Distribution of continuous random variable

frodo

New Member
Hi all, I have been struggling with this problem for quite a while now and I am getting nowhere

The problem is this:

I have been given a continuous random variable X with pdf=P, cdf=F and quantile function q.

I now consider the transformed random variable Y=F(X), and I have to show that Y is uniformly distributed on the interval [0,1].

My first thoughts was to use the formula for pdf of Y=t(X), that is

t(y)=p(t^-1(y))* d/dy t^-1(y)

which gives me

t(y)=p(q(y))*q´(y)

Is it correct to assume that I am trying to get this to equal 1? And how do I proceed from here?

Thank you!

BGM

TS Contributor
The most usual method for this question is to check the CDF of the resulting random variable $$F(X)$$ against the CDF of $$\text{Uniform}(0, 1)$$,
i.e. by considering

$$\Pr\{Y \leq y\} = \Pr\{F(X) \leq y \}$$

Playing around the inequality and the property of CDF shall give you the desired result.

Considering the derivative of CDF, i.e. the pdf will not be as trivial as that.