# Independence between error and regressor

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.