Geoscience Reference
In-Depth Information
Box 2.2: Probability Generating Function and Moments
If
X
is a discrete random variable with probability distribution
P
(
X
¼
k
)
¼
p
k
(
k
¼
0,
1,
2,
...
),
then
its
probability
generating
function
is:
h
i
d
k
p
2
x
2
1
k!
1
k!
g
ðÞ
0
gs
ðÞ
¼
p
0
þ
p
1
x
þ
þ ...
with
p
k
¼
ds
k
gs
ðÞ
¼
ðÞ
0
r
¼
∑
k
¼0
p
k
¼
g
(
r
)
(1). For the mean and
The
r
-th moment of
X
satisfies:
μ
[
g
0
(1)]
2
.
Suppose that
X
and
Y
are two independent discrete random variables with
probability distributions
P
(
X
g
0
(1);
2
(
X
)
g
00
(1) +
g
0
(1)
variance (
cf
.Feller
1968
,p.360):
E
(
X
)
¼
˃
¼
¼
k
)
¼
p
k
;
P
(Y
¼
k
)
¼
q
k
(
k
¼
0, 1, 2,
...
), then
the probability distribution of their sum
Z
¼
X
+
Y
satisfies:
g
z
(
s
)
¼
g
x
(
s
)
g
y
(
s
).
2.2.3 Binomial and Poisson Distributions
Suppose the outcome of an experiment, like observing presence or absence of
a rock type at a point, is either Yes or No. Presence or absence can be denoted as
1 or 0. The experiment is called a Bernoulli trial. If
X
is a Bernoulli variable with
P
(
X
p
)+
ps.
The binomial distribution results from
n
successive Bernoulli trials. Repeated
application of the Bernoulli distribution's generating function gives the generating
function:
¼
0)
¼
1
p
,
P
(
X
¼
1)
¼
p,
its generating function is:
g
(
s
)
¼
(1
n
gs
ðÞ
¼
½
ð
1
p
Þþ
ps
It follows that the mean and variance of a binomial variable are
E
(
X
)
¼
np
2
and
p
).
For
n
Bernoulli trials, there are (
n
+ 1) possible outcomes: n, (n
˃
¼
np
(1
,2,1,0.
The (
n
+ 1) probabilities for these outcomes are given by the successive terms of the
series obtained by expanding the expression (
p
+
q
)
n
. The probability that the
outcome is exactly
k
Yesses in
n
trials satisfies:
1),
...
!
p
k
q
nk
where
¼
P
X
n
n
!
n
k
n
k
X
i
¼
k
¼
k
!
ð
n
k
Þ!
i
¼1
This is the binomial frequency distribution. Petrographic modal analysis
(Chayes
1956
) provides an example of application of the binomial frequency distri-
bution. Suppose that 12 % by volume of a rock consists of a given mineral A.
The method of point-counting applied to a thin section under the microscope consists
of how many times a mineral is observed to occur at points that together form a
regular square grid. Suppose that 100 points are counted. From
p
¼
0.12 and
n
¼
100
it follows that
q
¼
μ
¼
np
¼
0.88, and for mean and standard deviation:
12 and
˃
¼ (
npq
)
0.5
¼3.25.
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