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].
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,
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"
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