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Math Help - Cumulative distribution function difficulty

  1. #1
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    Cumulative distribution function difficulty

    X is a continuous ramdom variable with a cumulative distribution function F such that Inverse-F is a well-defined function.

    Does well defined mean increasing over the interval and differentiatable? Any other requirements?
    What would be an example of a non-well-defined function? Just any function which isnt increasing over the interval?

    Thanks.
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  2. #2
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    A well defined function would have a unique value assigned to each input (i.e. the CDF must be 1-to-1). Since CDF's are non-decreasing, in order to be 1-to-1, this says the CDF must be strictly increasing.

    Nothing implies differentiable. In fact, suppose your CDF was:

    F(x) = \begin{cases} 0.1x & 0\leq x \leq 5 \\ 0.5(x-5)+0.5 & 5 \leq x \leq 6 \end{cases}

    This is certainly continuous but not differentiable everywhere (@ 5).

    What would be an example of a non-well-defined function? Just any function which isnt increasing over the interval?
    Yes. It would not be well-defined if the CDF was not increasing over some region. The CDF would not be 1-to-1. Suppose your CDF was constant c (c is between 0 and 1) from x=1 to x=2. What would the inverse function at c be? Well a function assigns one output to any input. Which number from x=1 to x=2 should the inverse function evaluated at c be? There should be 1 answer but there is a whole interval.
    Last edited by meymathis; March 25th 2009 at 10:17 AM. Reason: fixed error: monotonically to strictly
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  3. #3
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    Thanks a heap!
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  4. #4
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    Further help needed

    I'm assuming that stationary points that are not minimums or maximums are still ok to have in the interval of a cummulative distribution??
    Last edited by woollybull; March 25th 2009 at 02:13 AM. Reason: realise possible mistake
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  5. #5
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    Stationary points of the CDF are fine as long as it is still 1-to-1. 1-to-1 and continuous are the necessary conditions for a CDF to have an inverse. In general, 1-to-1 and onto are the necessary and sufficient conditions for a function to have an inverse. If a function is 1-to-1, then by suitably restricting the range to the image of the function, then that function has an inverse. For our purposes, 1-to-1 and continuous imply either strictly increasing or decreasing (CDF's always increase, though) and vice verse. But strictly increasing does not imply that there are no stationary points.

    For example F(x) = 4(x-0.5)^3+0.5 on 0\leq x \leq 1 has a stationary point at x=0.5. But it is strictly increasing. That is ok since the CDF is 1-to-1.

    Note: I said monotonically increasing in my previous post. It should be strictly increasing. I'll fix my previous post.
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  6. #6
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    I should also note that I found a paper that claims to have constructed a continuous strictly increasing function that has 0 derivative almost everywhere (except for on a set with measure zero - like the rational numbers). http://projecteuclid.org/DPubS/Repos...pja/1200672011
    I didn't read the paper. This says that you can construct a CDF that has stationary points almost everywhere, but is continuous and strictly increasing, and thus has a well-defined inverse.
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