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Nick Cox
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There is a small point of clarification here: We actually don't need the error term to be Gaussian (see the GausGauss-Markov Theorem), but if we assume it is then some nice connections can be made.

Linear regression is commonly motivated by minimizing the squared error loss. Squared error is perhaps the simplest loss function to motivate such a model with, so let's go with it for now. It turns out that minimizing the squared error loss is the same as assuming the likelihood for $y\vert x$ is gaussianGaussian, hence the normality of the error term.

There is a small point of clarification here: We actually don't need the error term to be Gaussian (see the Gaus-Markov Theorem), but if we assume it is then some nice connections can be made.

Linear regression is commonly motivated by minimizing the squared error loss. Squared error is perhaps the simplest loss function to motivate such a model with, so let's go with it for now. It turns out that minimizing the squared error loss is the same as assuming the likelihood for $y\vert x$ is gaussian, hence the normality of the error term.

There is a small point of clarification here: We actually don't need the error term to be Gaussian (see the Gauss-Markov Theorem), but if we assume it is then some nice connections can be made.

Linear regression is commonly motivated by minimizing the squared error loss. Squared error is perhaps the simplest loss function to motivate such a model with, so let's go with it for now. It turns out that minimizing the squared error loss is the same as assuming the likelihood for $y\vert x$ is Gaussian, hence the normality of the error term.

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Demetri Pananos
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There is a small point of clarification here: We actually don't need the error term to be Gaussian (see the Gaus-Markov Theorem), but if we assume it is then some nice connections can be made.

Linear regression is commonly motivated by minimizing the squared error loss. Squared error is perhaps the simplest loss function to motivate such a model with, so let's go with it for now. It turns out that minimizing the squared error loss is the same as assuming the likelihood for $y\vert x$ is gaussian, hence the normality of the error term.