I'm learning the book "Introduction to Statistical Learning" and in the Chapter 6 about "Linear Model Selection and Regularization", there is a small part about "Bayesian Interpretation for Ridge Regression and the Lasso" that I haven't understood the reasoning.

A Bayesian viewpoint for regression assumes that the coefficient vector $\beta$ has some prior distribution, say $p(\beta)$, where $\beta = (\beta_0, \beta_1, \dots, \beta_p)^\top$. The likelihood of the data can be written as $f(Y|X, \beta)$, where $X = (X_1, X_2, \dots, X_p)$. Multiplying the prior distribution by the likelihood gives us (up to a proportionality constant) the posterior distribution, which takes the form $$p(\beta|X, Y) \propto f(Y|X,\beta)p(\beta|X) = f(Y|X, \beta)p(\beta),$$ where the proportionality above follows from Bayes’ theorem, and the equality above follows from the assumption that X is fixed.

The authors assume the linear model: $$Y = \beta_0 + X_1\beta_1 + \dots + X_p\beta_p + \epsilon,$$ and suppose the errors are independent and drawn from a normal distribution. There is also assumption that $p(\beta) = \prod_{j=1}^p g(\beta_j)$ for some density function $g$.

Then from those points,

It turns out that ridge regression and the lasso follow naturally from two special cases of $g$:

  • If $g$ is a Gaussian distribution with mean zero and standard deviation a function of $\lambda$, then it follows that the posterior mode for $\beta$ $-$ that is, the most likely value for $\beta$ , given the data—is given by the ridge regression solution. (In fact, the ridge regression solution is also the posterior mean.)
  • If $g$ is a double-exponential (Laplace) distribution with mean zero and scale parameter a function of $\lambda$, then it follows that the posterior mode for $\beta$ is the lasso solution. (However, the lasso solution is not the posterior mean, and in fact, the posterior mean does not yield a sparse coefficient vector.)

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I don't understand why ridge regression and the lasso follow 2 special cases of $g$ like that and why they have such distribution. Would anyone please help me explain this? Thank you so much in advance for all your help!

  • 4
    $\begingroup$ Try this blog post bjlkeng.github.io/posts/… $\endgroup$ Jan 31, 2019 at 11:53
  • $\begingroup$ Thank you @DmitryMagas. I will try the blog post that you mentioned. $\endgroup$
    – Sophil
    Jan 31, 2019 at 15:03
  • $\begingroup$ @DmitriiMagas that's a great blog post and helped me a lot in my udnerstanding, thank you! Slightly offtopic question: may I ask how you find such nice blogs? $\endgroup$
    – oliver
    Jul 4, 2021 at 11:27

1 Answer 1


Least squares, Lasso and Rigde regression minimie the following objective functions respectively:

$\min ||y - X \beta||_2^2 $

$\min ||y - X \beta||_2^2 + \lambda ||\beta||_1 $,

$\min ||y - X \beta||_2^2 + \lambda ||\beta||_2 $,

until this point, this optimization has nothing to do with distribution(No assumption made on the distribution of y and parameter).

However, it would be preferred if we can add probability interpretation for the minimizer, this is why assume some distribution on y and parameters.

Now assume that $y|X,\beta \sim N(X \beta, \sigma I)$, then Least square minimizer is the Maximum likelihood estimator.

Further if assume $\beta \sim N(0, I)$, then rigde minimizer is the maximum a posterior probability (MAP) estimator while assume $\beta$ laplace distribution, then lasso minimizer is also the maximum a posterior probability (MAP) estimator.

In summary, we assume distribution on y and $\beta$ to give proability interpretation of the minimizer, However, these assumptions not necessarily hold in reality. Just for interpretation purpose only.

  • 2
    $\begingroup$ In ridge regression the regularization term is $\lambda \| \beta\|_2^2$. $\endgroup$
    – littleO
    May 19, 2021 at 9:00

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