poisson distribution functionlet xi x2 xamp b e

Question

Poisson distribution function:

Let X_I, 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 x_i'?

Solution:

Let Y = maximum (X_I, X₂, ..., X_n, )

Since Y ≤ y implies X_l ≤ y, X₂ ≤y, ..., X_n ≤ y, we have

F_y(y )= P(Y ≤ y) = P(X_I ≤ y,X₂ ≤ y, ..., X_n ≤ y )

= P(X₁ ≤ y) P(X₂ ≤ y) ... P(X_n ≤ y),

since X_I, X₂, ..., X_n, are independent

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

 Hence the pdf of Y is

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

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

6.2.2 The Method of Probability Density Function (Approach 2)

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

F'_x(x)= (d/dx ) dF_x(x) or dF_x(x) = f_x (x) dx

In other words, differential dF_x (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 dF_x ( x ) and dF_y ( y ) remain the same where F_y ( y ) is 1 cdf of Y. Hence

dF_y(y) = dF_x(x) = fx(g^-1(y)) ¦ dg^-1(y)/dy¦ dy

and

f_y(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.