Generalizations of Poisson Process i.e., P k (h) is independent of n as well as t. This process can...

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Generalizations of Poisson Process i.e., P k (h) is independent of n as well as t. This process can be generalized by considering λ no more a constant but a function of n or t or both. The generalized process is again Markovian in nature. (1)

Transcript of Generalizations of Poisson Process i.e., P k (h) is independent of n as well as t. This process can...

Generalizations of Poisson Process

i.e., Pk(h) is independent of n as well as t. This process can be generalized by considering λ no more a constant but a function of n or t or both. The generalized process is again Markovian in nature.

(1)

Generalizations of Poisson Process

This generalized process has excellent interpretations in terms of birth-death processes. Consider a population of organisms, which reproduce to create similar organisms. The population is dynamic as there are additions in terms of births and deletions in terms of deaths. Let n be the size of the population at instant t. Depending upon the nature of additions and deletions in the population, various types of processes can be defined.

Pure Birth Process Let λ is a function of n, the size of the population at instant t. Then

n ≥ 0 and λ0 may or may not equal to zero

(2)

Birth and Death Process Now, along with additions in the population, we consider deletions

also, i.e., along with births, deaths are also possible. Define

(3)

(2) and (3) together constitute a birth and death process. Through a birth there is an increase by one and through a death, there is a decrease by one in the number of “Individuals”. The probability of more than one birth or more than one death is O(h). We wish to obtain

Birth and Death Process

To obtain the differential-difference equation for Pn(i), we consider the time interval (0, t+h) = (0, t) + [t, t+h)

Since, births and deaths, both are possible in the population, so the event {N(t+h) = n , n ≥ 1} can occur in the following mutually exclusive ways:

Birth and Death Process

Birth and Death Process

(4)

(6)

(5) and (7) represent the differential-difference equations of a birth and death process which play an important role in queuing theory.

(5)As h → o, we have

(7)

(8)

Birth and Death Process We make the following assertion:

Births and Death Rates Depending upon the values of λ n and μn , various types of birth and death processes can be defined.

State (0) is absorbing state.

Birth and Death Process When the specific values of both λ n and μn are considered simultaneously, we get

the following processes:

Birth and Death Process

If the initial population size is i, i.e, X(0) = i, then we have the initial condition Pi(0) = 1 and Pn(0) = 0, n ≠ i.

(9)

(10)

From Equ.

(5) and (7)

(9) (10)

(11)

(12)

n =0

n =1

n

Sn

Some Notifications they may help

Birth and Death Process

constant

9 10

(13)

9

Birth and Death Process

The second moment M2(t) of X(t) can also be calculated in the same way.

(14)

(13)

Birth and Death Process

(12)

Birth and Death Process

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(16)<

Birth and Death Process

Birth and Death Process

Finally, the birth and death process is a special type of continuous time Markov process with discrete state space 0, 1, 2, … such that the probability of transition from state i to state j in (∆t) is O(∆t) whenever │i - j│≥ 2. In other words, changes take place through transitions only from a state to its immediate neighbouring state.

Thanks for your attention

Some Notifications they may help

In case we have:

1

2

a

b

c If we adding the part P1(t) for both sides as we have in our equation we will get:

BACK

Birth and Death Process

0 tt+ h

P{N(t+h)= n} = P{N(t)= n-i+j}* P{N(h)= i+j} =Pn-

i+j(t)*Pi(h)*Pj(h)

E00

E10

E01

E11

n

n

n-1

n+1

n

i

0

1

0

1

j

0

0

1

1

P{N(t)= n-i+j} = Pn-

i+j(t)

t

h

P{Eij(h)} = Pi(h)*Pj(h)

Eij

t h

P{N(t+h)= n} = P{N(t)= n-i+j} * P{Eij(h)} = Pn-i+j(t) * Pi(h)*Pj(h) = Pn(t+h)

P{N(t+h)= n} = Pn(t) {1-λn h + O(h)} {1- μn h + O(h)}

P{N(t+h)= n} = Pn-1(t) {λn-1 h + O(h)} {1- μn-1 h + O(h)}

P{N(t+h)= n} = Pn+1(t) {1- λn+1 h + O(h)} {μn+1 h + O(h)}

P{N(t+h)= n} = Pn(t) { λn h + O(h)} {μn h + O(h)}