A Error in (Multi)linear Regression

WWGD
Science Advisor
Homework Helper
Messages
7,746
Reaction score
12,954
Hi,
I keep reading varying accounts on conditions needed to " justify" the use of ( multi) linear regression to model data.

Specifically, I have seen several authors require errors to be normal, i.i.d , whilr others only require the errors be i.i.d with mean 0. Just where is the assumption of normality used to justify the use of linear models? I know of Gauss Mark of, but this seems too strong. I've heard that it is used to find the distribution of the coefficients and determine reliable confidence intervals for the coefficients? If so, do you suggest a source? If not, can you explain?
 
Physics news on Phys.org
I guess the question is what does "justify" mean. If the errors are zero mean iid normal and equally distributed, then I think the multi linear model is an unbiased estimator of the actual values. If the errors are otherwise distributed then a different model (i.e. different coefficients) might be a better choice.

But in practice, a multi linear model works pretty well in lots of other situations, and it's often hard to know that the errors actually look like.
 
Thank you. Just curious, as an aside, do you know of situations that cannot be modeled (log)linearly; with linearity meaning linearity in the coefficients?
 
You just mean give a scenario where a non linear model is better? Sure, suppose you drop an object, and write down the time and the height at those times as it falls. The height measurement has some noise to it. Then the right model for the height is going to be something like ##h(t)=-\frac{g}{2}t^2+h_0## if it starts at a height of ##h_0##.

I feel like there's a good chance I did not understand the question.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
Replies
3
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K