Bayesian Linear Regression I have the following question concerning Bayesian linear regression on my machine learning assignment:

Consider $f = w^Tx$, where $p(w) ∼ N(w | 0, Σ)$. Show that $p(f | x)$ is Gaussian. 

I believe that $p(f|x) ∼ N(w^Tx, \sigma^2)$ for some variance $\sigma^2$ but I'm not sure how to show it. I'm also not sure why the prior was mentioned for this question since we're deriving an expression for the likelihood. If anyone could give me some tips to push me in the right direction it'd be appreciated (please don't solve the entire problem).
 A: $p(f|\mathbf{x})$ is actually the prior for Gaussian process not the likelihood. Gaussian process by definition is assuming a particular prior over functions (represented as random variable $f|\mathbf{x}$) not parameters ($\mathbf{w}$). When the model is $y = f(x) + \epsilon$ where usually $\epsilon$ is considered to the noise in observation $y$, the likelihood (of $f$) is $p(y | f)$, the prior (of $f$) is $p(f|\mathbf{x})$ and the posterior (of $f$) is $p(f|y)$. $f$ is called the latent/hiden variable meaning that we cannot observe its values.  Therefore, we are usually not interested in $p(f|y)$. $y$ is the variable representing observation samples/training data.
By the assumption of linear regression model, $f$ is a linear function. Therefore, it can be expressed as $\mathbf{w}^T\mathbf{x}$.  Now, we want to estimate parameters $\mathbf{w}$ using a Bayesian approach. The likelihood (of parameters) would be $p(y|\mathbf{w})$, the posterior (of parameters) is $p(\mathbf{w}|y)$ and the prior (of parameters) is $p(\mathbf{w})$.
You are trying to be prove assuming $\mathbf{w}\sim\mathcal{N}(\mathbf{0}, \Sigma)$ and linearity of $f$ entails Gaussian process prior assumption.
You are proving when the Gaussian process assumption meets the linear regression assumption then the two priors (one over functions, the other over parameters) are equivalent. Mean is not necessarily zero though. That is to simplify algebra.
A: $p(f|x)$ can not be $\mathcal{N}(w^T x, \sigma^2)$ since $w$ is a random variable and you don't condition on it. If you did, then, $p(f|x,w)$ would be singular: for given $w$ and $x$ we know exactly what the outcome $f$ will be (since there's no noise in $f$). It's like having $\sigma=0$, though, $\mathcal{N}(\mu, 0)$ is not well-defined. All randomness in $f$ comes from $w$.
As mentioned in the comments, $f = w^T x$ turns out to be Gaussian, as well. This means $f \sim \mathcal{N}(\mu, \sigma^2)$ where
$$
\mu = \mathbb{E} [f] = \mathbb{E} [w^T x] \\
\sigma^2 = \mathbb{E} [f^2] - \mathbb{E} [f]^2 = \mathbb{E} [w^T x w^T x] - \mu^2
$$
The most difficult to compute here is $\mathbb{E} [w^T x w^T x]$, but using algebraic manipulations (like, rearranging terms) you can do it.
