Let the following classical linear regression:
$$y_i = x_i \theta + u_i, \quad E(u_i|x_i) \sim N(0, \sigma^2)$$
Can I conclude that $x$ and $u$ are independent?
I would like this because I want to prove that: $y_i|x_i \sim N(\theta x , \sigma^2)$. And I need the independence between $x_i$ and $u_i$.to use the linearity of the variance:
$$V(y_i|x_i) = V(x_i \theta|x_i) + V(u_i|x_i) $$
Some idea?
You don't need independence for the variance equation to hold. Given $x_i$, $x_i$ becomes a constant and is independent from $u_i$, i.e. $x_i$ is independent from $u_i$ given $x_i$.
$$V(y_i|x_i)=V(\overbrace{x_i\theta}^{\text{constant}}+u_i|x_i)=V(u_i|x_i)=\sigma^2$$
For the expectation, it's rather easier, you don't need any assumptions: $$E[y_i|x_i]=E[x_i\theta+u_i|x_i]=x_i\theta+E[u_i|x_i]=x_i\theta+0=x_i\theta$$
You have the correct mean and the variance. The only thing remains is to assert that $y_i|x_i$ is distributed normally, which is also easier since it is $u_i+x_i\theta$, i.e. $u_i+\text{constant}$. Adding constants to normal RVs creates normal RVs.
