#density function

[projecttopicsinnigeria]

PROJECT TOPIC ONEXPONENTIATED-EXPONENTIAL WEIBULL DISTRIBUTION: ITS PROPERTIES AND APPLICATIONS

PROJECT TOPIC ONEXPONENTIATED-EXPONENTIAL WEIBULL DISTRIBUTION: ITS PROPERTIES AND APPLICATIONS

PROJECT TOPIC ONEXPONENTIATED-EXPONENTIAL WEIBULL DISTRIBUTION: ITS PROPERTIES AND APPLICATIONS

ABSTRACT

A four-parameter distribution called Exponentiated-Exponential Weibull (EEW) distribution was proposed using a generator introduced in earlier research. Statis-tical properties of the distribution were derived along with its reliability functions and asymptotic behaviour. The distribution…

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[statstudent-blog]

Finding the pdf of a function of random variables: 1 dimensional case

In addition to the distribution function method talked about before, you can use a different method.

It starts out like this: X is some random variable with a density function (pdf) f(x). Then we say the set where f(x)>0 is called "A" (this is also known as the support of x. Now if we have a function of x, say Y=phi(x), then, if the following three conditions are met,

phi(x) is 1-1 on A

dy/dx is continuous on A

dx/dy does not =0

we can find f(y) (i.e. as the title says, we are trying to find the pdf of a function of random variables, and Y is a function of the random variable X) using this formula:

The first step in finding f(y) we solve the function of X for X. For example

if X is a random variable with pdf f(x)=e^(-x) for X>0, and Y=-3x+5, solving this function for X would give us x=(y-5)/-3

Note also that when x>0,  y<5

Then we use this find dx/dy, which is (-1/3)

Finally, we can develop the pdf of y by plugging in (y-5)/-3 in for x in the pdf at the beginning (e^-x), and multiply by absolute value of (dx/dy). The support of y is y<0, as we found out in step 1