Can I use all estimates in a first differenced regression to apply to levels? I have a time series y where I took the first differences, y’, and an independent variable x where I also took the first differences to get x’.
When I run a regression between y’ and independent variable x’, I get:
y’ = Bx’ and Pearson r = r
I am aware that B can be thought of as the B when thinking about levels because y’ = Bx’ is easily solved for y = Bx (so B is estimated to be the same for levels) , but I cannot figure out if I can do the same for r.
In general, I’m hoping someone can clear up what isn’t explicitly said often - the idea that if I want to find a relationship between y and x, I can take the first differences of both variables and get regression results. I am essentially wondering if all results from that differenced regression can be interpreted as if we did the regression on the levels aka the interpretation remains the same. Thank you.
 A: Suppose you have two random walks
$$
x_t:=\sum_{\tau=-\infty}^t u_\tau
$$
and
$$
y_t:=\sum_{\tau=-\infty}^t v_\tau
$$
where $u_\tau$ and $v_\tau$ are zero-mean i.i.d. sequences with $\text{Corr}(u_\tau,v_\tau)=\rho$ where $|\rho|\neq 1$. You would like to analyze the relationship between $y_t$ and $x_t$. The two random walks are not cointegrated, so you cannot regress one on the other and expect a sensible result.
You take first differences $\Delta x_t=u_t$ and $\Delta y_t=v_t$ and analyze them instead. You run a no-intercept regression* of $v_t$ on $u_t$ that yields an estimate $\hat\beta$ of the actual nonzero slope coefficient $\beta$. Now, regressions are about conditional expectations; you can legitimately say that if $u_t=c$, then the estimated expected value of $v_t$ is $\hat\beta c$ (that is, $\hat{\mathbb{E}}(v_t|u_t=c)=\hat\beta c$), which is generally different from the unconditional expectation of zero.
However, you cannot say that if $x_t=c$, then the estimated expected value of $y_t$ is $\hat\beta c$ (that is, $\hat{\mathbb{E}}(y_t|x_t=c)=\hat\beta c$). Thus inference from the first-difference model of $v_t$ on $u_t$ does not just carry over to the levels model of $y_t$ on $x_t$. It is not as simple as that.
*There is nothing special about the lack of intercept here. We just use the information that the true means of both $u_t$ and $v_t$ happen to be zero.
