In Bayesian linear regression, why do we assume the prior parameter w has zero mean?

Question

Likely question:In Bayesian linear regression, why do we assume parameter prior has zero mean?.

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But, I just focus on can we assume $p(w \mid \alpha) = \mathcal{N}(w \mid \mu, \alpha^{-1})$, and the reason of zero mean.

If mean of $w$ is $\mu$, we should minimum of $$\frac{\beta}{2} \sum_{n=1}^N\{y(x_n,w)-t_n\}^2+\frac{\alpha}{2}(w-\mu)^T(w-\mu)$$

so the result is not Ridge Regression(just my opinion).

My Question: can we assume $w$ is not zero mean? Does the result of whether the mean of $w$ is zero lead to different result?

Reason of ask the question: I didn't find the answer in the question above. I also want to know the reason \ undermeaning of assume zero mean.

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This problem is from book PRML 1.2.5 Curve fitting re-visited Page28.

Apologize for not being word-perfect in English.

Answer 1

Slightly too long for a comment, not sure if this consistutes a complete answer.

Philosophically, there is no reason you could not set $\mu \ne 0$. I think the reason people tend to use $\mu = 0$ is that there isn't any other natural default choice. By setting $\mu = 0$, you are shrinking towards a model in which $w = 0$, which has a more parsimonious feel to it. That is, we are shrinking towards a model in which $w$ is not included in the model. There is also something of a rotational invariance sort of argument: one way of expressing ignorance about the predictors is to say that any orthogonal transformation of the predictors should have the same prior. This implies we should take $\mu = 0$ and use the covariance $\Sigma = \alpha^{-1} I$.