Given two normal distribution $y_* | \mathbf{x}_*, \mathbf{w} \sim N( \mathbf{x}_*^T \mathbf{w}, \sigma_n^2)$ and $\mathbf{w} \sim N(\mathbf{\bar{w}}, A^{-1})$. I am trying to show that
$$y_* | \mathbf{x}_* \sim N(\mathbf{x}_*^T \mathbf{\bar{w}}, \mathbf{x}_*^TA^{-1} \mathbf{x}_*).$$
It will be great if we can show for vector $\mathbf{y}_*$. But I am trying the 1 demsion case. Below is my workings.
We want to show that
$$f(y_* | \mathbf{x}_*) = \frac{1}{\sqrt{2\pi \mathbf{x}_*^T A^{-1} \mathbf{x}_*}} \exp \left( -\frac{1}{2} \left( \frac{y_* - \mathbf{x}_*^T \mathbf{\bar{w}}}{\mathbf{x}_*^T A^{-1} \mathbf{x}_*} \right)^2 \right).$$
Consider
$$ \begin{align*} f(y_* | \mathbf{x}_*) &= \int f(y_* | \mathbf{x}_*, \mathbf{w}) \cdot f(\mathbf{w}) \, d\mathbf{w} \\ &= \int \left[ \frac{1}{\sqrt{2 \pi \sigma_n^2}} \exp\left( -\frac{1}{2} \left( \frac{y_* - \mathbf{x}_*^T \mathbf{w}}{\sigma_n} \right)^2 \right) \right] \\ & \qquad \cdot \left[ \frac{1}{\sqrt{(2\pi)^p \det(A^{-1})}} \exp \left( -\frac{1}{2} (\mathbf{w} - \mathbf{\bar{w}})^T A (\mathbf{w} - \mathbf{\bar{w}}) \right) \right] d\mathbf{w}. \end{align*} $$
Consider the terms inside the exponential,
$$ \begin{align*} I := -\frac12 \left( \frac{y_*^2}{\sigma_n^2} - \frac{2\mathbf{x}_*^T \mathbf{w} f_*}{\sigma_n^2} + \frac{(\mathbf{x}_*^T \mathbf{w})^2}{\sigma_n^2} + \mathbf{w}^TA\mathbf{w} - 2\mathbf{w}^TA\mathbf{\bar{w}} + \mathbf{\bar{w}}^T A \mathbf{\bar{w}} \right). \end{align*} $$
I am not sure if I should keep expending.