$\newcommand{\Cov}{\operatorname{Cov}}$Let $Y_i=a+bx_i+\varepsilon$ the simple regression model. The expression of the pearson coefficien is given by
$$\rho_{xY}=\frac{\Cov(x,Y)}{\sigma_x\sigma_Y}.$$
My question is about the interpretion of $\Cov(x,Y)$ and $\sigma_x$, since $x$ is not random.
I think that $$\Cov(x,Y)=\Cov(x,a+bx+\varepsilon)=\Cov(x,a)+\Cov(x,bx)+\Cov(x,\varepsilon)=b\Cov(x,x)$$ $$\Cov(x,Y)=b\cdot\operatorname{Var}(x).$$
Is this correct? What is the interpretation of this result?
$Cov(X,Y)$ is only defined for two random variables $X$ and $Y$. The notation $x$, as in $Cov(x,Y)$, implies that the random variable is degenerated, that is, $Prob(X=x)=1$. In such a case, $Cov(x,Y)=0$. On the other hand, it is not unheard of that people use (misuse) the notation of $Cov(x,y)$ to stand for a computational formula. I suspect that this may be the case here.