Proof that there is no autocorrelation I was wondering whether my reasoning is correct and whether I've written it down correctly:
Given the following equation $bread_i=\beta_{0} + \beta_{1}wage + u_{t}$, where $u_{t}$ is normally distributed with a mean of zero and a variancce of $\sigma^{2}$ and is independent of $wage$.
Furthermore we know that there is no autocorrelation, $bread$ and $wage$ are not cointegrated, however there is a unit root present in $bread$ and $wage$.
Given a first difference model $\Delta bread_{t} = \beta_1 \Delta wage_{t}+\epsilon_{t}$, where $\epsilon = \Delta u_{t}$. Is the error term $\epsilon_{t}$ in this equation uncorrelated over time? Please motivate your answer with a formal derivation.
Well I thought the following:
Since $\epsilon = u_{t} - u_{t-1}$ and we know that there is no autocorrelation between them, after first differencing there will be no autocorrelation afterwards.
Question is of course how to show it? Using COV maybe? I am currently trying to derive it, but I'm a bit stuck.
Thanks in advance!
 A: I will just use $y$ for bread and $x$ for wage. You write that "a unit root is present in both $y$ and $x$." Although this is rather vague, let's say that we have
$$y_t=y_{t-1}+v_t,\Rightarrow \Delta y_t = v_t$$
$$ x_t=x_{t-1}+\omega_t \Rightarrow \Delta x_t = \omega_t$$
with $v_t$ and $\omega_t$ independent white noises.
Then in the first difference specification
$$\Delta y_t=\beta_{1}\Delta x_t + \Delta u_{t}$$ 
we substitute using the expressions for $\Delta y_t$ and $\Delta x_t$ to obtain
$$v_t=\beta_{1}\omega_t + \Delta u_{t} \Rightarrow v_t-\beta_1\omega_t = \Delta u_t$$
So $\Delta u_t$ is a linear combination of independent white noises, and therefore white noise itself.
ADDENDUM 
$$\operatorname {Cov}(\Delta u_t, \Delta u_{t-1}) = E\left( \Delta u_t\Delta u_{t-1}\right)  = E\left( ( v_t-\beta_1\omega_t)( v_{t-1}-\beta_1\omega_{t-1})\right)$$
$$=E(v_tv_{t-1})-\beta_1E(v_t\omega_{t-1})-\beta_1E(\omega_t v_{t-1})+ \beta_1^2E(\omega_t\omega_{t-1})$$
$$=0+0+0+0=0$$
since we have white noises, and independent between them.
