Geoscience Reference
In-Depth Information
13.3
SOME PROBABILITY DISTRIBUTIONS FOR
DISCRETE VARIABLES
13.3.1
The geometric distribution
This distribution was already discussed briefly in connection with the return period. For
convenience, the reasoning can be briefly repeated here. Consider an experiment in which
there are two possible outcomes, namely success with a probability
p
, and failure with
a probability
q
p
). If the trials are independent of one another, then it follows
that the probability that (
k
=
(1
−
1) successes will be followed by one failure is given by
Equation (13.16), or as formulated here
−
p
k
−
1
(1
P
{
k
trials until first failure
}=
−
p
)
(13.29)
Equation (13.29) can be used next, for example, to calculate the probability that it will
take
k
or fewer trials to incur failure. This is simply the sum of the probabilities, or if
K
denotes the number of trials needed to experience failure, as a random variable,
k
p
i
−
1
(1
P
{
K
≤
k
}=
−
p
)
(13.30)
i
=
1
which yields immediately 1
−
p
k
. It can also be seen that Equation (13.30) yields unity,
that is certainty, when
k
. Actually, (13.30) could also have been obtained by
considering first the probability that failure would not occur for
k
trials; since the trials
are independent, this is equal to
p
k
. Its complement is the probability that it will take
k
or fewer trials for failure to occur.
→∞
Example 13.2. Probability of exceedance of a flood of a given return period
Assume that “success” means that the annual maximum flow
X
in a river, also called the
annual flood, does not exceed a given magnitude
x
, so that
p
=
F
(
x
). As an example,
consider the event for which
p
98; from Equation (13.15) it can be seen that this is a
50 y flood. Figure 13.4 shows the probability, that it will take exactly
k
years before that
event is exceeded as a function of
k
, calculated with Equation (13.29). The probability
that it will be exceeded the first year is 0.02. The probability that it will be exceeded after
exactly 50 y is 0
=
0
.
98
49
007 43. Figure 13.5 shows the probability calculated
with Equation (13.30), that it will take
k
or fewer years to exceed
x
, that is, the probability
of the occurrence of a flood in excess of the same magnitude
x
. It can again be seen that
the probability of this occurrence in the first year is 0.02; also, the probability that it will
be exceeded before 50 y have passed is 1
.
×
0
.
02
=
0
.
−
0
.
98
50
=
0
.
636. The probability approaches
one as
k
becomes large.
Example 13.3. Probability of sequences of dry or rainy days
The geometric distribution has been used by Hershfield (1970a, b; 1971) to study the
probability of periods of a given length with or without precipitation. In this type of
application
p
denotes the conditional probability of a success being followed by a success.
