I read some other topics on this but the concept still isn't clear to me. Given random variables $X$ and $Y$, I believe the correlation coefficient $\rho=\frac{Cov(X,Y)}{\sigma_X\sigma_Y}$ has an absolute value of 1 if and only if $Y$ is a linear function of $X$. I'm trying to get an intuitive understanding on why $\rho$ measures the strength of a specifically linear relationship and not something else? Is there a way to see this without involving parameters of least squares linear regression?


1 Answer 1


First You can see it by Using the Cauchy-Schwarz inequality. But I use another method.


$$h(t)=E\bigg((X-\mu_x)t+(Y-\mu_y)\bigg)^2\geq 0$$


$$h(t)=t^2 Var(X)+2t cov(X,Y)+Var(Y)=at^2+bt+c$$

since $h(t)\geq 0 $ so $\Delta\leq 0$ in hence

$$\big(2 cov(X,Y)\big)^2-4Var(X) Var(Y)\leq 0$$ so $$cov^2(X,Y)\leq \sigma^2_x \sigma^2_y$$



$$\Leftrightarrow$$ $$cov^2(X,Y)= \sigma^2_x \sigma^2_y$$ $$\Leftrightarrow$$ $$\Delta=0$$

$$\Leftrightarrow$$ $$h(t_1)=E\bigg((X-\mu_x)t_1+(Y-\mu_y)\bigg)^2=0$$ where $t_1=\frac{-b}{2a}$

$$\Leftrightarrow$$ $$P((X-\mu_x)t_1+(Y-\mu_y)=0)=1$$

$$\Leftrightarrow$$ almost surely $$Y=-t_1X+(\mu_y+t_1\mu_x)$$

  • 1
    $\begingroup$ I've taken cou to be a typo for cov. Expressions such as cov and Var are more readable if roman, not italic, in my view. $\endgroup$
    – Nick Cox
    Apr 4, 2020 at 13:02
  • $\begingroup$ @Nick Cox. You right. Thank you! $\endgroup$
    – Masoud
    Apr 4, 2020 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.