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I'm curious what would be the posterior distribution having prior dstribution as a mixture of two guassian with the likelihood dist as a gaussian. In other words:

  • Likelihood: $p(x|\mu,\sigma) = \mathcal{N}(\mu,\sigma^2)$
  • Prior: $p(\mu) = a\cdot\mathcal{N}(0,1)+b\cdot\mathcal{N}(1,1)$
  • Posterior: $p(\mu|x)=???$

Also, I would like to know the MAP objective of it. For example, giving a guassian prior amounts to the $L^2$ regularization, then which kind of regularization with prior with gaussian mixture model would induce?

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You can rewrite the prior as a Gaussian hyper prior: $$ p(\mu|i) = \mathrm{N}(i,1), \text{where } i \in \{0,1\} $$ and set a prior for $i$, say $$ p_0(i) = \begin{cases}a, & i=0, \\ b, & i=1. \end{cases} $$ First you could ignore the prior on $i$ and compute the posterior measure analytically parameterised in $i$. In this case, you would have just a Gaussian prior and a Gaussian likelihood, so you can compute the posterior $$ p(\mu|x,i) = \frac{p(x|\mu)p(\mu|i)}{p(x|i)} \propto p(x|\mu)p(\mu|i) $$ analytically using Bayes' formula (and the cheat sheet Wikipedia entry on conjugate priors).

Note that $p(\mu|x,i)$ is still Gaussian. To get to the actual posterior $p(\mu|x)$ we want to get rid of the $I$. This can be achieved using the following formula: $$ p(\mu|x) = \sum_{i=0}^1 p(i|x)\cdot p(\mu|x,i), \tag 1 $$ where $$ p(i|x) = \frac{p_0(i)\cdot p(x|i)}{\sum_{j=0}^1 p_0(j)\cdot p(x|j)}. $$ is the posterior measure over $i$ given the data $x$.

In any case, the posterior $p(\mu|x)$ will be again a Gaussian mixture with two components. This you can see in equation (1): We have a convex combination of two Gaussian measures there.

The question about the regularisation is somewhat more tricky. You get an $L^2$ regularisation in the Gaussian prior case, since the negative log-prior is an $L^2$-norm. Here, we have the regulariser $R(\mu)$ given by $$ R(\mu) = -\log(p(\mu)) = -\log\left(a \frac{1}{\sqrt{2\pi}} \exp(-\frac12\|\mu\|^2) + b \frac{1}{\sqrt{2\pi}} \exp(-\frac12\|\mu-1\|^2)\right). $$ I do not actually see a way to simplify this term. It is definitely something non-standard to do as a regularisation.

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