Uncorrelated vs independent
Uncorrelation means there is no linear dependence between the two random variables, while independence means no types of dependence exist between the two random variables.
Uncorrelated random variables may not be independent, but independent random variables must be uncorrelated.
For example, \(Z\sim N(0,1),X=Z,Y=Z^2\) \[ \begin{gather*} Cov(X,Y)=E[XY]-E[X]E[Y]\\ =E[Z^3]-E[Z]E[Z^2]\\ =E[Z^3]-0\\ =E[Z^3] \end{gather*} \] The moment generating function of distribution of \(Z\) is \[ \begin{gather*} M_Z(t)=E[e^{tz}]\\ =\int e^{tx}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz\\ =\int \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(z^2-2tz)}dz\\ =\int \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(z^2-2tz+t^2)+\frac{1}{2}t^2}dz\\ =\int \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(z-t)^2+\frac{1}{2}t^2}dz\\ =e^{\frac{1}{2}t^2}\int\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(z-t)^2}dz\\ =e^{\frac{1}{2}t^2} \end{gather*} \] Using the mgf we obtain the \(E[Z^3]\) as \[ \begin{gather*} E[Z^3]=M_Z'''(t=0)\\ =(0)e^{\frac{1}{2}(0)^2}+2(0)e^{\frac{1}{2}(0)^2}+(0)^3e^{\frac{1}{2}(0)^2}=0 \end{gather*} \] Therefore \[ Cov(X,Y)=E[Z^3]=0 \] This implies \(X,Y\) are uncorrelated, but they are dependent