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

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    $\begingroup$ 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. $\qquad$ $\endgroup$ Commented Jul 8, 2020 at 23:42
  • $\begingroup$ 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). $\endgroup$
    – User 2014
    Commented Jul 8, 2020 at 23:45
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    $\begingroup$ 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 $ $\endgroup$ Commented Jul 8, 2020 at 23:59
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    $\begingroup$ Is there any reference where I can check the proof of this asymptotic result? $\endgroup$
    – User 2014
    Commented Jul 9, 2020 at 0:04
  • $\begingroup$ 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. $\qquad$ $\endgroup$ Commented Jul 9, 2020 at 0:05

1 Answer 1

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$Cov(X,Y)$ 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.

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