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    Prove that the SD of an asymmetric function is larger than the SD of a symmetric one




    How can I prove that the standard deviation of an asymmetric function (mean>median>mode) is larger than the standard deviation of a symmetric one (mean=median=mode) of the same variable?

    I was asked if the statement "the standard deviation of an asymmetric function (mean>median>mode) is larger than the standard deviation of a symmetric one (mean=median=mode) of the same variable?" true or false and asked to prove the answer.
    Last edited by Scrooge; 06-04-2014 at 01:21 PM.

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    Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

    Is this homework?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

    Yes. It is a homework.

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    Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

    You edited the question! That makes a big difference.

    What do you think the answer is?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

    I think that the SD of an asymmetric function is larger than the SD of a symmetric one but can not prove it.

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    Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

    Let's say you have a A is an asymmetric function by your definition and S is a symmetric one. You claim A(X) > S(X).

    What can you tell me about another function f such that f(X) = A(X)/c for some constant c. Is it asymmetric? What will the standard deviation of this be?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

    Hi,
    this can not generally be true, can it? I mean for any assymetrical function one could construct a symmetrical one with a larger SD - is there some constraint on the functions that was not mentioned?

    regards
    rogojel

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    Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric

    Quote Originally Posted by rogojel View Post
    Hi,
    this can not generally be true, can it?
    I don't believe it is. Then again I've never heard the term asymmetric function as defined here before. But you point out the big reason I don't think it's true - there doesn't seem to be any constraint on the functions and as I was hinting to before you can arbitrarily increase or decrease the standard deviation just by multiplying the result by a constant.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Prove that the SD of an asymmetric function is larger than the SD of a symmetric


    Hi,
    I completely missed the hint. I thought that it shows how one can prove the statement and wondered what I am missing.

    regards
    rogojel

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