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Thread: Distribution of continuous random variable

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    Distribution of continuous random variable




    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!

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    Re: Distribution of continuous random variable


    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.

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