Expectation of a normal distrinution Given that, 
$y$ is generated according to $N(x^T_0w,\sigma^2)$ (N-Normal distribution)
Why the following true?
$$E[y_o^2]= \sigma^2 +(x^Tw)^2 $$

Starting from the definition of variance 
$$\sigma^2 = E[(x-E[x])^2]$$
we can write
$$\sigma^2  = E[x^2 - 2x \cdot E[x]-(E[x])^2] $$
and because of linearity of expectation
$$  \sigma^2= \\ E[x^2] - E[2x \cdot E[x]]+E[(E[x])^2] \\   = E[x^2] - 2(E[x])^2+(E[x])^2 \\  =  E[x^2] - (E[x])^2 $$
This can be written as 
$$E[x^2] = \sigma^2+(E [x])^2$$
