Covariance Does Not Imply Independence

Example

• With $X\sim N(0,1)$
• With $Y=X^2$
• Note that $X,Y$ are dependent
• But, $cov[X,Y]=E[XY]-E[X]E[Y]$
• $=E[X.X^2]-0.E[X^2]$
• $=E[X^3]$
• $=0$ For odd moments of any distribution symmetric around 0