Need help understanding this equality

[mkkrnfoo85]

Hello, I am really bothered by this because I can't seem to be able to prove the following is true.

$$\sum_{i=1}^n (x_i - u)^2 = \sum_{i=1}^n (x_i - x_{avg})^2 + n(x_{avg} - u)^2$$

for $$u = constant$$
and $$x_{avg} = \frac{\sum x_i}{n}$$

I just can't seem to grasp how I can prove that the right-side equals the left-side. Any push in the right direction would be extremely helpful.

Thanks in advance.

-Mark

 

[cristo]

What've you tried doing? Have you tried expanding the RHS and using the definition of x_avg? What do you get after doing this?

 

[HallsofIvy]

Your definition of "x_avg" is wrong. You need
$$x_{avg}= \frac{\sum x_i}{n}$$
so $\sum x_i= nx_{avg}$

Multiply it out on the left: $\sum (x_i^2- 2ux_i+ u^2)= \sum x_i^2- 2u\sum x_i+ u^2 \sum 1$
Of course, $\sum x_i= nx_{avg}$ and $\sum 1$= n so that is $\sum x_i^2- 2nux_{avg}+ nu^2$

Now do the same on the right and you should be able to cancel some things and arrive at the same answer.

 

[mkkrnfoo85]

oh that helped a lot. Thx, cristo and halls, and thanks for the correction. I got to the conclusion:

$$n(u^2-2ux_{avg}) == n(u^2-2ux_{avg})$$

=).

I had one more question. I need to know this relation to use in a proof in my 'probability' class. Is there an intuitive way to look at the equality to make it seem more obvious, and/or easier to memorize? Any takes on that would be helpful. Thanks again.

-Mark