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
230
6.1 Pure Birth Process (Yule-Furry Process)
Example. Consider cells which reproduce according to the following
rules:
i. A cell present at time t has probability λh + o(h) of splitting in two
in the interval (t, t + h)
ii. This probability is independent of age.
iii. Events between different cells are independent
Time>
231
Non-Probablistic Analysis
n(t) = no. of cells at time t
⇒ λn(t)∆(t) births occur in (t, t + ∆t)
where λ = birth rate per single cell.
n(t + ∆t) = n(t) + n(t)λ∆t
n(t + ∆t) − n(t)
∆t
→ n (t) = n(t)λ
or
n (t)
n(t)
=
d
dt
log n(t) = λ
log n(t) = λt + c
n(t) = Keλt
, n(0) = n0
n(t) = n0eλt
232
Probabilistic Analysis
N(t) = no. of cells at time t
P{N(t) = n} = Pn(t)
Prob. of birth in (t, t + h) if {N(t) = n} = nλh + o(h)
Pn(t + h) = Pn(t)(1 − nλh + o(h))
+ Pn−1(t)((n − 1)λh + o(h))
Pn(t + h) − Pn(t) = −nλhPn(t) + Pn−1(t)(n − 1)λh + o(h)
Pn(t + h) − Pn(t)
h
= −nλPn(t) + Pn−1(t)(n − 1)λ + o(h) as h → 0
Pn(t) = −nλPn(t) + (n − 1)λPn−1(t)
Initial condition Pn0
(0) = P{N(0) = n0} = 1
233
Pn(t) = −nλPn(t) + (n − 1)λPn−1(t); Pn0 (0) = 1
Solution:
(1) Pn(t) =
n − 1
n − n0
e−λn0t
(1 − e−λt
)n−n0
n = n0, n0 + 1, . . .
Solution is negative binomial distribution; i.e. Probability of obtaining
exactly n0 successes in n trials.
Suppose p = prob. of success and q = 1 − p = prob. of failure. Then in
first (n − 1) trials results in (n0 − 1) successes and (n − n0) failures
followed by success on nth
trial; i.e.
(2)
n − 1
n0 − 1
pn0−1
qn−n0
· p =
n − 1
n − n0
pn0
qn−n0
n = n0, n0 + 1, ...
If p = e−λt
and q = 1 − e−λt
⇒ (2) is same as (1).
234
Yule studied this process in connection with theory of evolution; i.e.
population consists of the species within a genus and creation of new
element is due to mutations. Neglects probability of species dying out and
size of species.
Furry used same model for radioactive transmutations.
Notes on Negative Binomial Distribution
The geometric distribution is defined as the number of trials to achieve
one 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
φN (s) = E(e−sN
) = p
∞
n=1
e−sn
qn−1
=
p
q
∞
n=1
(e−s
q)n
.
235
But
∞
n=1
(e−s
q)n
= e−s
q/(1 − e−s
q)
φN (s) =
p
q
·
e−s
q
1 − e−sq
=
pe−s
1 − e−sq
=
pz
1 − qz
if z = e−s
φN (z) = pz(1 − qz)−1
φN (z) = p{(1 − qz)−1
+ z(1 − qz)−2
q}
φN (1) = p{(1 − q)−1
+ q(1 − q)−2
} = 1 +
q
p
=
p + q
p
=
1
p
Similarly φN (1) =
2
p2
−
2
p
⇒ V (n) =
1
p2
−
1
p
236
Theorem: If Ni (i = 1, 2, . . . , n0) are iid geometric random variables
with parameter p, then N = N1 + N2 + . . . + Nn0
is a negative binomial
distribution having generating function
φN (z) =
pz
1 − qz
n0
, z = e−s
˙.. E(N) = n0/p, V (N) = n0
1
p2
−
1
p
If p = e−λt
and N(t) is a pure birth process
E[N(t)] = n0eλt
V [N(t)] = n0[e2λt
− eλt
]
237
6.2 Generalization
In Poisson Process, the prob. of a change during (t, t + h) is independent
of 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 λnh + o(h).
The probability of more than one change is o(h). Then
Pn(t + h) = Pn(t)(1 − λnh) + Pn−1(t)λn−1h + o(h), n = 0
P0(t + h) = P0(t)(1 − λ0h) + o(h)
⇒ Pn(t) = −λnPn(t) + λn−1Pn−1(t)
P0(t) = −λ0P0(t).
Equations can be solved recursively with P0(t) = P0(0)e−λ0t
.
238
If the initial condition is Pn0
(0) = 1, then the resulting equations are:
Pn(t) = −λnPn(t) + λn−1Pn−1(t), n > n0
Pn0
(t) = −λn0
Pn0
(t)
Pure birth process assumed λn = nλ.
Change of Language
If n transitions take place during (0, t), we may refer to the process as
being in state En. Changes occur En → En+1 → En+2 → . . . for the
pure birth process.
239
6.3 Birth and Death Processes
Consider transitions En → En−1 as well as En → En+1 if n ≥ 1.
If n = 0, we only allow E0 → E1.
Assume that if the process at time t is in En, then during (t, t + h) the
transitions En → En+1 have prob. λnh + o(h), En → En−1 have
prob. µnh + o(h) and Prob. more than 1 change occurs = o(h)
Pn(t + h) = Pn(t){1 − λnh − µnh}
+ Pn−1(t){λn−1h} + Pn+1(t){µn+1h} + o(h)
⇒ Pn(t) = −(λn + µn)Pn(t) + λn−1Pn−1(t) + µn+1Pn+1(t)
240
For n = 0
P0(t + h) = P0(t){1 − λ0h} + P1(t)µ1h + o(h)
⇒ P0(t) = −λ0P0(t) + µ1P1(t)
If the initial conditions Pn0
(0) = 1 in which case 0 in above is replaced
by n0.
If λ0 = 0, then E0 → E1 is impossible and E0 is an absorbing state.
If λ0 = 0, then P0(t) = µ1P1(t) ≥ 0 so that P0(t) increases
monotonically.
Note: lim
t→∞
P0(t) = P0(∞) = Probability of being absorbed.
241
P0(t) = −λ0P0(t) + µ1P1(t)
Pn(t) = −(λn + µn)Pn(t) + λn−1Pn−1(t) + µn+1Pn+1(t)
As t → ∞, Pn(t) → Pn (limit) hence P0(t) = 0 for large t and
Pn(t) = 0 for large t. Therefore
0 = −λ0P0 + µ1P1
⇒ P1 =
λ0
µ1
P0
0 = −(λ1 + µ1)P1 + λ0P0 + µ2P2
⇒ P2 =
λ0λ1
µ2µ1
P0
⇒ P3 =
λ0λ1λ2
µ1µ2µ3
P0 etc.
Note that dependence on initial conditions has disappeared.
242
6.4 Relation to Markov Chains
Define for t → ∞
P(En+1|En) = Prob. of transition En → En+1
= Prob. of going to En+1 conditional on being in En.
Similarly define P(En−1|En).
P(En+1|En) ∝ λn, P(En−1|En) ∝ µn
⇒ P(En+1|En) =
λn
λn + µn
, P(En−1|En) =
µn
λn + µn
Same conditional probabilities hold if it is given that a transition will take
place during (t, t + h) conditional on being in En.
243
6.5 Linear birth and death processes
λn = nλ , µn = nµ
⇒ P0(t) = µP1(t)
Pn(t) = −(λ + µ)nPn(t) + λ(n − 1)Pn−1(t) + µ(n + 1)Pn+1(t)
Steady state behavior is characterized by
lim
t→∞
P0(t) = 0 ⇒ P1(∞) = 0
Similarly as t → ∞ Pn(∞) = 0
If P0(∞) = 1 ⇒ Probability of ultimate extinction is 1.
If P0(∞) = P0 < 1, the relations P1 = P2 = P3 . . . = 0 imply with
prob. 1 − P0 the population can increase without bounds. The population
must either die out or increase indefinitely.
244
Pn(t) = −(λ + µ)nPn(t) + λ(n − 1)P(n−1)(t) + µ(n + 1)Pn+1(t)
Define Mean by M(t) =
∞
n=1
nPn(t)
and consider M (t) =
∞
1
nPn(t).
M (t) = −(λ + µ)
∞
1
n2
Pn(t) + λ
∞
1
(n − 1)n P(n−1)(t)
+ µ
∞
1
(n + 1)nPn+1(t)
Write (n − 1)n = (n − 1)2
+ (n − 1), (n + 1)n = (n + 1)2
− (n + 1)
245
M (t) = − (λ + µ)
∞
1
n2
Pn(t)
+ λ
∞
1
(n − 1)2
Pn−1(t) + µ
∞
1
(n + 1)2
Pn+1(t) + 1 · P1(t)
+ λ
∞
1
(n − 1)Pn−1(t) − µ
∞
1
(n + 1)Pn+1(t) + P1(t)
⇒ M (t) = λ
∞
1
nPn(t) − µ
∞
1
nPn(t)
= (λ − µ)M(t)
M(t) = n0e(λ−µ)t
if Pn0
(0) = 1
246
M(t) = n0e(λ−µ)t
M(t) → 0 or ∞ depending on λ < µ or λ > µ .
Similarly if M2(t) =
∞
1
n2
Pn(t) one can show
M2(t) = 2(λ − µ)M2(t) + (λ + µ)M(t)
and when λ > µ, the variance is
n0e2(λ−µ)t
1 − e(µ−λ)t λ + µ
λ − µ
247