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Math Help - Transformation of probability density function

  1. #1
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    Transformation of probability density function

    Let X be a continuous random variable with a strictly increasing (and therefore 1-1) cumulative distribution function F(x) = P [X <= x] and probability density function f(x).

    (a) [6 marks] What is the probability density function of Y = F(X)?
    (b) [2 marks] Evaluate P [Y <= 0.5].

    So far, I got that:

    F(y) = P[Y <= y] = P[F(X) <= y] = P[X <= F^-1(y)]

    I'm not sure how to proceed from there?
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  2. #2
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    Quote Originally Posted by My Little Pony View Post
    Let X be a continuous random variable with a strictly increasing (and therefore 1-1) cumulative distribution function F(x) = P [X <= x] and probability density function f(x).

    (a) [6 marks] What is the probability density function of Y = F(X)?
    (b) [2 marks] Evaluate P [Y <= 0.5].

    So far, I got that:

    F(y) = P[Y <= y] = P[F(X) <= y] = P[X <= F^-1(y)]

    I'm not sure how to proceed from there?
    Read this: http://www.ece.virginia.edu/~mv/edu/...generation.pdf
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by My Little Pony View Post
    Let X be a continuous random variable with a strictly increasing (and therefore 1-1) cumulative distribution function F(x) = P [X <= x] and probability density function f(x).

    (a) [6 marks] What is the probability density function of Y = F(X)?
    (b) [2 marks] Evaluate P [Y <= 0.5].

    So far, I got that:

    F(y) = P[Y <= y] = P[F(X) <= y] = P[X <= F^-1(y)]=F(F^-1(y))=y

    which is a UNIFORM (0,1)
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