Moment distributions of phasetype Mogens Bladt Institute for - PowerPoint PPT Presentation

Moment distributions of phase–type Mogens Bladt Institute for Applied Mathematics and Systems National Autonomous University of Mexico Sandbjerg, 2/8/2011 Joint work with Bo Friis Nielsen

Outline - In this talk we consider moment distributions which underlying distribution is of either phase–type or matrix–exponential. - We show that moment distributions of any order are again phase–type or matrix–exponential. - Moment distributions have applications in various fields like demography and engineering. - We especially focus on demographic applications relating to the Lorenz curve and Gini index in which case explicit formulas may be obtained.

Moment distributions - Let f be the density a distribution on [0 , ∞ ). Let � ∞ 0 x n f ( x ) dx be its n ’th moment. µ n = - Then g n ( x ) = x n f ( x ) µ n is a density and is called the n ’th moment distribution of f . - We shall now assume that f is either of phase–type or matrix–exponential. - We shall start with the first order moment distribution.

First order moment distribution - Let f be a density on [0 , ∞ ) and let F denotes its corresponding distribution function. - Consider a stationary renewal process with inter–arrival distribution F . - Hence the renewal process is delayed with initial arrival distribution given by the density ¯ f e ( x ) = 1 − F ( x ) F ( x ) = . µ 1 µ 1 - Let F e denote the corresponding distribution function. - Let A t be the age of the process at time t (time from previous arrival) and R t the residual life–time (time until next arrival).

First order moment distributions - Then P ( A t > x , R t > y ) = ¯ F e ( x + y ) . - Differentiating twice w.r.t. x and y , f ( A t , R t ) ( x , y ) = f ( x + y ) . µ 1 - From this formula we read that A t and R t have the same marginal distribution. - The spread S t = A t + R t has density � x f ( A t , R t ) ( x − t , t ) dt = xf ( x ) f S t ( x ) = . µ 1 0

Phase–type distributions X t p + 1 p 5 4 3 2 1 t τ

Phase–type distributions - A Phase–type distribution is the time until absorption in a Markov jump process with finitely many states, one of which is absorbing and the rest being transient. - We write τ ∼ PH( π, T ) (note that t = − Te , where e is the column vector of ones. - Phase-type distributions is a flexible tool in applied probability. Allows for many closed form solutions to complex problems. - They are dense in the class of distributions on the positive reals. - They are, however, light tailed.

An example of use A general and easy to prove result states that � � e Ts e − e Ts e P s = e Λ s = . 0 0 Let τ ∼ PH( π, T ). Let f denote density of τ . Then f ( x ) dx = P ( τ ∈ ( x , x + dx ]) p � π i p x = ij t j dx i , j =1 � e Tx � � = π i ij t j dx i , j =1 π e Tx tdx . = Hence f ( x ) = π e Tx t

Renewal theory Consider a renewal process with inter–arrival times T 1 , T 2 , ... being i.i.d. ∼ PH( π, T ). X t p 3 2 1 t τ 1 τ 1 + τ 2 Renewal density: u ( x ) =probability of an arrival in [ x , x + dx ). The concatated process is a Markov process { J t } t ≥ 0 with intensity matrix R = T + t π : r ij dx = t ij dx + t i dx π j .

Renewal theory Transition probabilities of { J t } t ≥ 0 : P s = exp (( T + t π ) s ) . Hence p � π i p x u ( x ) dx = ij t j dx i , j =1 p � e ( T + t π ) x � � = π i ij t j dx i , j =1 π e ( T + t π ) x tdx . = Hence u ( x ) = π e ( T + t π ) x t .

Stationary renewal process - A stationary renewal process with phase–type inter–arrival times T 2 , T 3 , ... i.i.d. ∼ PH( α, T ) is a delayed renewal process with T 1 ∼ PH( π 1 , T ), where π 1 = α T − 1 α T − 1 e . - π 1 is the stationary distribution of the Markov jump process with intensity matrix T + t α . - π 1 is also the stationary distribution for the time reversed process. - Time reversing a PH distribution essentially works as for Markov jump processes.

Reversed Markov jump initial Markov jump process process generating A t distribution generating R t π 1 A t R t arrival arrival • • time t π 1 Reversed Markov jump exit=initial Reversed Markov jump process generating A t distribution process generating R t

First moment distribution of phase–type - Let f be the density of a PH( π, S ). - Let π 1 = π S − 1 /π S − 1 e . Then A t , R t ∼ PH( π 1 , S ) or A t , R t ∼ PH( π 1 , ˆ S ) where ˆ S = ∆( m 1 ) − 1 S ′ ∆( m 1 ) and m 1 = − α S − 1 . - We time reverse A t or R t . If R t ∼ PH( π 1 , S ), then we time reverse A t with the choice of representation PH( π 1 , ˆ S ). If we time reverse R t ∼ PH( π 1 , S ), then we use A t ∼ PH( π 1 , ˆ S ). - The exit distribution the A t –process is π 1 , the same as the initial distribution of R t . Hence we may generate the initial distribution of the R t process by realizing the time–reversed of A t and then realize the process of R t . The total time it takes for both processes to exit is just the spread S t .

First moment distribution of phase–type - If we reverse R t we get a representation is for the first α 1 , ˆ moment distribution (ˆ S 1 ), where � s ′ ∆( m 2 ) , 0 � α 1 = ˆ � ρ − 1 � ∆ − 1 ( m 2 ) S ′ ∆( m 2 ) 1 ∆ − 1 ( m 2 )∆( m 1 ) ˆ S 1 = , ∆ − 1 ( m 1 ) S ′ ∆( m 1 ) 0 with ρ i = α ( − S − i ) e and m i = ρ − 1 i − 1 α ( − S ) − i . - If we reverse A t we get a representation ( α 1 , S 1 ) with � � ρ − 1 α 1 = 1 α ∆( r ) , 0 � � ∆ − 1 ( r ) S ∆( r ) ∆ − 1 ( r ) S 1 = , 0 S where r = ( − S ) − 1 e .

Moment distributions based on matrix–exponentials - Let f be the density of a matrix–exponential distribution with representation ( α, S , s ) with s = − Se (the latter only being notationally convenient). - Then its n ’th moment distribution is again matrix–exponential with representation ( α n , S n , s n ), where     − S 0 ... 0 0 S � � α S − n 0 S − S ... 0 0     α n = α S − n e , 0 , ..., 0 S n =  , s n =     ... ... ... ... ... ..        0 0 0 0 S s

Moment distribution based on matrix–exponentials - This easily follows from α n e S n x s n = α S − n ( − 1) n ( − 1) n n ! α S − n e = x n α e Sx s x n α e Sx s x n S n e Sx s = . α S − n e n ! µ n - To obtain the corresponding distribution function F n we integrate partially and obtain n F n ( x ) = 1 − α S − n ( − xS ) i � e Sx e α S − n e i ! i =0 - In particular for n = 1 we get F 1 ( x ) = 1 − α S − n � � e Sx e + xe Sx s α S − n e

Higher order moment distributions of phase–type - In principle we now conclude that moment distributions of any order are again phase–type if the original distribution is. - This follows trivially from the n ’th order moment distributions is the first order moment distribution of the n − 1’th order moment distribution! - Hence, in principle, there is an algorithm for generating a PH representation. - The order, however, will blow up unnecessarily. - The following result provides a lower order representation of the n ’th moment distribution, but we lack a probabilistic proof :-(

Higher order moment distributions of phase–type Consider a phase–type distribution with representation ( α, S ). Then the n ’th order moment distribution is again of phase–type with representation ( α n , S n ), where � ρ n +1 � α n = π n +1 • s , 0 , . . . , 0 ρ n  C n +1 D n +1 0 . . . 0 0  0 . . . 0 0 C n D n     0 0 C n − 1 . . . 0 0     S n = . . . .. . .. . .  . . . . . . . .  . . . . . .       0 0 0 . . . C 2 D 2   0 0 0 . . . 0 C 1 and ρ i = µ i / i ! = α ( − S ) − i e are the reduced moments, i α ( − S ) − i , C i = ∆( π i ) − 1 S ′ ∆( π i ) , D i = ρ i − 1 π i = ρ − 1 ∆( π i ) − 1 ∆( π i − 1 ) . ρ i

Lorenz curve and Gini index - If F is a distribution function and F 1 the corresponding first moment distribution, then the parametric curve t → ( F ( t ) , F 1 ( t )) is called the Lorenz curve or concentration curve. - By definition, dF 1 ( x ) dF ( x ) = x > 0 µ 1 and d 2 F 1 ( x ) µ 1 dF ( x ) = 1 1 dx dF 2 ( x ) = f ( x ) > 0 . µ 1 - Hence the Lorenz curve is convex. For the ME (and PH) we get � �� 1 − α e St e , 1 − α S − 1 � e St e + te St s t → . α S − 1 e

Gini index F 1 ( x ) y = x 1 1 2 G F ( x ) 1

Gini index - The Gini index is defined as twice the area enclosed by the Lorenz curve and the line y = x . - The Lorenz curve starts in (0 , 0) and ends in (1 , 1). Since the curve is convex it “lies under” y = x . - The area under the y = x for x = 0 to x = 1 is 1 / 2. - The area A under the Lorenz curve is � ∞ F ′ ( t ) F 1 ( t ) dt A = 0 � ∞ � � α e St s 1 − α 1 e S 1 t e = dt 0 � ∞ α e St s α 1 e S 1 t edt = 1 − 0 1 + ( α ⊗ α 1 ) ( S ⊕ S 1 ) − 1 ( s ⊗ e ) =