Poisson distribution function, Civil Engineering

Poisson distribution function:

Let XI, X2, ..., X,& b e n independently and identically distributed random variables each having the same cdf F ( x ). What is the pdf of the largest of the xi'?

Solution:

Let Y = maximum (XI, X2, ..., Xn, )

Since Y ≤ y implies Xl ≤ y, X2 ≤y, ..., Xn ≤ y, we have

Fy(y )= P(Y ≤ y) = P(XI ≤ y,X2 ≤ y, ..., Xn ≤ y )

= P(X1 ≤ y) P(X2 ≤ y) ... P(Xn ≤ y),

since XI, X2, ..., Xn, are independent

= {F(y)}n, since the cdf of each Xi is F(x).

 Hence the pdf of Y is

fy (y) = F'(y) = n{F(y)}n-1 f(y),

where f ( y ) = F'( y ) is the pdf of Xi.

6.2.2 The Method of Probability Density Function (Approach 2)

For a univariate continuous random variable x having the pdf' fx ( x ) and the cdf Fx (x), we have

F'x(x)= (d/dx ) dFx(x) or dFx(x) = fx (x) dx

In other words, differential dFx (x) represents the element of probability that X assumes a value in an infinitesimal interval of width dx in the neighbourhood of X = x.

For a one-to-one transformation y = g ( x ), there exists an inverse transformation x = g - 1 ( y ), so that under the transformation as x changes to y, dx changes to dg-1(y)/dy and

dF (x) = f(x) dx = fx (g-1(y))¦dg-1(y)/dy¦ dy

The absolute value of dg-1(y)/dy is taken because may be negative and fx ( x ) and d Fx ( x ) are always positive. As X, lying in an interval of width dx in the neighbourhood of X = x, changes to y, that lies in the corresponding interval of width dy in the neighbourhood of Y = y, the element of probability dFx ( x ) and dFy ( y ) remain the same where Fy ( y ) is 1 cdf of Y. Hence

dFy(y) = dFx(x) = fx(g-1(y)) ¦ dg-1(y)/dy¦ dy

and

fy(y) = d/dy Fy(y) = fx (g-1 (y)) ¦ dg-1(y)/dy¦                                         (6.2)

Equation (6.2) may be used to find the pdf of a one to one function of a random variable. The method could be generalized to the multivariate case to obtain the result that gives the joint pdf of transformed vector random variable Y under the one to one transformation Y - G ( X ) , in terms of the joint pdf of X The generalized result is stated below

fy(y) = fx(G-1 (y))1/¦J¦

where the usual notations and conventions for the Jacobian J = ¦ð y/ð x¦ are assumed

Remarks:

This technique is applicable hust to continuous random variables and only if the functions of random variable Y = G (X) define a one to one transformation of the region where the pdf of X is non zero.

Posted Date: 1/30/2013 7:04:30 AM | Location : United States







Related Discussions:- Poisson distribution function, Assignment Help, Ask Question on Poisson distribution function, Get Answer, Expert's Help, Poisson distribution function Discussions

Write discussion on Poisson distribution function
Your posts are moderated
Related Questions
Los Angeles Abrasion Test: In which the aggregates are placed in a rotating drum in which a certain number of cast iron cylindrical balls are also placed, and the drum rotated



Define the Influence line-beam shear?   1.  Replace point of interest with a link element of nominal length.  If the point of interest is a reaction, place a hinge at that poin

State the term - Terminology There are many technical terms whose understanding provides us with a good insight of  the subject. Some of  these terms are, indeed, general and t

III. Calculate the magnitude and sense of the member forces BC, BG and HG using the method of sections

Didtribution of shear stree over a triangular section

Supplementary or Untensioned Reinforcement It is that reinforcement, provided in prestressed concrete member, which is not provided in a stretched condition like tendons.

what do u mean by effective span and its derivation?

Define the Precedence Diagramming Methods a) Activity on Node (AON) with arrows for dependencies b) Uses all forms of precedence relationships