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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- Apr 2nd 2009, 06:59 PMantmanminimizing 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].

- Apr 2nd 2009, 11:00 PMMoo
Hello,

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" - Apr 2nd 2009, 11:31 PMCaptainBlack
Consider $\displaystyle \{a,b\},\ b>a$, for any $\displaystyle c \in [a,b]$

$\displaystyle

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

$

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

$\displaystyle |a-c|+|b-c|>b-a$

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

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

$\displaystyle |1-c|+|2-c|+|3-c|+|15-c|+|25-c|+|50-c|$

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

$\displaystyle

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

$

CB