48    Applications of hypergeometric and Poisson distributions Exercise 6) of l/k is
it " n i=i
An estimate of k is thus made and hence, if the total area A is known, the total number of trees may be estimated as kA. For further details see Diggle (1975, 1983), Ripley (1981) and Upton and Fingleton (1985).
Method 2-Counting
Another method of testing the hypothesis of a Poisson forest is to subdivide the area of interest into N equal smaller areas called cells. The numbers Nk of cells containing k plants can be compared using a x2-test with the expected numbers under the Poisson assumption using (3.10), with n = the mean number of plants per cell.
Extensions to three and four dimensions
Suppose objects are randomly distributed throughout a 3-dimensional region. The above concepts may be extended by defining a Poisson point process in R3. Here, if A is a subset of R3, the number of objects in A is a Poisson random variable with parameter k\A |, where k is the mean number of objects per unit volume and \A\ is the volume of A. Such a point process will be useful in describing distributions of organisms in the ocean or the earth's atmosphere, distributions of certain rocks in the earth's crust and of objects in space. Similarly, a Poisson point process may be defined on subsets of U* with a view to describing random events in space-time.
3.7 COMPOUND POISSON RANDOM VARIABLES
Let Xk,k = 1,2,... be independent identically distributed random variables and let N be a non-negative integer-valued random variable, independent of the Xk. Then we may form the following sum:
Ss^Xt+X^-'+X,!, (3.12)
where the number of terms is determined by the value of N. Thus SN is a random sum of random variables: we take SN to be zero if N = 0. If N is a Poisson random variable, SN is called a compound Poisson random variable. The mean and variance of SN are then as follows.
Theorem 3.8 Let E(Xl) = fi and Var(A",) = <r2, |/i | < oo, a < go. If N is
Compound Poisson random variables 49
Poisson with parameter A, then SN defined by (3.12) has mean and variance
£(SN) = Xft Var(5N) = A0»2 + <r2).
Proof The law of total probability applied to expectations (see p. 8) gives E(SN)= £ E{SN\N = k)Pr{N = k}.
k = 0
But conditioned on N = k, there are k terms in (3.12) so E(SN\N = k) = kjx. Thus
£(SW)= £ nkPr{N = k}
Je = 0
= fiE(N) = kfi.
Similarly,
£(S2) = £ £(S2|N = fc)Pr{iV = k}
= £ [Var(SN|N = fc) + £2(SJN = fc)]Pr{Ar = /c} * = o
= £ (fcff2 + ^V>>Pr{JV = fe} t = o
= <r2£(JV) + ^2£(N2) = a2A + /i2[Var(iV) + £2(N)] = a2/l + /i2(A + A2). The result follows since Var(SN) = E(Sl) - X2fi2.
Example The number of seeds (N) produced by a certain kind of plant has a Poisson distribution with parameter A. Each seed, independently of how many there are, has probability p of forming into a developed plant. Find the mean and variance of the number of developed plants (ignoring the parent).
Solution Let Xk = 1 if the kth seed develops into a plant and let Xk = 0 if it doesn't. Then the Xk are i.i.d. Bernoulli random variables with
Pr     = 1} = p = 1 - Pr {Xi = 0}
and
£(*i) = P Var(X1) = P(l-P)-
50    Applications of hypergeometric and Poisson distributions
The number of developed plants is
sN = xl + x2 + ... + xN
which is therefore a compound Poisson random variable. By Theorem 3.8, with fi = p and a2 - p(l — p) we find
E(SN) = Xp Var(S)¥) = A(p2 + p(l-p)) = kp.
As might be suspected from these results, in this example SN is itself a Poisson random variable with parameter Xp. This can be readily shown using generating functions - see Section 10.4.
3.8 THE DELTA FUNCTION
We will consider an interesting neurophysiological application of compound Poisson random variables in the next section. Before doing so we find it convenient to introduce the delta function. This was first employed in quantum mechanics by the celebrated theoretical physicist P.A.M. Dirac, but has since found application in many areas.
Let Xt be a random variable which is uniformly distributed on (x0 — e/2, x0 + e/2). Then its distribution function is
Fx(x) = Pr{Xt<x} =
x s£ x0 - e/2,
[x - (x0 - e/2)],
, x ^ x0 + e/2 = He(x-xQ).
		H / c
1-		/
		
x -e/2 x 0	0	x +e/2 0
l/£
x -e/2 o
|x-x0| <e/2,
x +C/2 o
Figure 3.8
The delta function 51
The density of Xe is
fx(x) = dFI:=,\1/£'     lx ~ *ol < «A dx     [0, otherwise.
= <5s(x - x0).
The functions Ht and 5t are sketched in Fig. 3.8.
As s->0, HJ^x — Xq) approaches the unit step function, H(x — x0) and St{x — x0) approaches what is called a delta function, 5(x — x0). In the limit as e -»0, <5e becomes 'infinitely large on an infinitesimally small interval' and zero everywhere else. We always have for all e > 0,
<5e(x-x0)dx = 1.
We say that the limiting object 8(x — x0) is a delta function or a unit mass concentrated at x0.
Substitution property
Let / be an arbitrary function which is continuous on (x0 — e/2, x0 4- e/2). Consider the integrals
When e is very small,
/(x)<5t(x-x0)dx = -
/•xo + t/2
f(x)dx.
xo-e/2
/.^-6/(Xo)=/(x0).
We thus obtain the substitution property of the delta function:
/(x)«5(x-x0)dx = /(x0). (3.13)
Technically this relation is used to define the delta function in the theory of generalized functions (see for example Griffel, 1985). With /(x)= 1, (3.13) becomes
j:
S(x — x0) dx = 1 Furthermore, since 5(x) = 0 for x # 0,
<5(x' — x0) dx' = H(x — x0) ■■
I:
0, x < x0,
1, X ^ Xq.
52    Applications of hypergeometric and Poisson distributions
1._
1-P
0 1
Figure 3.9
Thus we may informally regard 8(x - x0) as the derivative of the unit step function H(x — x0). Thus it may be viewed as the density of the constant x0.
Probability density of discrete random variables
Let AT be a discrete random variable with Pr{X = 1) = 1 — Pt(X - 0) = p. Then the probability density of X is written
fx(x) = (l-p)ö(x) + pö(x-l).
This gives the correct distribution function for X because
Fx(x) = Pr(X<x)= f /1(xOdx' = (l-p)H(x) + pH(x-l)
J — 00
CO, x<0, = \ 1 - p,     0 < x < 1, 11, x>l,
as is sketched in Fig. 3.9.
Similarly, the probability density of a Poisson random variable with parameter X is given by
fAx) = e~i £ ^ö(x-k).
3.9 AN APPLICATION IN NEUROBIOLOGY
In Section 3.5 we mentioned the small voltage changes which occur spontaneously at nerve-muscle junctions. Their arrival times were found to be well described by a Poisson point process in time. Here we are concerned with their magnitudes. Figure 3.10 depicts the anatomical arrangement at the nerve-muscle junction. Each cross represents a potentially active site.
The small spontaneous voltage changes have amplitudes whose histogram is fitted to a normal density-see Fig. 3.11. When a nerve impulse, having travelled out from the spinal cord, enters the junction it elicits a much bigger
Nerve fibre Porrion of nerve fibre
O        0.1        0.2        0.3       0.4       0.5       0.6       0.7        0.8 0.9
Amplihjde of sponfaneous potentials (nV)
Figure 3.11 Histogram of small spontaneous voltage changes and fitted normal density. From Martin (1977). Figures 3.11-3.13 reproduced with permission of the American Physiological Society and the author.
54    Applications of hypergeometric and Poisson distributions
response whose amplitude we will call V. It was hypothesized that the large response was composed of many unit responses, the latter corresponding to the spontaneous activity.
We assume that the unit responses are Xu X2,... and that these are normal with mean ft and variance a2. A large response consists of a random number N of the unit responses. If N = 0, there is no response at all. Thus
V = Xl+X2 + ~- + XN,
which is a random sum of random variables. A natural choice for N is a binomial random variable with parameters n and p where n is the number of potentially active sites and p is the probability that any site is activated. However, the assumption is usually made that N is Poisson. This is based on the Poisson approximation to the binomial and the fact that a Poisson distribution is characterized by a single parameter. Hence V is a compound Poisson random variable. The probability density of V is then found as follows:
Pr{Ke(p,» + d»)}= £ Pv{Ve(v,v + dv)\N = k}PT{N = k}
k = 0
Number of
Amplirude of end-plafe potentials (mV)
Figure 3.12 Decomposition of the compound Poisson distribution. The curve marked I corresponds to plt the curve marked II to p2, etc., in (3.14).
Application in neurobiology 55
■e~x Pr {Ve(v, v + dv)\N = 0}
+ I
k!
Pr{Ke(i>,t> + di;)|N = /c}
Ä fcl.
I
'2nkff2
exp
dr
= e-*ô(v)+ £ pk(t>)
*: = i
where is a delta function concentrated at the origin. Hence the required density is
fv(v) = e-xS(v)+ £ pk(v)
k=l
(3.14)
The terms in the expansion of the density of V are shown in Fig. 3.12. The density of V is shown in Fig. 3.13 along with the empirical distribution. Excellent agreement is found between theory and experiment, providing a validation of the 'quantum hypothesis'. For further details see Martin (1977).
0.4
2.8
0.8        1.2 1.6        2.0 2.4
Amplirudeof end-plare potentials (mV)
Figure 3.13 Histogram of responses. The curve is the density for the compound Poisson distribution, the column at 0 corresponding to the delta function in (3.14).