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Math Help - minimizing value

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    minimizing value

    Let x be a random variable with support {1, 2, 3, 5, 15, 25, 50}, each point of which has the same probability 1/7. Argue that c=5 is the value that minimizes h(c)=E(|X-c|). Compare this to the value of b that minimizes g(b)=E[(X-b)^2].
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    Hello,
    Quote Originally Posted by antman View Post
    Let x be a random variable with support {1, 2, 3, 5, 15, 25, 50}, each point of which has the same probability 1/7. Argue that c=5 is the value that minimizes h(c)=E(|X-c|). Compare this to the value of b that minimizes g(b)=E[(X-b)^2].
    You can have a look here : http://www.medicine.mcgill.ca/epidem...n-elevator.pdf
    and here for a "similar" problem : http://www.mathhelpforum.com/math-he...blem-50-a.html (I think that the relevant messages are from post #15)

    For b, it's the problem of least squares. You can google for "least squares"
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    Quote Originally Posted by antman View Post
    Let x be a random variable with support {1, 2, 3, 5, 15, 25, 50}, each point of which has the same probability 1/7. Argue that c=5 is the value that minimizes h(c)=E(|X-c|). Compare this to the value of b that minimizes g(b)=E[(X-b)^2].
    Consider \{a,b\},\ b>a, for any c \in [a,b]

     <br />
|a-c|+|b-c|=b-a<br />

    and if c \not\in [a,b] :

    |a-c|+|b-c|>b-a

    Hence any c \in [a,b] minimises |a-c|+|b-c|

    Apply this sequentially to \{1,50\},\ \{2,25\},\ \{3,15\} shows that any c \in [3,15] minimises:

    |1-c|+|2-c|+|3-c|+|15-c|+|25-c|+|50-c|

    and as the last point not accounted for is in this interval and |5-c| is minimised when c=5 we have c=5 minimises:

     <br />
E(|X-c|)=(1/7) \sum |x_i-c|<br />

    CB
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