Gaussian noise model derivation I have the following linear regression model, $y = f(x;w) + n$, where $y$ is the vector of true labels, $x$ is the observed data, $f(x;w) = w^Tx$, and $n$ ~ $N(0, \sigma^2)$ is the noise.  Why then does $p(y|x;w) = N(w^Tx;\sigma^2)$ (taken from page 12 of these notes: http://cs229.stanford.edu/notes/cs229-notes1.pdf)?  Could someone please derive this for me?
 A: I think whuber boils the questions down to its essentials with

adding a constant $μ$ to a Normal$(0,σ^2)$ variable gives a Normal$(μ,σ^2)$ variable      

So how do we know this is true? How we show that depends on what you're prepared to start with. To talk about the normal at all we must begin with some kind of definition, so it depends on what you define the general normal to be. (As whuber points out, Wikipedia defines $N(\mu,\sigma^2)$ as a scaled, shifted standard normal, from which the pdf can be derived).
We can go the other way; we could begin with say a $N(0,\sigma^2)$ (your $n$ variable) and investigate the effect of adding $\mu$ to it. Here's the density of $n$:
$$\frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{n^2}{2 \sigma^2}}$$
Let $Y=n+\mu$. 
Now, how do we work out the density of the transformed (shifted) random variable? Again, it depends on what you're prepared to assume. It can be done from first principles ($F_Y(y)=P(Y\leq y)=P(n+\mu\leq y)=P(n\leq y-\mu)=F_n(y-\mu)$ and then differentiate $F_Y(y)$ to obtain the density $f_Y(y)$).
Or we could use the transformation result $f_Y(y) = f_X(g^{-1}(y))|\frac{d}{dy}g^{-1}(y)|$
So $n=Y-\mu$, hence $g=y-\mu$ and $dy = dn$
Hence $f_Y(y)=f_n(y-\mu).|1|$
$$=\frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(y-\mu)^2}{2 \sigma^2}}$$
which is a $N(\mu,\sigma^2)$
A: To complement what whuber said in  the comments. You can do some simple manipulations to derive the distribution of $y^{(i)}$. In the regression equation:
$$y^{(i)} = \theta^T x^{(i)} + \epsilon^{(i)}$$
We isolate $\epsilon^{(i)} $ and than replace it in the density formula:
$$p(\epsilon^{(i)}) = \frac{1}{\sqrt{2\pi\sigma}}exp{\left(-\frac{(\epsilon^{(i)})^2}{2\sigma^2} \right)} $$ 
$$p(y^{(i)} - \theta^T x^{(i)}) = \frac{1}{\sqrt{2\pi\sigma}}exp{\left(-\frac{(y^{(i)} - \theta^T x^{(i)})^2}{2\sigma^2} \right)} $$ 
Then we rewrite the probability term in a more convenient way, considering $\theta$ as a fixed parameter and $x^{(i)}$ as a given variable. So
$$p(y^{(i)}| x^{(i)}; \theta) = \frac{1}{\sqrt{2\pi\sigma}}exp{\left(-\frac{(y^{(i)} - \theta^T x^{(i)})^2}{2\sigma^2} \right)} $$ 
