# $\operatorname{Cov}(x,Y)$ in linear regression model

$$\newcommand{\Cov}{\operatorname{Cov}}$$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?

• If you're wondering about the interpretation of the expression $\rho_{xY} = \dfrac{\operatorname{Cov}(x,Y)}{\sigma_x\sigma_Y}, \vphantom{\frac.{\displaystyle\sum}}$ that seems to suggest that someone other than yourself used that expression. In what context did it come up? Sometimes $X$ and $Y$ are both random but $X$ is treated as non-random because what is of interest is only the conditional distribution of $Y$ given $X.$ In other contexts, an experimenter can judiciously choose the $x$ values and then nature provides the $Y$ values. $\qquad$ Commented Jul 8, 2020 at 23:42
• The context is that $Y|x\sim N(a+bx,\sigma^2)$ and I'm trying to prove that $\rho_{xY}^2=R^2$ (coefficient of determination). Commented Jul 8, 2020 at 23:45
• If $X$ is random and $\operatorname E(Y\mid X) = a+bX$ and $\operatorname{var}(Y\mid X) = \sigma^2,$ then the value of $R^2$ arising from an i.i.d. sample $(X_i,Y_i),\,\,i=1,\ldots,n$ is in general not equal to the square of the correlation between $X$ and $Y,$ although it approaches that as $n\to\infty. \qquad$ Commented Jul 8, 2020 at 23:59
• Is there any reference where I can check the proof of this asymptotic result? Commented Jul 9, 2020 at 0:04
• On the other hand, if you have fixed, not random, values of $x_i,$ then you can examine the distribution of $R^2,$ (which is a random variable if $Y_1,\ldots,Y_n$ are random), but I don't know what $\text{“}\rho^2\text{''}$ would mean in that case. $\qquad$ Commented Jul 9, 2020 at 0:05

$$Cov(X,Y)$$ 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.