Exponential distribution - Poisson distribution function:
Let X follow exponential distribution with parameter θ. Find the pdf of
Y = X1/β where β( ≥ 1 ) is a constant.
The function Y = X1/β is one to one in the interval ( 0,∞) where pdf of X is non zero. The inverse of the transformation Y = X1/β is X - Yβ which gives dX/dY =β Yβ-1. Also
fx(x) = 1/θ e-x/θ , x> 0.
f(y) = f(g-1(y))¦ dg-1(y)/dy ¦
=1/θ e -yβ/θ ¦β yβ-1¦, y>0
and the pdf is
f(y) = β y β -1/θ e -yP/θ , y>0 with parameters θ >0,β ≥ 1
The distribution of Y, just derived Example 4, is known as Weibull distribution. That has application is life testing. Remember that if Y follows Weibull distribution with parameters θ and β then Yβ follows the exponential distribution with parameter θ.
The treatment of distribution of functions of more than one variable is similar. However, care should be taken to determine the correct region of variation of the variables under the transformation and use it properly for obtaining the marginal distributions. We assume an example to illustrate the point.