Solving the Equation: Epsilon = 0

  • Thread starter Thread starter Axll
  • Start date Start date
  • Tags Tags
    Epsilon
Click For Summary
SUMMARY

The discussion focuses on solving the equation for epsilon, defined as the residual value in the linear regression formula: epsilon = Yi - (b1 + b2*Xi). The coefficients b1 and b2 are calculated using the mean values of Y and X, with b2 derived from the formula b2 = (sum of Xi*Yi - n*mean value of X*mean value of Y)/(squared sum of Xi - n*squared mean value of X). The key conclusion is that the sum of all residuals (epsilon) equals zero, confirming the properties of least squares regression.

PREREQUISITES
  • Understanding of linear regression concepts
  • Familiarity with statistical notation and symbols
  • Knowledge of mean and variance calculations
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation of linear regression coefficients in detail
  • Learn about the properties of residuals in regression analysis
  • Explore the implications of the least squares method
  • Investigate the use of statistical software for regression analysis, such as R or Python's statsmodels
USEFUL FOR

Students and professionals in statistics, data analysis, and machine learning who are looking to deepen their understanding of linear regression and residual analysis.

Axll
Messages
1
Reaction score
0
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?
 
Physics news on Phys.org
Let's clean up these formulas a bit first. I think your epsilon caries an index:
<br /> \begin{align}<br /> \epsilon_i &amp;= y_i - (b_1 + b_2 x_i)\\<br /> \intertext{where}<br /> b_1 &amp;= \bar{y} - b_2 \bar{x} \, ,\\<br /> b_2 &amp;= \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}<br /> \end{align}<br />
Expression (3) looks ugly, but you will not need to use it explicitly.
Sum over all N of the epsilons:
<br /> \sum_i \epsilon_i = \sum_i y_i - (b_1 + b_2 x_i) \, .<br />
Now show that the sum goes to zero. Hint:
<br /> \sum_i x_i = N \bar{x}<br />
 
Last edited:

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
Replies
3
Views
2K
Replies
10
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K