# Algebric proofs

1. Nov 21, 2005

### Axll

I need help solving this

the sum of the residual values is equal to 0, epsilon used as symbol to represent the residual value in this formula here

epsilon = Yi - b1 - b2*Xi

b1 = mean value of Y - b2*mean value of X
and
b2 = (sum of Xi*Yi - n*mean value of X*mean value of Y)/(squared sum of Xi - n*squared mean value of X)

i checked out a few books out of the library, but none we're helpful at all and the textbook we use in class is pretty much useless, the teacher doesn't even use it

Can anyone help me out please?

2. Dec 1, 2005

### bigplanet401

Let's clean up these formulas a bit first. I think your epsilon caries an index:
\begin{align} \epsilon_i &= y_i - (b_1 + b_2 x_i)\\ \intertext{where} b_1 &= \bar{y} - b_2 \bar{x} \, ,\\ b_2 &= \frac{\sum_i x_i y_i - N\bar{x}\bar{y}}{\left(\sum_i x_i \right)^2 - N \bar{x}^2} \, .\label{e:ugly} \end{align}
Expression (3) looks ugly, but you will not need to use it explicitly.
Sum over all $$N$$ of the epsilons:
$$\sum_i \epsilon_i = \sum_i y_i - (b_1 + b_2 x_i) \, .$$
Now show that the sum goes to zero. Hint:
$$\sum_i x_i = N \bar{x}$$

Last edited: Dec 1, 2005