Minimizing a Function: Finding the Optimal Value of a in f(a)

[autre]

Homework Statement

Let $f(a)=\frac{1}{n-1}\sum_{i=1}^{n}(x_{i}-a)^{2}$

Find the value of a that minimizes f(a) by replacing $(x_{i}-a)$ by $((x_{i}-\bar{x})+(\bar{x}-a))$.


2. The attempt at a solution
$f(a)=\frac{1}{n-1}\sum_{i=1}^{n}((x_{i}-\bar{x})+(\bar{x}-a))^{2}=\frac{1}{n-1}\sum_{i=1}^{n}((x_{i}-\bar{x})^{2}+(\bar{x}-a){}^{2}+2(x_{i}-\bar{x})(\bar{x}-a))=\frac{1}{n-1}[\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}+\sum(\bar{x}-a){}^{2}+\sum2(x_{i}-\bar{x})(\bar{x}-a)]$

I'm a little stuck here. Any ideas?

 

[autre]

Any ideas, guys?

 

[Dick]

Any ideas, guys?

Try using $\sum_{i=1}^{n} x_{i}=n \bar{x}$.

 

[autre]

Thanks!

 

[SammyS]

Homework Statement

Let $f(a)=\frac{1}{n-1}\sum_{i=1}^{n}(x_{i}-a)^{2}$

Find the value of a that minimizes f(a) by replacing $(x_{i}-a)$ by $((x_{i}-\bar{x})+(\bar{x}-a))$.


2. The attempt at a solution
$f(a)=\frac{1}{n-1}\sum_{i=1}^{n}((x_{i}-\bar{x})+(\bar{x}-a))^{2}=\frac{1}{n-1}\sum_{i=1}^{n}((x_{i}-\bar{x})^{2}+(\bar{x}-a){}^{2}+2(x_{i}-\bar{x})(\bar{x}-a))$

$=\frac{1}{n-1}\left[\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}+\sum(\bar{x}-a){}^{2}+\sum2(x_{i}-\bar{x})(\bar{x}-a)\right]$​

I'm a little stuck here. Any ideas?

In your final expression, the first sum doesn't depend on a. It's easy to make the other two equal to zero. Is that the minimum possible?

Is the third sum equal to zero for all values of a ?

 

[autre]

It's easy to make the other two equal to zero. Is that the minimum possible?

I know that the minimum should be the mean a = 1\n$\sum x_i$. How do I make the other two equal to zero?

 

[Dick]

I know that the minimum should be the mean a = 1\n$\sum x_i$. How do I make the other two equal to zero?

What is $\Sigma^n_1 (x_i-\bar{x})$?

 

[SammyS]

I know that the minimum should be the mean a = 1\n$\sum x_i$. How do I make the other two equal to zero?

Do you realize that $\bar{x}$ is:

$\displaystyle\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i \,?$​

So you're saying that the solution is $a=\bar{x}\,.$