Renewal processes

Interarrival times {0,T 1,T 2,..} is an i.i.d. sequence with a common distribution fct. F S i =  j=1 i T j {S i } is a nondecreasing, positive sequence of reneval times (point) The distribution of S i 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) = max i {S i · 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(S n · t and S n+1 > t) = P(S n+1 > t) - P(S n > t and S n+1 > t)

The counting process P(N(t)=n)=P(S n · t and S n+1 > t) = P(S n+1 > t) - P(S n > t and S n+1 > t) Now S n > t => S n+1 > t so P(S n > t and S n+1 > t) = P(S n > t) Altogether P(S n · t and S n+1 > t) = P(S n+1 > t) - P(S n > t) = P(S n · t) - P(S n+1 · t) = F (n) (t) – F (n+1) (t)

The counting process M(t) =  n n P(N(t)=n) =  n n P(S n · t and S n+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(h 2 ) 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( s 0 t M(t-u) f(u) du) = f(t) + s 0 t m(t-u) f(u) du = f(t) + (f * m)(t)

Recurrence times Backward recurrence time (age): A(t) = t – S N(t) Forward recurrence time (excess): Y(t) = S N(t)+1 –t F A,t (a) = P(A(t) · a) F Y,t (y) = P(Y(t) · y)

Distribution of age F 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) = s 0 1 P(A(t) · a | S 1 =s) f(s) ds = s 0 t P(A(t) · a | S 1 =s) f(s) ds + s t 1 P(A(t) · a | S 1 =s) f(s) ds = s 0 t P(A(t-s) · a) f(s) ds + s t 1 P(A(t) · a | S 1 =s) f(s) ds = s 0 t F A,t-s (a) f(s) ds + s t 1 I(t · a) ¢ f(s) ds = s 0 t F A,t-s (a) f(s) ds + I (t · a} R(t) = (F A,. (a) * f)(t) + I(t · a) R(t)

Distribution of excess F Y,t (y) = P(Y(t) · y) We condition on the first renewal, i.e. P(Y(t) · y) = s 0 1 P(Y(t) · y | S 1 =s) f(s) ds = s 0 t P(Y(t) · y | S 1 =s) f(s) ds + s t 1 P(Y(t) · y | S 1 =s) f(s) ds = s 0 t P(Y(t-s) · y) f(s) ds + s t 1 P(Y(t) · y | S 1 =s) f(s) ds = s 0 t F Y,t-s (y) f(s) ds + s t 1 I(s-t · y) ¢ f(s) ds = (F Y,. (y) * f)(t) + s t 1 I(s · (t+y)) ¢ f(s) ds = (F Y,. (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 density in general m(t) = f(t) + (f * m)(t)  m(s) = f(s) + f(s) m(s)  m(s)=f(s)/(1-f(s)) lim t -> 1 m(t) = lim s -> 0 s m(s) = lim s -> 0 s f(s)/(1-f(s)) = (l’Hospital) lim s -> 0 d/ds (s f(s)) / lim s -> 0 d/ds (1-f(s)) = f(0)/ ((d/ds -f(s))| s=0 ) = 1/E(T i ) !!!

Example (Poisson) m(t) = ¸ F A,t (a) = (F A,. (a) * f)(t) + I(t · a) R(t) F Y,t (y)= (F Y,. (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) F A,t (a) = I(t · a) R(t) + ¸ s 0 t I(s · a) R(s) ds = I(t · a) R(t) + ¸ s 0 min(t,a) R(s) ds = I(t · a) exp(- ¸ t)+ (1-exp(-min(t,a))) F Y,t (y) = (F Y,. (y) * f)(t) + F(t+y)-F(t) = F(t+y)-F(t) + ¸ s 0 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 S n-1 SnSn T n = Z n + Y n ZnZn YnYn ONOFF

Alternating renewal process I(t) = I(S N(t) < t · S N(t) +Z N(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) = s 0 1 P(I(t)=1 | S 1 =s) f(s) ds = s 0 t P(I(t)=1 | S 1 =s) f(s) ds + s t 1 P(I(t)=1 | S 1 =s) f(s) ds = s 0 t P(I(t-s)=1) f(s) ds + s t 1 P(t · Z 1 | S 1 =s) f(s) ds = (O * f)(t) + s t 1 P(Z 1 ¸ t| S 1 =s) f(s) ds

Alternating renewal process O(t) = (O * f)(t) + s t 1 P(Z 1 ¸ t| S 1 =s) f(s) ds = (O * f)(t) + s 0 1 P(Z 1 ¸ t| S 1 =s) f(s) ds = (O * f)(t) + P(Z 1 ¸ t) = (O * f)(t) + 1-F Z (t) O(s)=1-F Z (s) + O(s)*f(s)

Example (2 state Markov) 2 exponential distributions F Z (t)=1-exp(- ¸ t) F Y (t)=1-exp(- ¹ t) f(t) = ¸ ¹ s 0 t exp(- ¸ (t-s)) exp(- ¹ s) ds E(T)=E(Y)+E(Z)=1/ ¹ + 1/ ¸ lim t -> 1 =1/E(T)=1/(1/ ¹ +1/ ¸ ) O(s)=1-F Z (s) + O(s)*f(s)  O(s)=(1-F Z (s))/(1-f(s)) or O(s) = 1-F Z (s) + (1-F Z (s)) m(s) lim t -> 1 O(t) = lim s -> 0 s O(s) = lim s -> 0 s(1-F Z (s)) + s(1-F Z (s)) m(s) = lim s -> 0 (1-F Z (s)) lim s -> 0 s m(s) = s R Z (t) dt / E(T) s R Z (t) dt = s 1 ¢ R Z (t) dt = tR Z (t) + s t ¢ f Z (t) dt -> s t ¢ f Z (t) dt = E(Z) lim t -> 1 O(t) = E(Z)/E(T) !!!

Autocorrelation C II (s) = E((I t -E(I))(I t+s -E(I))) = E((I t I t+s ) – E 2 (I) E(I) = lim t -> 1 O(t) = E(Z)/E(T) E(I t I t+s )=P(I t and I t+s ) T n = Z n + Y n S N(t) =S N(t)-1 +T N(t) A(t)=t-S N(t)

Autocorrelation C II (s) = E((I t I t+s ) – E 2 (I) t lies in the 1st renewal period E(I) = E(Z)/E(T) E(I t I t+s )=P(I t and I t+s ) P(I t and I t+s ) = s P(t+s · Z 1 | S 1 =x) + P(t · Z 1 and t+s ¸ S 1 | S 1 =x) O(t+s-x) f(x) dx

Autocorrelation P(I t and I t+s ) = s P(t+s · Z 1 | S 1 =x) + P(t · Z 1 and t+s ¸ S 1 | S 1 =x) O(t+s-x) f(x) dx = s P(t+s · Z 1 | S 1 =x) + P(t · Z 1 and t+s ¸ x | S 1 =x) O(t+s-x) f(x) dx = s P(t+s · Z 1 | S 1 =x) + I(t+s ¸ x) P(t · Z 1 | S 1 =x) O(t+s-x) f(x) dx = s P(t+s · z | Z 1 =z, S 1 =x) f Z,S (z,x) dzdx + s I(t+s ¸ x) P(t · z| Z 1 =z, S 1 =x) O(t+s-x) f Z,S (z,x) dzdx = s I(t+s · z) f S|Z (z,x) f Z (z) dzdx + s I(t+s ¸ x) I(t · z) O(t+s-x) f S|Z (z,x) f Z (z) dzdx = s I(t+s · z) f Y (x-z) f Z (z) dzdx + s I(t+s ¸ x) I(t · z) O(t+s-x) f Y (x-z) f Z (z) dzdx

Autocorrelation P(I t and I t+s ) = s I(t+s · z) f Y (x-z) f Z (z) dzdx + s I(t+s ¸ x) I(t · z) O(t+s-x) f Y (x-z) f Z (z) dzdx = ss t+s x f Y (x-z) f Z (z) dzdx + s t t+s s t x O(t+s-x) f Y (x-z) f Z (z) dzdx = s t+s s z f Y (x-z) dx f Z (z) dz + s t t+s O(t+s-x) s t x f Y (x-z) f Z (z) dz dx = s t+s f Z (z) dx + s t t+s O(t+s-x) s t x f Y (x-z) f Z (z) dz dx = R Z (t+s) + s t t+s O(t+s-x) s t x f Y (x-z) f Z (z) dz dx · R Z (t+s) + s s t x f Y (x-z) f Z (z) dz dx = R Z (t+s) + P(Z 1 ¸ t) = R Z (t+s)+ R Z (t) \leq 2R Z (2s) (for large s) C II (s) = E((I t I t+s ) – E 2 (I) \leq E((I t I t+s ) · 2R Z (2s)

Example - Pareto distributions (power/heavy tails) Let f Z (z)=K z - ® I(z ¸ z 0 ) ® >1 F Z (z)= K/( ® -1) (z 0 1- ® – z 1- ® ) I(z ¸ z 0 ) K= ( ® -1)/z 0 1- ® F Z (z)= (1 – (z/z 0 ) 1- ® ) I(z ¸ z 0 ) R Z (z)=1-F Z (z) = (z/z 0 ) 1- ® + I(z · z 0 ) · (z/z 0 ) 1- ® C II (s) ¼ 2R Z (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 ® =3-2H LRD  ® < 2

Sample means E T = 1/T s 0 T I(t) dt I(t) indicates on-state Var(E T )=E(E T 2 )= 1/T 2 E(( s 0 T I(t) dt) 2 ) = 1/T 2 E( s 0 T I(t) dt s 0 T I(t) dt) = 1/T 2 E( s 0 T s 0 T I(t) I(s) ds dt) = 1/T 2 s 0 T s 0 T E(I(t) I(s)) ds dt = 1/T 2 s 0 T s 0 T C II (t-s) ds dt ¼ 1/T 2 s 0 T s 0 T 2R Z (2|t-s|) ds dt = 4/T 2 s 0 T s 0 t R Z (2(t-s)) ds dt

Sample means for 2 state Markov process Var(E T ) = 4/T 2 s 0 T s 0 t R Z (2(t-s)) ds dt R Z (z) =1-F Z (t)=exp(- ¸ t) Var(E T ) · 4/T 2 s 0 T s 0 t exp(-2 ¸ (t-s)) ds dt = 4/T 2 s 0 T exp(-2 ¸ t) s 0 t exp(2 ¸ s) dx dt = 4/T 2 / ¸ s 0 T exp(-2 ¸ t) (exp(2 ¸ t)-1) dt =4/T 2 / ¸ s 0 T (1-exp(-2 ¸ t)) dt =4/T 2 / ¸ (T+1/2 ¸ (1-exp(- ¸ T)) =4/ ¸ (1/T+1/2T 2 ¸ (1-exp(- ¸ T)) ¼ 4/ ¸ /T

Sample means for white noise w is white noise B(t)= s 0 t w(t) dt B(t) is Brownian motion (Wiener process) Var(B(t))= ´ t (by definition) E T =1/T s 0 t w(t) dt = 1/T B(T) Var(E T )=1/T 2 var(B(T))=1/T 2 ´ T = ´ /T 2 state Markov like white noise

Sample means for Brownian motion B(s)=B(t)+ s t s w(x) dx = B(t)+b s ¸ t b and B(t) are independent C BB (t,s) = E(B(t)B(s)) = E(B(t) (B(t)+b)) =E(B 2 (t))= ´ t = ´ min{t,s} !!! E T =1/T s 0 t B(t) dt Var(E T )=1/T 2 s 0 T s 0 t C BB (t,s) ds dt =1/T 2 s 0 T s 0 T ´ min{t,s} ds dt =2/T 2 s 0 T s 0 t ´ s ds dt =1/T 2 s 0 T ´ t 2 dt =1/T 2 /3 ´ T 3 = 1/3 ´ T

Sample means for renewal with Pareto distributions Var(E T ) = 4/T 2 s 0 T s 0 t R Z (t-s) ds dt R Z (z) = C z 1- ® Var(E T ) = 4C/T 2 s 0 T s 0 t (t-s) 1- ® ds dt = -4C/T 2 s 0 T s t 0 x 1- ® dx dt = 4C/T 2 /(2- ® ) s 0 T t 2- ® dt = 4C/T 2 /(2- ® )/(3- ® ) T 3- ® = 4C/(2- ® )/(3- ® ) T 1- ® For ® ¼ 1 right between white noise (s 0 ) and Brownian motion (s -1 ) fractional Brownian motion (s -1/2 ) B H (t)= s 0 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) C XX (s) = E(X(0)X(s))= (1/s) -2H E(X(0/s)X(s/s)) = s 2H C XX (0,1) C XX (t,s)= E(X(t)X(s))= (1/s) -2H E(X(t/s)X(s/s))= = s 2H C XX (t/s,1) -> s 2H C XX (0,1) for t/s -> 0 C XX (t,t+s)= E(X(t)X(t+s))= E(X(t/(t+s))X((t+s)/(t+s)))= = (t+s) 2H C XX (t/(t+s),1) -> (t+s) 2H C XX (1,1) for t -> 1

Self similarity Y(n)=X(n)-X(n-1) C YY (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)) = m 2H (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))) = m 2H (C XX (1/m,1/m+1)+ C XX (0,1)- C XX (1/m,1)- C XX (0,1/m+1)) = m 2H (C XX (1/m,1/m+1) - C XX (0,1/m+1) + C XX (0,1)- C XX (1/m,1)) ¼ m 2H (C XX (0,1)+1/m D 1 +1/m D 2 + D 12 /2 1/m 2 + D 21 /2 1/m 2 + D 11 /2 1/m 2 + D 22 /2 1/m 2 -(C XX (0,1)+1/m D 2 + D 22 /2 1/m 2 ) + C XX (0,1) -(C XX (0,1)+1/m D 1 + D 11 /2 1/m 2 )) = m 2H (D 12 /2 1/m 2 + D 21 /2 1/m 2 ) = m 2H-2 (D 12 + D 21 )/2

Frequency Domain R Z (z) = C z 1- ® log(R Z (z))=log(C) + (1- ® ) log(z) C YY (1,m) = K m 2H-2 S YY ( ! ) = C ! 1-2H log(S YY ( ! ))=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 f Z (z)=K z - ® I(z ¸ z 0 ) ® >1 R Z (z) ¼ (z/z 0 ) 1- ® C II (s) ¼ 2R Z (2s) · (2s) 2(H-1) (H - Hurst parameter, I indicates on period) H>1/2 : Long Range Dependence (LRD) LRD  ® < 2 log(R Z (z))=log(C) + (1- ® ) log(z)

Summary M/G/1 M/M/1: Q= ½ /(1- ½ ) M/G/1: (Pollachek-Kinchine) Q= ½ + ( ½ 2 + ¸ 2 var(S))/2/(1- ½ ) f S (s)=K s - ® I(s ¸ s 0 ) ® >1 E(S 2 ) = K s s0 1 s 2 s - ® ds = K s s0 1 s 2- ® ds = [s 3- ® ] 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) C YY (1,m) ¼ m 2H-2 (D 12 + D 21 )/2 C YY (1,m) = K m 2H-2 S YY ( ! ) = C ! 1-2H log(S YY ( ! ))=log(C) + (1-2H) log( ! )

Summary (Sample means) E T = 1/T s 0 T I(t) dt I(t) indicates on-state Var(E T )= 4/T 2 s 0 T s 0 t R Z (2(t-s)) ds dt ¼ 4/ ¸ /T (2 state Markov) = ´ /T (White noise) = 1/3 ´ T (Brownian motion) = 4C/(2- ® )/(3- ® ) T 1- ® (Power tail)