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Math Help - Distribution function and Densities

  1. #1
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    Distribution function and Densities

    These question seemed simple to me, however I believe I am getting confused between distribution functions and densities.

    Question

    Find the distribution function, and hence the density, of (1/x)


    Any help would be great. The reason why im confused is because i thought 1/x was already a density, so why would it need to be found?

    Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    Do you have an orginal distribution?
    I can tranform any rv via 1/x, if I know what distributon we are starting with.
    In other words.
    Give me the density of X and I will give you the density of Y=1/X.
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  3. #3
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    Yes it is a random variable, and I have been told that the range is from

    0 < x < 2

    Does this help?
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  4. #4
    MHF Contributor matheagle's Avatar
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    nope
    how about an f(x) or an F(x)?
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  5. #5
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    OK, here is the complete question:-

    X has density x/2 over 0 < x < 2, and 0 elsewhere. Sketch its density and distribution function. Find the distribution function, and hence the density, of X/2, e^X, 1/X, and |X - 1|.

    So all I know is that it has something to do with random variables. Please dont explain all of them as that is time consuming, if you can do one then Im sure I can work out the rest with the same technique.
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by ryanhorne View Post
    OK, here is the complete question:-

    X has density x/2 over 0 < x < 2, and 0 elsewhere. Sketch its density and distribution function. Find the distribution function, and hence the density, of X/2, e^X, 1/X, and |X - 1|.

    So all I know is that it has something to do with random variables. Please dont explain all of them as that is time consuming, if you can do one then Im sure I can work out the rest with the same technique.

    that is just calculus one

    Let Y=1/X

    the density of Y is

    f_Y(y)=f_X(x)/y^2 since we need the absolute value around the -y^{-2}

    Thus f_Y(y)={1\over 2y}y^{-2}={1\over 2y^3}

    for y>1/2
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  7. #7
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    Just had a thought, do you reckon the final answer should have been negative?
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by ryanhorne View Post
    Just had a thought, do you reckon the final answer should have been negative?

    probabilities are never negative
    same with probability functions
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  9. #9
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    haha, good point.
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