I am trying to estimate the regression model, say standard linear model, with the error term having a Pareto distribution instead of normal. Although it is fairly straightforward to construct the maximum likelihood function, it is practically not easy because we do know how to obtain the estimate for the minimum of the error term as it is an unobserved variable. What I can think of to do it is to estimate the coefficient and Pareto parameters with the grid of predetermined the minimum threshold Pareto parameters and choose the one with the largest likelihood value. Is that how people practically do that?

The model I have in mind is:

$y_i=\beta_0 + \beta_1 x_i + \epsilon_i$ where $\epsilon_i \sim Pareto(k,\alpha)$

where $\epsilon_i>0$, and $\epsilon_i >k>0$. So, you would estimate $\beta_0$, $\beta_1$, $k$ and $\alpha$.

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    $\begingroup$ It is strange to see "error term" and "Pareto distribution" in the same sentence, because it is hard to conceive of the Pareto as a meaningful distribution for an additive error. Are you perhaps using it multiplicatively? (BTW, is there a "not" missing from "we do ... know how"?) $\endgroup$
    – whuber
    May 31, 2016 at 16:57
  • $\begingroup$ The error is added additively not multiplicatively. $\endgroup$
    – Ikuyasu
    May 31, 2016 at 18:15
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    $\begingroup$ Thank you. I'm still trying to understand what you mean by "practically not easy." I'm comparing your situation to ordinary least squares regression, where the error variance is unobservable (but easily estimated) and to GLMs whose parameters are unobservable but also readily estimated with MLE. What, then, is so special about this particular parameter (the "minimum of the error term")? $\endgroup$
    – whuber
    May 31, 2016 at 18:20
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    $\begingroup$ It is well-known that the MLE for the lower bound $k$ of a Pareto distribution is the minimum value from the sample: $\hat{k} = \min x_i$. The MLE for the exponential parameter $\alpha$ is also easily found: $\hat{\alpha} = n / \sum \log (x_i/\hat{k})$. $\endgroup$
    – jbowman
    May 31, 2016 at 18:32
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    $\begingroup$ I can't quite determine what model you really have in mind. For a random response $Y$ and non-random regressor $x$, would it be something like $$Y\sim\text{Pareto}(\beta_0+\beta_1x,\alpha)$$ or might it be something like $$Y - (\beta_0+\beta_1x) \sim\text{Pareto}(k,\alpha)$$ or maybe even $$Y - (\beta_0+\beta_1x)\sim\text{Pareto}(1,\alpha)$$? (The first parameter $k$ is the scale parameter and $\alpha$ is the shape parameter.) Could you be more specific about the model you want to fit? $\endgroup$
    – whuber
    May 31, 2016 at 18:39


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