In relation to the question I have posted earlier.
I tried to solve the integral: $$p(t|x,\mathbf{x},\mathbf{t}) = \int p(t|x,\mathbf{w})p(\mathbf{w}|\mathbf{x},\mathbf{t}) d\mathbf{w} \tag{1.68}$$ to show that: $$p(t|x,\mathbf{x},\mathbf{t}) = \mathcal{N}(t|m(x),s^2(x)) \tag{1.69}$$ where the mean and variance are: $$m(x) = \beta \phi(x)^T \mathbf{S}\sum_{n=1}^N \phi(x_n)t_n \tag{1.70}$$ $$s^2(x) = \beta^{-1}+\phi(x)^T\mathbf{S}\phi(x) \tag{1.71}$$ And matrix $\mathbf{S}$ is given by: $$\mathbf{S}^{-1} = \alpha \mathbf{I} + \beta\sum_{n=1}^N \phi(x_n)\phi(x_n)^T \tag{1.72}$$ and vector $\phi(x)$ is defined with elements: $\phi_i(x)=x^i$ for $i=0,...,M$.
What I am really interested in is whether I have made any mistakes in the below, because I believe once you reach the last line in the below it just becomes a matter of applying the rule of multiplication of multivariate Gaussians (and it is quite tedious)?
My workings: $$p(t|x,\mathbf{x},\mathbf{t}) = \int p(t|x,\mathbf{w})p(\mathbf{w}|\mathbf{x},\mathbf{t}) d\mathbf{w} \\ = \int \frac{\mathcal{N}(t|y(x,\mathbf{w},\beta^{-1})) \prod_{i: x_i \in \mathbf{x}} \mathcal{N}(t_i|y(x_i,\mathbf{w},\beta^{-1}) p(\mathbf{w}|\alpha)}{\int p(\mathbf{t}|\mathbf{x},\mathbf{w},\beta) p(\mathbf{w}|\alpha) d\mathbf{w}} d\mathbf{w} $$ Let $\tilde{t} = [t,t_1,t_2,..,t_M]^T$ and $\tilde{x} = [x,x_1,...,x_M]^T$, also note that $\int p(\mathbf{t}|\mathbf{x},\mathbf{w},\beta) p(\mathbf{w}|\alpha) d\mathbf{w} =c \in \mathbb{R}$, therefore: $$= c \int \prod_{i: x_i \in \tilde{x}} \mathcal{N} (t_i|y(x_i,\mathbf{w}),\beta^{-1})p(\mathbf{w}|\alpha) d\mathbf{w} \tag{last line}$$ where $p(\mathbf{w}|\alpha)=\left( \frac{\alpha}{2\pi} \right)^{(M+1)/2} \exp \Big\{ -\frac{\alpha}{2} || w ||^2 \Big\}$