Thread: 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].

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"

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 , for any



and if :



Hence any minimises

Apply this sequentially to shows that any minimises:



and as the last point not accounted for is in this interval and is minimised when we have minimises:



CB