1. ## 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].

2. Hello,
Originally Posted by antman
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"

3. Originally Posted by antman
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 $c \in [a,b]$

$
|a-c|+|b-c|=b-a
$

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:

$
E(|X-c|)=(1/7) \sum |x_i-c|
$

CB

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# 1, 2, 3, 5, 15, 25, 50

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