Poisson distribution function:
Let X_{I}, X_{2}, ..., 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}, X_{2}, ..., X_{n}, )
Since Y ≤ y implies X_{l} ≤ y, X_{2} ≤y, ..., X_{n} ≤ y, we have
F_{y}(y )= P(Y ≤ y) = P(X_{I} ≤ y,X_{2} ≤ y, ..., X_{n} ≤ y )
= P(X_{1} ≤ y) P(X_{2} ≤ y) ... P(X_{n} ≤ y),
since X_{I}, X_{2}, ..., 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.