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We all know that we can fit a linear regression model even in cases when our predictors are not linearly related to the dependent. Like the example below (polynomial regression) -

(y = a + b X^2) can be written as (y = a + b Z) where Z = X^2

My question is, can there be changes to the form of the Beta coefficients as well?

More over, I am thinking of a few peculiar cases like this -

y = a + e^b X (where e is the Euler number)

would this still qualify for a linear regression or these mutations cannot be applied on the coefficients?

Thank you.

Anaconda_named_Python
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1 Answers1

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Is e the Euler number? If so then let B=e^b and you can find b by taking the natural logarithm of whatever coefficient you got

The linear regression will attempt to fit a linear model to the data, so even if the relationship is non linear it will only give you a coefficient for $X$ and not $X^2$. If you wanted the coefficient for $X^2$ I guess you could square all your data points and it would give you the coefficient. This is never done though, you would specify a quadratic term instead.

I do not think it makes sense to express the coefficient as a function of some true coefficient. It may help in interpretability, as in logistic regression with a binary outcome variable, you must exponentiate your coefficients to determine the effect a one unit increase will have on the odds of the response variable being $1$. It increases the odds by a factor of that much.

Vons
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  • yes, e is the Euler number; have edited the same. So per my understanding, this can be challenging, my only point being, if X^2 can be transformed to a polynomial regression, why not the apply the same logic for the Betas. say the e^b in y = a + e^b X be represented like c and create the new equation like y = a + cX, sorry if this is not adding much info, iam a bit confused here. – Anaconda_named_Python May 01 '21 at 08:15
  • No there is no point in doing that. You could transform the response variable though. log y = a+ bx – Vons May 01 '21 at 13:16