Going through the Wikipedia article on Maximum a posteriori estimation, it got confusing after reading this:

It is closely related to the method of maximum likelihood (ML) estimation, but employs an augmented optimization objective which incorporates a prior distribution (that quantifies the additional information available through prior knowledge of a related event) over the quantity one wants to estimate. MAP estimation can therefore be seen as a regularization of ML estimation.

How can the MAP estimation be seen as a regularization of ML estimation?


My understanding of regularization being penalizing high weights in the context of Machine learning. That being done through modifying the optimization problem by adding a term in the loss function which contains the weights to be learned. And the objective being minimization of loss, the parameters with higher values get penalized more.

An intuitive explanation is very welcome.


1 Answer 1


Maximum likelihood method aims at finding model parameters that best match some data:

$$ \theta_{ML}=\mathrm{argmax}_\theta \,p(x|y,\theta) $$

Maximum likelihood does not use any prior knowledge about the expected distribution of the parameters $\theta$ and thus may overfit to the particular data $x$, $y$.

Maximum a-posteriori (MAP) method adds a prior distribution of the parameters $\theta$:

$$ \theta_{MAP}=\mathrm{argmax}_\theta \, p(x|y,\theta)p(\theta) $$ The optimal solution must still match the data but it has also to conform to your prior knowledge about the parameter distribution.

How is this related to adding a regularizer term to a loss function?

Instead of optimizing the posterior directly, one often optimizes negative of the logarithm instead:

$$ \begin{align} \theta_{MAP}&=\mathrm{argmin}_\theta \, -\log p(x|y,\theta)p(\theta) \\ &=\mathrm{argmin}_\theta \, -(\log p(x|y,\theta) + \log p(\theta)) \end{align} $$

Assuming you want the parameters $\theta$ to be normally distributed around zero, you get $\log p(\theta) \propto ||\theta||_2^2$.

  • $\begingroup$ While I understand the latter part of your answer about regularisation, the thing that's bugging me is this statement: Maximum likelihood does not use any prior knowledge about the expected distribution of the parameters θ and thus may overfit to the particular data x, y. When is this situation more likely? Don't we run the risk of degrading the model by making an assumption on the prior? $\endgroup$
    – naive
    Commented Sep 19, 2018 at 10:54
  • 1
    $\begingroup$ By "degrading the model" you mean limiting its flexibility? Of course you do - but you are limiting its flexibility to prefer plausible solutions. The prior can be very weak (when you are not certain how the parameters should look like) and should be strong if you have a very good idea what you are looking for. Also, note that when you have lots of data, the regularization term becomes negligible compared to the likelihood term. $\endgroup$ Commented Sep 19, 2018 at 11:17

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