In Bayes inference, we define p(w) and p(y|X, w) as following. $$ p(w) = N(w | 0, \sigma_w^2 I) \\ p(y | X, w) = N(y | Xw, \sigma_y^2 I) $$
and we can calculate p(w|X, w) as following.
$$ p(w | X, y) = N(\mu, \Sigma) \\ \Sigma = (\frac{1}{\sigma_y^2} X^T X + \frac{1}{\sigma_w^2})^{-1} \\ \mu = \frac{1}{\sigma_y^2}\Sigma X^T y $$
and we can predict as following. $$ p(y^* | x^*, X, w) = N(y^* | \mu^*, \sigma^*{^2}) = N(y^* | x^*{^T} \mu, \sigma_y^2 + x^*{^T} \Sigma x^*) $$
and many textbooks say that $\mu^*$ means trust of prediction.
When mean of X is 0, it sounds correct. Because $\sigma^*{^2}$ is minimum when $x^* = 0$.
However when mean of X is $\overline{X}$ ($\overline{X} \not= 0$), $\sigma^*{^2}$ is minimum when $x^* = 0$ too.
If we consider $\sigma^*$ as trust of prediction, $\sigma^*{^2}$ should be minimum when $x^* = \overline{X}$.
In derivation of Bayes inference, we does not assume that mean of X is 0.
So I can't understand why $\sigma^*{^2}$ is considered as trust of prediction. is it necessary that mean of X is 0 ?
Could anyone help me.