ(z) dzdx + s I(t+s ¸ x) I(t · z) O(t+s-x) fY(x-z) fZ(z) dzdx

= sst+sx fY(x-z) fZ(z) dzdx +

stt+s st

x O(t+s-x) fY(x-z) fZ(z) dzdx

= st+s sz fY(x-z) dx fZ(z) dz + st

t+s O(t+s-x) stx fY(x-z) fZ(z) dz dx

= st+s fZ(z) dx + stt+s O(t+s-x) st

x fY(x-z) fZ(z) dz dx

= RZ(t+s) + stt+s O(t+s-x) st

x fY(x-z) fZ(z) dz dx· RZ(t+s) + s st

x fY(x-z) fZ(z) dz dx= RZ(t+s) + P(Z1 ¸ t) = RZ(t+s)+ RZ(t) \leq 2RZ(2s) (for large s)

CII(s) = E((It It+s) – E2(I) \leq E((It It+s) · 2RZ(2s)

Example - Pareto distributions (power/heavy tails)

• Let fZ(z)=K z-® I(z ¸ z0) ®>1• FZ(z)= K/(®-1) (z0

1-® – z1-®) I(z ¸ z0)• K= (®-1)/z0 1-®

• FZ(z)= (1 – (z/z0)1-®) I(z ¸ z0)

• RZ(z)=1-FZ(z) = (z/z0)1-® + I(z · z0) · (z/z0)1-®

CII(s) ¼ 2RZ(2s) = K · (2s)2(H-1) (H - Hurst parameter)H>1/2 : Long Range Dependence (LRD)

1-® = 2(H-1) H=(1-®)/2+1=3/2-®/2 or ®=3-2HLRD ® < 2

Sample means

• ET = 1/T s0T I(t) dt

• I(t) indicates on-state• Var(ET)=E(E T 2)= 1/T2 E((s0

T I(t) dt)2)= 1/T2 E(s0

T I(t) dt s0T I(t) dt)

= 1/T2 E(s0

T s0T I(t) I(s) ds dt)

= 1/T2 s0T s0

T E(I(t) I(s)) ds dt= 1/T2 s0

T s0T CII(t-s) ds dt

¼ 1/T2 s0T s0

T 2RZ(2|t-s|) ds dt= 4/T2 s0

T s0t RZ(2(t-s)) ds dt

Sample meansfor 2 state Markov process

• Var(ET) = 4/T2 s0T s0

t RZ(2(t-s)) ds dt

• RZ(z) =1-FZ(t)=exp(-¸ t)• Var(ET) · 4/T2 s0

T s0t exp(-2¸ (t-s)) ds dt

= 4/T2 s0T exp(-2¸ t) s0

t exp(2¸ s) dx dt= 4/T2/¸ s0

T exp(-2¸ t) (exp(2¸ t)-1) dt=4/T2/¸ s0

T (1-exp(-2¸ t)) dt=4/T2/¸ (T+1/2¸ (1-exp(-¸ T))=4/¸ (1/T+1/2T2¸ (1-exp(-¸ T)) ¼ 4/¸/T

Sample meansfor white noise

• w is white noise• B(t)=s0

t w(t) dt• B(t) is Brownian motion (Wiener process)• Var(B(t))=´ t (by definition)• ET=1/T s0

t w(t) dt = 1/T B(T)

• Var(ET)=1/T2 var(B(T))=1/T2 ´ T = ´/T• 2 state Markov like white noise

Sample meansfor Brownian motion

• B(s)=B(t)+sts w(x) dx = B(t)+b s ¸ t

• b and B(t) are independent• CBB(t,s) = E(B(t)B(s)) = E(B(t) (B(t)+b)) =E(B2(t))=´ t =

´ min{t,s} !!!• ET=1/T s0

t B(t) dt• Var(ET)=1/T2 s0

T s0t CBB(t,s) ds dt

• =1/T2 s0T s0

T ´ min{t,s} ds dt• =2/T2 s0

T s0t ´ s ds dt

• =1/T2 s0T ´ t2 dt

• =1/T2/3 ´ T3 = 1/3 ´ T

Sample meansfor renewal with Pareto distributions

• Var(ET) = 4/T2 s0T s0

t RZ(t-s) ds dt

• RZ(z) = C z1-®

• Var(ET) = 4C/T2 s0T s0

t (t-s)1-® ds dt= -4C/T2 s0

T st0 x1-® dx dt

= 4C/T2/(2-®) s0T t2-® dt

= 4C/T2/(2-®)/(3-®) T3-®

= 4C/(2-®)/(3-®) T1-®

For ® ¼ 1 right between white noise (s0) and Brownian motion (s-1) fractional Brownian motion (s-1/2) BH(t)=s0

t (t-s)H-1/2 w(s) ds

Self similarity• A process X is self similar with Hurst parameter H iff:

a-H X(at) is equivalent to X(t)(up to finite joint distributions)

CXX(s) = E(X(0)X(s))= (1/s)-2H E(X(0/s)X(s/s)) = s2H CXX(0,1) CXX(t,s)= E(X(t)X(s))= (1/s)-2H E(X(t/s)X(s/s))=

= s2H CXX(t/s,1) -> s2H CXX(0,1) for t/s -> 0

CXX(t,t+s)= E(X(t)X(t+s))= E(X(t/(t+s))X((t+s)/(t+s)))= = (t+s)2H CXX(t/(t+s),1) -> (t+s)2H CXX(1,1) for t -> 1

Self similarity• Y(n)=X(n)-X(n-1)• CYY(1,m) = E((X(1)-X(0))(X(1+m)-X(m))) =E(X(1)X(1+m))+E(X(0)X(m))-E(X(1)X(m))-E(X(0)X(1+m)) = m2H (E(X(1/m)X(1/m+1))+E(X(0)X(1))-E(X(1/m)X(1))-E(X(0)X(1/m+1))) = m2H (CXX(1/m,1/m+1)+ CXX(0,1)- CXX(1/m,1)- CXX(0,1/m+1)) = m2H (CXX(1/m,1/m+1) - CXX(0,1/m+1) + CXX(0,1)- CXX(1/m,1))

¼ m2H (CXX(0,1)+1/m D1+1/m D2 + D12/2 1/m2 + D21/2 1/m2 + D11/2 1/m2 + D22/2 1/m2 -(CXX(0,1)+1/m D2 + D22/2 1/m2) + CXX(0,1) -(CXX(0,1)+1/m D1+ D11/2 1/m2 ))

= m2H (D12/2 1/m2 + D21/2 1/m2) = m2H-2 (D12+ D21)/2

Frequency Domain

• RZ(z) = C z1-®

• log(RZ(z))=log(C) + (1-®) log(z)

• CYY(1,m) = K m2H-2

• SYY(!) = C ! 1-2H

• log(SYY(!))=log(C) + (1-2H) log(!)

Distribution of files sizes

Time averages (aggregated)

Time averages (cont’d)

Aggregated statistics

Estimating the Hurst parameter

Miniproject• Make a statistic on the filesizes of your file system.• Check for power tailed behaviour.• Simulate an M/G/1 queue with power tailed service times.• Compare with results for an M/M/1 queue with the same load:

½ = mean service time/mean interarrival time

• Simulate an alternating renewal process with power tailed ”ON” distribution.

• Compute an autocorrelation estimate.• Compute estimates of the 1-step increments of sample means.• Compute a power spectrum estimate.

Summary LRD

• Let fZ(z)=K z-® I(z ¸ z0) ®>1

• RZ(z) ¼ (z/z0)1-®

• CII(s) ¼ 2RZ(2s) · (2s)2(H-1) (H - Hurst parameter, I indicates on period)

• H>1/2 : Long Range Dependence (LRD)• LRD ® < 2• log(RZ(z))=log(C) + (1-®) log(z)

Summary M/G/1

• M/M/1:Q=½/(1-½)

• M/G/1: (Pollachek-Kinchine)Q=½ + (½2 + ¸2 var(S))/2/(1-½)

• fS(s)=K s-® I(s ¸ s0) ®>1

• E(S2) = K s s01 s2 s-® ds = K s s0

1 s2-® ds = [s3-®]s0

1 /(3-®)

Summary SS

• A process X is self similar with Hurst parameter H iff: a-H X(at) is equivalent to X(t)(up to finite joint distributions)

• Y(n)=X(n)-X(n-1)• CYY(1,m) ¼ m2H-2 (D12+ D21)/2

• CYY(1,m) = K m2H-2

• SYY(!) = C ! 1-2H

• log(SYY(!))=log(C) + (1-2H) log(!)

Summary (Sample means)

• ET = 1/T s0T I(t) dt

• I(t) indicates on-state• Var(ET)= 4/T2 s0

T s0t RZ(2(t-s)) ds dt

• ¼ 4/¸/T (2 state Markov)• = ´/T (White noise)• = 1/3 ´ T (Brownian motion)• = 4C/(2-®)/(3-®) T1-® (Power tail)