The linear in linear least squares regression

[joshthekid]

It is my understanding that you can use linear least squares to fit a plethora of different functions (quadratic, cubic, quartic etc). The requirement of linearity applies to the coefficients (i.e B in (y-Bx)^2). It seems to me that I can find a solution such that a coefficient b_i^2=c_i, in other words I can just express the squared b_i with a linear c_i. So can't I always find a linear solution?

My gut feeling is that linearity is required because normality is required for the orthoganality principle to hold? But I am not sure.

Thanks

 

[SteamKing]

It is my understanding that you can use linear least squares to fit a plethora of different functions (quadratic, cubic, quartic etc). The requirement of linearity applies to the coefficients (i.e B in (y-Bx)^2). It seems to me that I can find a solution such that a coefficient b_i^2=c_i, in other words I can just express the squared b_i with a linear c_i. So can't I always find a linear solution?

My gut feeling is that linearity is required because normality is required for the orthoganality principle to hold? But I am not sure.

Thanks

I'm not sure what your question is here.

The least squares method for fitting a set of data to a general polynomial function (linear, quadratic, cubic, etc.) results in a set of linear equations which must be solved to determine the unknown coefficients of the polynomial. The number of equations to be solved is equal to the degree of the polynomial plus 1.

 

[BWV]

Normality is only required in OLS for accurate testing of error terms. As long as you have finite variance and a non-singular covariance matrix you can get the beta terms, nothing is required to be orthogonal.

 

[Number Nine]

It has nothing to with normality; linearity in the parameters just let's you derive a solution using linear algebra. It's entirely possible to fit non-linear functions by least squares, it just requires numerical methods.

The model
y = \alpha + \beta^2x + \epsilon
is linear in the parameters (\alpha, \beta^2), so there's no problem. Conversely, the model
y = \alpha e^{\beta x} + \epsilon
is clearly non-linear in \beta.

Normality is only required in OLS for accurate testing of error terms. As long as you have finite variance and a non-singular covariance matrix you can get the beta terms.

This is mostly true. Strictly speaking, the Gauss-Markov theorem guarantees that the least-squares estimates have good properties assuming only zero-mean, equal variance, and uncorrelated errors, so the normality assumption isn't so important if you just want to curve fit, but modelling is easier assuming normal errors. Under the usual normality assumptions, the least-squares estimates are the maximum likelihood estimates, which are better understood than the more general least-squares estimates. It also makes in easier test hypotheses about the coefficients.

 

[Stefan Degonda]

I am not completely sure about my answers, but I try my best.

You do not need normally distributed error terms for an OLS estimator to be unbiased, nor is that the reason that you have linearity in coefficients. According to the central limit theorem your estimates will be normally distributed in the limit anyway. However, you want to have normally distributed error terms for hypothesis testing in small samples. Overall the assumption of normality has something to do with the assumption of the classical linear model in general an not with the linearity of its estimates (for an elaboration on the ols assumption see for example http://economictheoryblog.com/2015/04/01/ols_assumptions).

I think the answer to you question is far more technical, in the sense that what you get for your coefficient is a scalar which is constant per se. In that sense it cannot be a function of an non linear form. However, that is my own understanding of this subject and I would love to hear another answer to this question if there is one.

Cheers,
Stefan