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Chapter IV
Probability distributions and their applications
4.1
DISCRETE DISTRIBUTIONS
4.1.1
Binomial distribution
Consider a discrete time scale. At each point on this time scale, an event may either
occur or not occur. Let the probability of the event occurring be p for every point on
the time scale. Thus, the occurrence of the event at any point on the time scale is
independent of the history of any prior occurrences or non-occurrences. The
probability of an occurrence at the i th point on the time scale is p for i = 1,2, K . A
process having these properties is said to be a Bernoulli process.
As an example of a Bernoulli process consider that during any year the probability of
the maximum flow exceeding 10,000 cubic feet per second (cfs) on a particular river
is p . Common terminology for a flow exceeding a given value is an exceedance.
Further consider that the peak flow in any year is independent from year to year (a
necessary condition for the process to be a Bernoulli process). Let q = 1 − p be the
probability of not exceeding 10,000 cfs. We can neglect the probability of a peak
exactly 10000 cfs since the peak flow rates would be a continuous process so the
probability of a peak exactly 10000 cfs would be zero. In this example, the time scale
is discrete with the points nominally 1 year in time apart. We can now make certain
probabilistic statements about the occurrence of a peak flow in excess of 10000 cfs
(an exceedance).
For example, the probability of an exceedance occurring in year 3 and not in year 1 or
2 is qqp since the process is independent from year to year. The probability of
(exactly) one exceedance in any 3-year period is pqq + qpq + qqp since the
exceedance could occur in either the first, second, or third year. Thus the probability
of (exactly) one exceedance in three years is 3 pq 2 . In a similar manner, the
probability of 2 exceedances in 5 years can be found from the summation of the
terms ppqqq , pqqpq , K , qqqpp . It can be seen that each of theses terms is equivalent
to p 2 q 3 and that the number of terms is equal to the number of ways of arranging 2

Chapter IV: Probability Distributions and Their Applications
items (the p 's) among 5 items (the p 's and q 's). Therefore the total number of terms
⎛5⎞
is ⎜⎜ ⎟⎟ or 10 so that the probability of exactly 2 exceedances in 5 years is 10 p 2 q 3 .
⎝ 2⎠
This result can be generalized so that the probability of X = x exceedances in n
⎛n ⎞
years is ⎜⎜ ⎟⎟ p X q n − X . The result is applicable to any Bernoulli process so that the
⎝X⎠
probability of X = x occurrences of an event in n independent trials if p is the
probability of an occurrence in a single trial is given by:
⎛n⎞
f X ( x : n, p ) = ⎜⎜ ⎟⎟ p x q n − x
⎝ x⎠
x = 0,1,2, K , n
This equation is known as the binomial distribution. The binomial distribution and
the Bernoulli process are not limited to a time scale. Any process that may occur with
probability p at discrete points in time or space or in individual trials may be a
Bernoulli process and follow the binomial distribution.
The cumulative binomial distribution is
X ⎛n⎞
F X ( x; n, p) = ∑i = 0 ⎜⎜ ⎟⎟ p i q n − i
⎝i ⎠
x = 0,1,2, K , n
and gives the probability of x or fewer occurrences of an event in n independent trials
if the probability of an occurrence in any trial is p.
Continuing the above example, the probability of less than 3 exceedances in 5 years is
2 ⎛ 5⎞
FX (2;5, p ) = ∑i =0 ⎜⎜ ⎟⎟ p i q 5−i
⎝i ⎠
= f X (0;5, p) + f X (1;5, p) + f X (2;5, p)
The mean and variance of the binomial distribution are
E ( X ) = np
var( X ) = nqp
The coefficient of skew is (q − p) / npq so that the distribution is symmetrical for
p = q , skewed to the right for q > p and skewed to the left for q < p .
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Chapter IV: Probability Distributions and Their Applications
The binomial distribution has an additive property. That is if X
has a binomial
distribution with parameters n1 and p and Y has a binomial distribution with
parameters n 2 and p , then Z = X + Y has a binomial distribution with parameters
n = n1 + n2 and p .
The binomial distribution can be used to approximate the hyper-geometric distribution
if the sample selected is small in comparison to the number of items N from which
sample is drawn. In this case the probability of a success would be about the same for
each trial.
Example: In order to be 90 percent sure that a design storm is not exceeded in a 10
year period. What should be the return period of the design storm?
Solution: Let p be the probability of the design storm being exceeded. The
probability of no exceedances is given by
⎛10 ⎞
f X (0;10, p) = ⎜⎜ ⎟⎟ p 0 q 10
⎝0 ⎠
0.90 = (1 − p)10
p = 1 − (0.90)1 / 10 = 1 − 0.9895 = 0.0105
T = 1 / p = 95 years.
Comment: To be 90 percent sure that a design storm is not exceeded in a 10-year
period a 95-year return storm must be used. If a 10-year return period storm is used,
the chances of it being exceeded is 1 − f X (0;0,1) = 0.6513 . In general the chance of at
least one occurrence of a T-year event in T-years is 1 − f X (0; T ,1 / T ) = 1 − (1 − 1 / T ) T .
Therefore, for a long design life, the chance of at least one occurrence of an event
with return period equal to the design life approaches 1 − 1 / e or 0.632.Thus if the
design life of a structure and its design return period are the same, the chances are
very great that the capacity of the structure will be exceeded during its design life.
4.1.2
Poisson distribution
The Poisson distribution is like the binomial distribution in that it describes
phenomena for which the average probability of an event is constant, independent of
the number of previous events. In this case, however, the system undergoes transitions
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Chapter IV: Probability Distributions and Their Applications
randomly from one state with n occurrences of an event to another with ( n + 1)
occurrences, in a process that is irreversible. That is, the ordering of the events cannot
be interchanged. Another distinction between the binomial and Poisson distributions
is that for the Poisson process the number of possible events should be large.
The Poisson distribution may be inferred from the identity
e −µ e µ = 1
where the most probable number of occurrences of the event is µ . If the factorial is
expanded in a power series expansion, the probability P(r) that exactly r random
occurrences will take place can be inferred as the r th term in the series, i.e.,
e −µ µ r
p (r ) =
r!
(4.1.2.1)
This probability distribution leads directly to the interpretation that:
e − µ = the probability that an event will not occur,
µ e − µ = the probability that an event will occur exactly once,
( µ 2 / 2! ) e − µ = the probability that an event will occur exactly twice, etc,
The mean and the variance of the Poisson distribution are:
E( X ) = µ
Var ( X ) = µ
The coefficient of skew is µ −1 / 2 so that as µ gets large, the distribution goes from a
positively skewed distribution to a nearly symmetrical distribution. The cumulative
Poisson probability that an event will occur x times or less is:
x
p (≤ x ) = ∑ p ( r )
r =0
Of course, the probability that the event will occur ( x + 1) or more times would be the
complement of P(x).
The Poisson distribution is useful for analyzing the failure of a system that consists of
a large number of identical components that, upon failure, cause irreversible
transitions in the system. Each component is assumed to fail independently and
randomly. Then µ is the most probable number of system failures over the life time.
To summarize:
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