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Is a generalized linear model with a Gaussian distribution the same as a linear model?

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    $\begingroup$ If using an identity link, yes, I think it is but not confident enough to post as an answer! $\endgroup$
    – tristan
    Apr 2, 2015 at 13:02
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    $\begingroup$ @tristan: Not necessarily, think e.g. about a linear model that optimizes L1 loss instead of squared loss. The Gaussian GLM with identity link is essentially identical to the corresponding least squares model. $\endgroup$
    – Michael M
    Apr 2, 2015 at 13:22
  • $\begingroup$ @MichaelM good point, that's why I didn't add as an answer because I wasn't confident of caveats etc $\endgroup$
    – tristan
    Apr 2, 2015 at 13:23

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Essentially yes, if you think about a gaussian generalized linear model (glm) with identity link function, then that is the same as what is usually meant with a linear model.

But, "linear model" could be used with other shades of meaning, like estimating it with some other loss function than sum of squares. And a gaussian glm could use some other link function than the identity.

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  • $\begingroup$ What about having unknown variance $\sigma^2$ in the "linear model"? $\endgroup$ Jul 14, 2021 at 1:27
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In the generalized linear model (specially meant for classification problem), the source of non-linearity is activation funtion with linear argument of the form $\omega^\top x +\omega_0$. Here, we just need to estimate the parameters $\omega, \omega_0$ only and the activation might be user defined.

While, in regression problem, the using least square estimation, the prediction function happens to be a conditional expectation of the form $E(y|x)$. If the joint distribution of x and y is Gaussian, then the prediction function $E(y|x)=\omega_0 + \omega^\top x$ linear both in the parameters $\omega, \ \omega_0$ and in the predictor x.

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