0

What means : covariance are measure of linear dependance ?

  • For example, if $Correlation(X,Y)=1$, do we have that $X=\alpha Y+c$ ?

  • Now, if $Correlation(X,Y)=\frac{1}{2}$, what that will say ? That $X=\frac{1}{2}Y+???$

  • For example, the skewness of a random variable (suppose the mean is null and the variance is $1$) is $\mathbb E[X^3]=\mathbb E[X^2X]$. In somehow, we could see $X^2$ as the quadratic distance from the mean and $X$ as the algebraic distance from the mean. So, if $X$ is symmetric, then $\mathbb E[X^3]=0$. Now, if the tail at right is bigger, then $\mathbb E[X^3]>0$. In somewho, I would interpret it as the quadratic distance $X^2$ and the distance $X$ from the mean "moves" in the same direction, so in somehow, $X$ will be at right from the mean with higher probability than at left. Same if $\mathbb E[X^3]$, then $X^2$ and $X$ moves in opposite direction, so since $X^2\geq 0$, in at the end, $X$ will be at left from the mean with higher probability, and thus the tail at leaf will be bigger.

  • Does these interpretation works or not really ? Also, at the end, I don't really see what mean $\mathbb E[XY]$ measure linear dependance...

John
  • 858

1 Answers1

1

Let me answer the first two questions. Suppose that $Y=aX+b$ for some constants $a\ne 0$ and $b$. It's clear that $\rho_{X,Y}:=\operatorname{Corr}(X,Y)=\operatorname{sgn}(a)$ (when $a=0$ the correlation coefficient is undefined). Now assume that $|\rho_{X,Y}|=1$. Define \begin{align} \varphi(x):&\!=\mathsf{E}[(X-\mathsf{E}X)x+(Y-\mathsf{E}Y)]^2 \\ &\!=x^2\sigma_X^2+2x\operatorname{Cov}(X,Y)+\sigma_Y^2. \end{align} The determinant of the quadratic equation in the second line is $(2\operatorname{Cov}(X,Y))^2-4\sigma_X^2\sigma_Y^2\le 0$, which implies that $$ |\operatorname{Cov}(X,Y)|\le \sigma_X\sigma_Y \quad\Leftrightarrow\quad |\rho_{X,Y}|\le 1. $$ Also by the properties of expectations $\varphi(x)=0$ iff $(X-\mathsf{E}X)x+(Y-\mathsf{E}Y)=0$ a.s., i.e. $$ Y=aX+b, $$ where $a=\sigma_Y/\sigma_X$ and $b=\mathsf{E}Y-a\mathsf{E}X$.

When $|\rho_{X,Y}|<1$, there can be any relation between $X$ and $Y$, e.g. $Y=X^3$. In the last case suppose that $X\sim N(0,1)$. Then $$ \rho_{X,Y}=\frac{\mathsf{E}XY}{\sigma_X\sigma_Y}=\frac{3}{\sqrt{15}}. $$

  • Thank you. So, at the end if $|\sigma _X\sigma _Y|<1$ what can we say about $X$ and $Y$ ? It's not really clear in your answer... – John Sep 24 '19 at 20:02
  • It's not clear because this information doesn't say anything about the dependence between $X$ and $Y$. –  Sep 24 '19 at 21:40