Download Use of Compound Poisson Processes in Biological Modeling.

Use of Compound Poisson Processes in Biological Modeling.
(Lecture notes prepared in GSI, winter 1998)
Ewa Gudowska-Nowak 1;2
1 GSI, Plankstr. 1, D-64291 Darmstadt, Germany
2 Institute of Physics, Jagiellonian University, 30-059 Krakow, Poland.
(30. November 2000)
A brief discussion of the use of Poisson distributions is presented starting
with the denition of random occurrences. Criteria for random occurrences to
be Poisson distributed are presented and followed by a discussion of Poisson
stochastic processes and their generalizations. Examples cover simple applications of the compound Poisson processes in biological modeling.
1
I. BASIC DEFINITIONS
Random events are those which can have more than one possible outcome. One can thus
associate a probability with each outcome. The outcome of a random event is not predicable,
only probabilities of possible outcomes may be specied. With a random event A one can
associate a random variable X which takes dierent possible numerical values x1 ; x2:::xn :::
depending on dierent outcomes of A. The distribution function of a random variable is
determined by assigning corrsponding probabilities P (x1); P (x2):::P (xn):::. Its derivative
P (x + x)
p(x) lim
x!1
x
(1.1)
is called a probability density function. Random variables X; Y are called statistically independent, if they joint probability density p(x; y) is completely factorizable, i.e. p(x; y) =
p(x)p(y).
A stochastic process [2{4] fx(t; !)g is composed by a family of random variables which are
indexed by time, i.e. for each time t, the random variable x(t) takes on value ! = xt with some probability. The most popular example of a stochastic process is a Brownian movement,
discovered by botanician R. Brown in 1827 [5]. Brown has observed under the miscroscope a
strong irregular motion of pollen particles on a surface of water. The trajectories of particles
in a \Brownian process" are irregular and a displacement of a Brownian particle at time t
is a probabilistic, random variable.
Given a random variable X with density p(x), the characteristic function is dened as
Z1
isX
p(x)eisxdx for X continuous
X (s) = E (e ) =
?1
X
X (s) = E (eisX ) = pk eisx
for X discrete
(1.2)
k
k
where E (Y ) stands for the expectation or mean value of a random variable Y . Function X (s)
dtermines completely the probability distribution of the random variable. In particular, if
P (X ) is continuous everywhere and dP (x) = p(x)dx, then
Z1
1
p(x) = 2
X (s)e?ixsds
(1.3)
?1
2
The characteristic function of the sum of independent random variables is the direct product
of the characteristic functions for each one of those variables. The corresponding probability
density function is given by a convolution
Z1
p(z) p(Z = X + Y ) =
(1.4)
pX (z ? s)pY (s)ds
?1
If it exists, the characteristic function X (s) generates the moments of the distribution of
a random variable X . Substituting for Z = eis in eq.(1.2), we get for the discrete random
variable X
G(Z ) = E (Z k ) =
1
X
k=0
pk Z k
(1.5)
where
pk 0;
1
X
k=0
pk = 1
(1.6)
Derivatives of G(Z ) evaluated at Z = 1 are related to the moments:
1
X
0
G (1) = kZ k?1p
G00 (1) =
1
X
k=0
k=0
k jZ =1
= E (k)
k(k ? 1)Z k?2pk jZ =1 = E [k(k ? 1)] = E (k2) ? E (k)
(1.7)
Example 1.
Poisson distribution gives the probability of nding exactly k events in a given length time,
if the events occur independently, at a constant rate. Let us place at random n points in
the time interval (0; T ) with P fk in ti g being the probability that k points will lie in the
interval ti included in (0; T ):
!
n
(1.8)
P fk in tig = k pk qn?k
where p = tT . For n ! 1, T ! 1 in such a way that
converges to a Poisson distribution:
i
P fk in tig = e?t (tki!)
i
3
k
n
T
! , the above distribution
(1.9)
Note, that the interval ti has to be small compared to T . Probability generating function
for the Poisson distribution is
G(Z ) = e(Z ?1)
(1.10)
from which direct calculations of moments gives E (k) = 2(k) = E (k2) ? E 2(k) = .
II. SUMS OF RANDOM NUMBER OF RANDOM VARIABLES
Consider now a sum SN of N independent random variables X
SN =
N
X
i=1
Xi
(2.11)
where N is a random variable with a probability generating function g(s)
g(s) =
1
X
i=0
gisi
(2.12)
Let us assume that each Xi has the same probability generating function f (s) (that means
that Xi0s are sampled from the same probability distribution function):
f (s) =
1
X
j =1
fj s j
(2.13)
By use of the Bayes rule of conditional probabilities
A and B g
ProbfA given B g P (AjB ) = ProbfProb
fB g
(2.14)
the probability that SN takes value j can be then written as
P (SN = j ) hj =
1
X
n=0
P (SN = j jN = n)P (N = n)
(2.15)
For xed value of n, using the statistical independence of Xi's, the sum SN has a probability
generating function being a direct product of f (s):
F (s) = f (s)n =
4
1
X
j =0
Fj sj
(2.16)
from which it follows that P (SN = j jN = n) = Fj . The formula (2.15) can be then rewritten
as
hj =
1
X
n=0
Fj gn
(2.17)
The compound probability generating function of SN is then
h(s) =
=
=
1
X
n=0
1
X
j =0
1 X
1
X
j =0 n=0
hj sj =
Fj gnsj =
gnf (s)n gf (s)
(2.18)
Example 2
A hen lays N eggs, where N has a Poisson distribution with mean . The weight of the nth
egg is Wn 2 f0; 1; 2:::g, where W1; W2 ::: are idependent and identically distributed random
variables with common probability generating function f . By virtue of the above analysis and
use of the formulae (1.10), (2.18), the generating function of the total weight W = PNi=1 Wi
is given by:
gW = exp(?(1 ? f (s)))
(2.19)
W is said to have a compound Poisson distribution. Note that the random observations Wi
can be sampled from any distribution; the crucial point in derivation of (2.18) is that the
number of elements in the their sum N constitutes a Poisson distributed random variable.
If Wi's are distributed according to a Poisson law
f (s) = exp(? + s)
(2.20)
the total weight W is a random variable with a compound Poisson-Poisson (Neyman type
A) distribution:
p(w) =
1 (N)r e?n N e?
X
r!
N!
N =0
5
(2.21)
The compound Poisson distribution (CPD) has a wide application in ecology, nuclear
chain reactions and queing theory. It is sometimes known as the distribution of a \branching process" [3] and as such has been also commonly used to describe radiobiological
eects in cells.
Example 3
In their cluster theory of the eects of ionizing radiation, Tobias et al. [15] have used so
called Neyman [7] distribution (see above, Example 2) which is nothing but a compound
Poisson-binomial (or in a limitting case Poisson-Poisson) distribution. In the derivation the
following reasoning has been used:
when a single heavy ion crosses a cell nucleus, it may produce DNA strand breaks and
chromatin scissions wherever the ionizing track structure overlaps chromatin structure
the multiple yield of such lesions depends on the radial distribution of deposited energy
and on the microdistribution of DNA in the cell nucleus
the number of crossings strongly depends on the geometry of DNA coiling in the cell
nucleus (in a human cell nucleus, the total length of doubled-stranded DNA is more
than one meter and in the nucleus the DNA is packed in coiled strands). For a given
cell line, a \typical" average number n of possible crossings per a particle is assumed.
if p is a probability that a chromatin break occurs at each particle crossing (and q is
the probability that it does not), the distribution of the number of chromatin breaks
in the cluster per one-particle traversal is binomial:
!
n
(2.22)
P (ijn) = i piq(n?i)
with the probability generating function
gs = [sp + (1 ? p)]n
6
(2.23)
the probability that j particles cross the nucleus is given by a Poisson distribution
P = (Fj ! ) e?F
j
(2.24)
with the probability generating function
hs = exp(?F + Fs)
(2.25)
the overall probability that i lesions will be observed after m particles traversed the
nucleus is given by a Neyman distribution
P(ij; F; n) =
1 (nm)!pi q (nm?i) (F )m e?F
X
i!(nm ? i)!m!
m=1
(2.26)
with a compound probability generating function
G(s) = exp(? + [sp + 1 ? p]nm)
(2.27)
where = F .
From the latter, by direct dierentiation one gets expected (mean) value and variance
< i >= np
< i2 > ? < i >2 = n(n ? 1)p2 + np
(2.28)
which explains overdispersion of the probability density function eq.(2.26). By assuming a
repairless cell line, we are able to derive the surviving fraction as a zero class of the initial
distribution, i.e. the proportion of cells with no breaks.
P(0j; F; n) =
1 (nm)!q nm (F )m e?F
X
=
(nm)!m!
m=1
exp[?F (1 ? qn)]
(2.29)
(2.30)
Note the dierence between Neyman and Poisson distribution, for which
P(0j; F; n) = exp[?F ] = exp[? < i >]
7
(2.31)
III. MARKOV PROPERTY
Stochastic processes and Markov processes, in particular, serve as a powerfull tool
to describe and understand biological phenomena at various levels of complexity-from the
molecular to the ecological level. The modeling of biological systems via stochastic processes
allows to incorporate the eects of secondary factors for which the detailed knowledge is
missing. The technique has been widely used to model population growth and extinction
[1,8], population genetics [9,10], chemical kinetics [2,14], ring of neurons [11,12], opening
and closing of biological channels [13] or cell survival after irradiation [15,16]. In what follows. we will briey recall denition of Markovianity.
One says that a real stochastic process is statistically determined if one knows its nth
order or n-point distribution function [2]
P (x1 ; t1; x2 ; t2; x3 ; t3:::xn?1 ; tn?1; xn; tn)
(3.32)
for any n and t, where P (x1; t1 ; x2; t2 ; x3; t3 ; :::; xn?1; tn?1; xn; tn) stands for the probability
that the process fx(t)g is in the state xn (takes the value xn) at time tn and in the state
xn?1 at time tn?1... and in the state x1 at time t1 . These functions are not arbitrary but
they must satisfy certain conditions. A distribution of a given order is determined from a
distribution of lower order by use of the Bayes rule for conditioned probabilities:
P (x1; t1; x2 ; t2; x3 ; t3:::xn?1 ; tn?1; xn; tn) =
P (xn; tnjxn?1; tn?1; :::x1 ; t1):::P (x2 ; t2jx1 ; t1)P (x1; t1)
(3.33)
n?1 ; tn?1 ; xn ; tn )
P (xn; tnjxn?1; tn?1; :::x1 ; t1 ) = P (xP1;(tx1;;xt2 ;; tx2;;xt3 ;; tx3:::x
; t :::x ; t
(3.34)
where
1 1 2 2 3 3
n?1 n?1)
denes the conditioned probability that the process takes on value xn at time tn provided the
sequence of events fxn?1; tn?1; :::x2 ; t2; x1 ; t1g took place at earlier times. A Markov process
8
is a stochastic process fx(t)g which can be fully characterized by a conditioned probability
and a one-point probability functions [2].
The basic denition of Markovianity of the process can be expressed as
P (xn; tnjxn?1; tn?1 ; :::x1; t1 ) = P (xn; tnjxn?1; tn?1)
(3.35)
This criterion is to be complemented by the so-called Smoluchowski-Chapman-Kolmogorov
(SCK) equation:
Z1
P (xn; tnjxk ; tk ) = P (xn; tnjxm ; tm )P (xm; tmjxk ; tk )dxm; n > m > k
(3.36)
1
which follows directly from the denitions eqs.(3.34) and (3.35). Therefore the process which
does not satisfy either the basic denition eq.(3.35) or the SCK equation (3.36)is not Markovian. A non-Markovian process may satisfy one of these relations but both are necessary
conditions of Markovianity (i.e. neither is a sucient one).
9
IV. RANDOM OCCURRENCES
An important example of a discontinuous Markov process is the statistics of the
number of occurrences of an event in time interval (0; t). If we denote that number by x(t),
then fx(t)g is a discontinuous stochastic process taking integer values 0; 1; 2:::n; :::. If the
number of occurrences in an interval (t0; t), assuming x(t0 ) is independent of the past of
x(t), then the process is Markov, according to the above denition. Since x(t) can increase
only, hence
Pij (t; t0 ) P fx(t) = ijx(t0 ) = j g = 0
for j > i
(4.37)
Consider now a small interval t. Neglecting probabilities of order (t)2 and higher, we assume that in the interval we can have only one occurrence and its probability is proportional
to t, namely
P fx(t + t) = ijx(t) = ig 1 ? qi(t)t
P fx(t + t) = i + 1jx(t) = ig = qi(t)t
(4.38)
If qi(t) = (t) is independent of i, we say that the occurrences form random points in
time and the resulting fx(t)g process is Poisson. To see it, let as assume for simplicity 1
that (t) = =constant. Let Pn(t) Pn;0(t) denotes the probability that we have n points
in the interval (0; t), that is, x(t) = n, provided we have started with zero points at time 0,
x(0) = 0. The SCK equation (3.36) can be now written in the form
Pn;0(t + t) = tPn?1;0 (t) + (1 ? t)Pn;0 (t)
(4.39)
from which the following dierential equation results
d P (t) = lim Pn(t + t) ? Pn(t) =
t!0
dt n
t
Pn?1;0(t) ? Pn;0(t)
(4.40)
1 If (t) is not a constant, the process is nonhomogeneous in time but can be transformed to a
R
homogeneous process by a linear time transformation = 0 (s)ds
10
For n = 0
d P = ?P (t)
0
dt 0;0
(4.41)
with the initial condition that P0(0) = 1 and Pn(0) = 0. After using the Laplace transform,
the solution to these equation can be easily found
P~n;0(s) = s + P~n?1;0(s)
P~0(s) = s +1 (4.42)
which leads to
Pn
(t) = e?t (t)
n
(4.43)
n
So far, we have discussed the points random in time, but the same arguments apply to
characterize points randomly distributed on a line. We shall now derive the formula [4] for
the probability distribution function for the distance between n random points on a line.
Example 4
Consider a xed point t0 and its two closest \neigbouring"points to the left and right,
respectively (cf. Figure 1):
z ? y1 ?! x1 !
t?1
t0
t1
Figure 1
Given x1 > 0, the distribution function P (x1) of the random variable X1 representing
the distance between the points t1 and t0 equals the probability that x1 is less than x, and
this is the probability that at least one point has been found in the interval (t0 ; t0 + x):
P (x1) = P fn(t0 ; t0 + x) 1g = 1 ? P fn(t0; t0 + x) = 0g = 1 ? e?x
11
(4.44)
The probability density distribution function for x follows by taking derivative of the above
expression:
p(x) = e?x(x)
(4.45)
where stands for the Heaviside step function:
(x) = 1
for x 0
(x) = 0
otherwise
(4.46)
We can now form a random variable z being the sum of x1 and y1 which represents the
distance from t?1 to t1 . Since both random variables x1 and y1 are independent, the probability density function for their sum equals the convolution of their respective densities
[2{4]:
Z1
Z1
p(z) =
p(z ? y)p(y)dy =
p(z ? x)p(x)dx
(4.47)
?1
?1
Since
p(x) = e?x(x)
p(y) = e?y (y)
and
(4.48)
We obtain for z
Zz
2
p(z) = e?(x?y) e?y dy = 2ze?z
0
(4.49)
In a similar way the density distribution function for n independent variables having exponential distributions with the same parameter can be derived leading to the gamma
density distribution function.
p(z) = N e?z zn?1 n(z)
(4.50)
with a normalisation factor N
N = ?(n)
n
12
(4.51)
The gamma distribution is a basic statistical tool for describing variables bounded at one
side (in our example, x1 ; y1; z 2 [0; 1)).
Let us analyze further a problem of cutting the interval [0; 1) into k fragments by
placing on it randomly n breaking points. The fragments are supposewd to be greater than a
minimum detectable length .Our task is to derive the probability density function p(k; jn).
The problem posed in this way is an adequate model of a chromatin break model discussed
by Schmidt [17].
Example 5.
Seeking the distribution of the number of chromatin fragments detected given the number
of chromatin breaks produced on an arbitrary chromosome. The total number of chromatin
breaks produced in a cell nucleus after irradiation is a random number equal to the sum of
the breaks produced on each chromosome by a random number of particles passing through
the individual chromosomes.
The probability density function for the length xi of the ith fragment is assumed in the
exponential form (note that a mean spacing between the random points have been set up
to unity)
p(xi ) = e?x
i
(4.52)
so that the density of the sum of all n + 1 fragments is
(4.53)
p(z) = ?(n1+ 1) e?z zn
The probability that exactly k of those fragments is greater than some value y is given by
!
n
+
1
p(k; yjn + 1) = k e?ky (1 ? e?y )n+1?k =
Z
p(k; yjn + 1; z)p(z)dz
(4.54)
Let us assume further that z = 1, so that the length y is expressed in units of z, z=y = a
!
Z1
n
+
1
?
ky
?
y
n
+1
?
k
= p(k; a1 jn + 1; ya) ?(n1+ 1) e?ya (ya)nyda (4.55)
p(k; yjn + 1) = k e (1 ? e )
0
13
The binomial coecient on the right hand side of the above equation can be expanded
leading to the identity
!
! X?k
n
+
1
?
k
n + 1 n+1
m
?
(
m
+
k
)
y
=
m
k m=0 (?1) e
Z1
(4.56)
= p(k; 1 jn + 1; ya)??1(n + 1)e?ya (ya)nyda
a
0
By rearranging the terms, the above equation can be rewritten as a Laplace transform of
the function p(k; a1 jn + 1; ya)an:
L[p(k; a1 jn + 1; ya)an] =
!
! n+1
X?k
n
+
1
?
k
n
+
1
?(
n
+
1)
m
?
(
m
+
k
)
y
=
=
m
(y)n+1
k m=0 (?1) e
!
n+1
X?k
n
+
1
?
k
m
L
[(a ? (m + k))n(a ? (m + k))]
(?1)
=
m
m=0
(4.57)
where we have used
L[(a ? (m + k))n(a ? (m + k))] = ?(ynn++1 1) e?(m+k)
From eq.(4.57) the form of the desired probability density function follows:
! n+1
X?k
1
n
+
1
p(k; a jn + 1; z = 1) = k
(?1)m [1 ? a1 (m + k)]n(1 ? m a+ k )
m=0
(4.58)
(4.59)
The density eq.(4.59) refers then to the situation when k random fragments greater than
the minimum \detectable" length a1 are produced by placing randomly n breaks on the line
of a unit length.
14
Probability density
50
40
30
20
10
0
0
200
400
600
800
1000
600
800
1000
x
Probability density
80
60
40
20
0
0
200
400
x
15
REFERENCES
[1] N. Goel and N. Richter-Dyn, Stochastic Processes in Biology, (Academic Press, New York,
1974).
[2] N.G. Van Kampen, Stochastic Processes in Physics and Chemistry, (North Holland, Amsterdam, 1981).
[3] S. Karlin and H. Taylor, First Course in Stochastic Processes, (Academic Press, New York,
1976).
[4] A. Papoulis, Probability, Random Variables and Stochastic Processes, (McGraw-Hill, Tokyo,
1981).
[5] R. Brown, Philos. Mag. 4 (1828) 161.
[6] C.A. Tobias, E. Goodwin and E. Blakely, in Quantitative Mathematical Models in Radiation
Biology, J. Kiefer, ed., Springer Verlag, Berlin 1988, p.135.
[7] J. Neyman, Am. Math. Stat. 10 (1939) 35.
[8] D.R. Nelson and N.M. Shnerb, Phys. Rev. E58 1998 1384.
[9] T. Maruyama, Mathematical Modeling in Genetics, (Springer Verlag, Berlin, 1981).
[10] W. Horsthemke and R. Lefever, Noise-induced transitions. Theory and applications in physics,
chemistry and biology, (Springer Verlag, Berlin, 1984).
[11] J.J. Hopeld, Proc. Natl. Acad. Sci. USA 79 (1982) 2554.
[12] A. Crisanti and H. Sompolinsky, Phys. Rev. A36 (1987) 4922.
[13] B. Hille, Ionic channels of excitable membranes, (Sinauer Inc., Sunderland, MA, 1992).
[14] S. Larsson, Bioch. et Biophys. Acta 1365 (1998) 294.
[15] C.A. Tobias, Radiat. Res. 104 (1985) S77.
[16] N. Albright, Radiat. Res. 118 (1989) 1.
16
[17] J. Schmidt, Heavy ion induced lesions in DNA, Ph.D. thesis, University of California at Berkeley, 1993.
17