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.

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.
