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Why is $Y=\beta_0 x^{\beta_1} e$ a linear model? When we apply the transform, it becomes $lnY = ln\beta_0+\beta_1 lnx +lne$, and why is it still linear when the $\beta_0$ part is under ln?

user2793618
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    Just give $\log\beta_0$ another name, such as $\gamma_0.$ The two models aren't equivalent, though: think about what happens when $\beta_0$ is negative. – whuber Sep 22 '20 at 22:14

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$Y=\beta_0 x^{\beta_1} e$ is not a linear model, it's a power model that can be linearized to a linear model, $\ln Y = \ln \beta_0+\beta_1 \ln x +\ln e$.

Doing that, and estimating it with least-squares, entail an assumption on the original model: $\ln e \sim \mathcal N (0, \sigma^2)$. In other words, you assume that the $e$ terms are lognormal-distributed.

Firebug
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The meaning of linear is not on based on the predictors. It's based on the beta's. In your example the predictor is on the power of beta 1 not the beta itself.

Hamid Sakhi
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  • Your characterization of "linear" seems overly restrictive. See https://stats.stackexchange.com/a/148713/919 for a detailed discussion of various possible meanings of "linear model." – whuber Sep 23 '20 at 13:43