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Thread: (Simple) proof for the arithmetic mean?

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    (Simple) proof for the arithmetic mean?




    Hello. I am having trouble trying to figure out a way to answer this homework exercise:

    "Prove that the arithmetic mean cannot be greater than the largest element in the set of numbers from which it was calculated".

    Here's how I began:

    Let X={x1, x2, x3,....xn} be a set of numbers out of which the arithmetic mean is x-bar.

    Suppose that x-bar is greater than some xi which is the greatest element of the set X.

    Then my intuition is that, maybe, I can use the fact that {(x1-xbar)+(x2-xbar)+(x3-xbar)+...+(xi-xbar)+...._(xn-xbar)} *should* add up to 0 but in this case it will not. But I'm not sure how to proceed.

    I know the fact that the arithmetic mean cannot be greater (or smaller) than the greatest(or smallest) element of the set is a well-known and obvious fact, which is why I am frustrated that I cannot show this more explicitly.... particularly because it is so obvious!

    Any help is appreciated!

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    Re: (Simple) proof for the arithmetic mean?

    Denote the greatest element of the set you are considering so that . Then,



    Can you continue?

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    Re: (Simple) proof for the arithmetic mean?

    Well, if I re-express (and assume all x's are ordered from x1 the smallest to xM the greatest, following your notation):

    \frac{1}{n}\sum_{i=1}^{n}x_{i} = \frac{x_{1}}{n}+\frac{x_{2}}{n}+\frac{x_{3}}{n}+...+\frac{x_{M}}{n}

    It becomes immediatley obvious that:

    \frac{x_{M}}{n}\leq x_{M}

    Which implies that any other (individual) element of \frac{x_{1}}{n} , \frac{x_{2}}{n} , \frac{x_{3}}{n}, ..., \frac{x_{M}}{n} is less than x_{M}

    But I cannot quite see how their sum must be less than x_{M}, even though I know it is true

    Should be using properties of finite sums to find this out?

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    Re: (Simple) proof for the arithmetic mean?

    Ok, we'll do it another way. Assume you have ordered your set so that :



    Then you can write :



    ...



    What happens if you sum all these inequalities?

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    Re: (Simple) proof for the arithmetic mean?

    By definition you have the following n bounds / inequalities hold simultaneously:

    x_i \leq x_M ~~ \forall i = 1, 2, \ldots, n

    Therefore you can apply these into the arithmetic mean which immediately yields the desired result.

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    Re: (Simple) proof for the arithmetic mean?


    OH GOD! THANK YOU BOTH!

    I think I got it now. I will make it all explicity just to see if there are any problems on my reasoning.

    As you two pointed out, the following constraints hold:

    x_{1} \leq x_{n}
    x_{2} \leq x_{n}
    x_{3} \leq x_{n}
    \dots

    Which means that:

    x_{1} + x_{2} + x_{3} \dots \leq x_{n}+x_{n}+x_{n} \dots

    So I can do:

    \frac{1}{n}\sum_{i=1}^{n}x_{i} \leq \frac{1}{n}\sum_{i=1}^{n}x_{n}
    \frac{1}{n}\sum_{i=1}^{n}x_{i} \leq \frac{1}{n}  n\cdot x_{n}

    \bar{x} \leq x_{n}

    Thank you! I knew it had to be easy!

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