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6. Birth and Death Processes

6.1 Pure Birth Process (Yule-Furry Process)

6.2 Generalizations

6.3 Birth and Death Processes

6.4 Relationship to Markov Chains

6.5 Linear Birth and Death Processes

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6.1 Pure Birth Process (Yule-Furry Process)Example. Consider cells which reproduce according to the followingrules:

i. A cell present at time t has probability h + o(h) of splitting in twoin the interval (t, t + h)

ii. This probability is independent of age.

iii. Events between different cells are independent

Time>

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Non-Probablistic Analysis

n(t) = no. of cells at time t

n (t)( t) births occur in (t, t + t)

where = birth rate per single cell.

n(t + t) = n(t) + n(t) t

n(t + t) n(t)

t n (t) = n(t)

orn (t)n(t)

=ddt

log n(t) =

log n(t) = t + c

n(t) = Ke t , n(0) = n0

n(t) = n0et

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Probabilistic AnalysisN (t) = no. of cells at time tP {N (t) = n} = P n (t)Prob. of birth in (t, t + h) if {N (t) = n} = nh + o(h)

P n (t + h) = P n (t)(1 nh + o(h))

+ P n 1(t)(( n 1)h + o(h))

P n (t + h) P n (t) = nhP n (t) + P n 1(t)(n 1)h + o(h)

P n (t + h) P n (t)h

= nP n (t) + P n 1(t)(n 1) + o(h) as h 0

P n (t) = nP n (t) + ( n 1)P n 1(t)

Initial condition P n 0 (0) = P {N (0) = n 0} = 1

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P n

(t) = nP n (t) + ( n 1)P n 1(t); P n 0 (0) = 1

Solution:

(1) P n (t) =n 1

n n0e n 0 t (1 e t )n n 0 n = n0 , n 0 + 1 , . . .

Solution is negative binomial distribution; i.e. Probability of obtainingexactly n0 successes in n trials.

Suppose p = prob. of success and q = 1 p = prob. of failure. Then inrst (n 1) trials results in (n 0 1) successes and (n n 0) failuresfollowed by success on n th trial; i.e.

(2)

n 1n0 1 pn 0 1qn n 0 p = n 1n n0 p

n 0 qn n 0 n = n0 , n 0 + 1 ,...

If p = e t and q = 1 e t

(2) is same as (1).

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Yule studied this process in connection with theory of evolution; i.e.population consists of the species within a genus and creation of newelement is due to mutations. Neglects probability of species dying out andsize of species.

Furry used same model for radioactive transmutations.

Notes on Negative Binomial Distribution

The geometric distribution is dened as the number of trials to achieveone success for a series of Bernoulli trials; i.e.

Geometric Distribution: P {N = n} = pqn 1 , n = 1 , 2, . . .

N is number of trials for 1 success

N (s) = E (e sN ) = p

n =1e sn qn 1 =

pq

n =1(e s q)n .

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But

n =1

(e s q)n = e s q/ (1 e s q)

N (s) = pq e

s q

1 e s q= pe

s

1 e s q= pz

1 qzif z = e s

N (z) = pz(1 qz) 1

N (z) = p{(1 qz) 1 + z(1 qz) 2q}

N (1) = p{(1 q) 1 + q(1 q) 2} = 1 +

q

p=

p + q

p=

1

pSimilarly N (1) =

2 p2

2 p V (n) =

1 p2

1 p

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Theorem: If N i (i = 1 , 2, . . . , n 0) are iid geometric random variableswith parameter p, then N = N 1 + N 2 + . . . + N n 0 is a negative binomialdistribution having generating function

N (z) =pz

1 qz

n 0

, z = e s

.. E (N ) = n0 /p, V (N ) = n01

p2 1 p

If p = e t and N (t) is a pure birth process

E [N (t)] = n0et

V [N (t)] = n0[e2t et ]

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6.2 Generalization

In Poisson Process, the prob. of a change during (t, t + h) is independentof number of changes in (0, t ). Assume instead that if n changes occur in(0, t ), the probability of new change to n + 1 in (t, t + h) is n h + o(h).The probability of more than one change is o(h). Then

P n (t + h) = P n (t)(1 n h) + P n 1(t) n 1h + o(h), n = 0P 0(t + h) = P 0(t)(1 0h) + o(h)

P n (t) = n P n (t) + n 1P n 1(t)

P 0(t) = 0P 0(t).

Equations can be solved recursively with P 0(t) = P 0(0)e 0 t .

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If the initial condition is P n 0 (0) = 1 , then the resulting equations are:

P n (t) = n P n (t) + n 1P n 1(t), n > n 0

P n 0 (t) = n 0 P n 0 (t)

Pure birth process assumed n = n .

Change of Language

If n transitions take place during (0, t ), we may refer to the process asbeing in state E n . Changes occur E n E n +1 E n +2 . . . for thepure birth process.

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For n = 0

P 0(t + h) = P 0(t){1 0h} + P 1(t)1h + o(h)

P 0(t) = 0P 0(t) + 1P 1(t)

If the initial conditions P n 0 (0) = 1 in which case 0 in above is replaced

by n0 .If 0 = 0 , then E 0 E 1 is impossible and E 0 is an absorbing state.If 0 = 0 , then P 0(t) = 1P 1(t) 0 so that P 0(t) increasesmonotonically.

Note: limt

P 0(t) = P 0( ) = Probability of being absorbed.

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P 0(t) = 0P 0(t) + 1P 1(t)P n (t) = ( n + n )P n (t) + n 1P n 1(t) + n +1 P n +1 (t)

As t , P n (t) P n (limit ) hence P 0(t) = 0 for large t and

P n (t) = 0 for large t . Therefore

0 = 0P 0 + 1P 1

P 1 =0

1P 0

0 = (1 + 1)P 1 + 0P 0 + 2P 2

P 2 =01

21P 0

P 3 =012123

P 0 etc.

Note that dependence on initial conditions has disappeared.

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6.4 Relation to Markov Chains

Dene for t

P (E n +1 |E n ) = Prob. of transition E n E n +1= Prob. of going to E n +1 conditional on being in E n .

Similarly dene P (E n 1 |E n ).

P (E n +1 |E n ) n , P (E n 1 |E n ) n

P (E n +1 |E n ) = n

n + n, P (E n 1 |E n ) =

n n + n

Same conditional probabilities hold if it is given that a transition will takeplace during (t, t + h) conditional on being in E n .

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P n (t) = ( + )nP n (t) + (n 1)P (n 1) (t) + (n + 1) P n +1 (t)

Dene Mean by M (t) =

n =1nP n (t)

and consider M (t) =

1

nP n (t).

M (t) = ( + )

1

n2P n (t) +

1

(n 1)n P (n 1) (t)

+

1 (n + 1) nP n +1 (t)

Write (n 1)n = ( n 1)2 + ( n 1), (n + 1) n = ( n + 1) 2 (n + 1)

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M (t) = ( + )

1n2P n (t)

+

1(n 1)2P n 1(t) +

1(n + 1) 2P n +1 (t) + 1 P 1(t)

+

1

(n 1)P n 1(t)

1

(n + 1) P n +1 (t) + P 1(t)

M (t) =

1

nP n (t)

1

nP n (t)

= ( )M (t)

M (t) = n0e( ) t if P n 0 (0) = 1

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M (t) = n0e( ) t

M (t) 0or

depending on

< or

> .

Similarly if M 2(t) =

1n2P n (t) one can show

M 2(t) = 2( )M 2(t) + ( + )M (t)

and when > , the variance is

n0e2( ) t 1 e( ) t +

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