I have a question on how it would look the linear regression model given that $\epsilon_{i}\sim Laplace(0,\lambda)$ with a reparametrization $b=\frac{1}{\lambda}$.

$Y_{i}=\alpha+\beta x_{i}+\epsilon_{i} \hspace{.3cm}\forall \hspace{.3cm} i=1,...,n$

That would also imply that the $Y_{i}$ follow also a Laplace distribution?


1 Answer 1


You can't know the distribution of $Y$ without assuming a prior on $x$ (and $w$ if conducting a fully Bayesian analysis). However, given $x$, the target (i.e. $Y$) will be distributed according to Laplace distribution with its mean shifted by $\alpha+\beta x$. This will result in a more robust regression if your data has outliers because compared to Normal, Laplace distribution has heavier tails, which assigns more probability mass to large noise values. Also, the closed-form solution does not exist, you may need to solve iteratively.

  • $\begingroup$ Sure, with that $Laplace(\alpha+\beta x,\lambda)$ i can get the Likelihood to make Bayesian Analysis to get a posterior, supposing i already got the prior on $x$ no?. $\endgroup$ Commented Nov 17, 2019 at 19:41
  • $\begingroup$ Yes, you can if you have $p(x)$. $\endgroup$
    – gunes
    Commented Nov 17, 2019 at 19:43
  • $\begingroup$ Just one final question, under this assumption the Laplace model would have homoscedasticity (the scale parameter will be constant for all the $Y_{i}$? $\endgroup$ Commented Nov 17, 2019 at 20:31

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