# Simple Linear Regression With Laplace Distribution (Double Exponential)

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?

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.
• 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?. Commented Nov 17, 2019 at 19:41
• Yes, you can if you have $p(x)$. Commented Nov 17, 2019 at 19:43
• Just one final question, under this assumption the Laplace model would have homoscedasticity (the scale parameter will be constant for all the $Y_{i}$? Commented Nov 17, 2019 at 20:31