The fourth edition of Zhejiang University probability proves that the uncorrelation of normal distribution random variables is equivalent to independence

Proof proposition ：
$\forall fullfootjuststatebranchclothOfalong\; withmachinechangeThe\; amountX,Y,AndItsUnitedclosebranchclothalsofullfootjuststatebranchcloth,YesX,YNophaseTurn\; off\iff X,Ysinglestate$

1. If $X,Y$ Independent , Then there will be $X,Y$ Unrelated
   because $X,Y$ Independent , So there are $E(XY)=E(X)E(Y)$, Then its covariance $Cov(X,Y)=E(XY)-E(X)E(Y)$ be equal to $0$, The correlation coefficient $\rho =\frac{Cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}$, So the correlation coefficient is $0$, So it's not relevant .

2. Next, we need to prove the normal distribution $X,Y$ Unrelated , Then it is independent
   From the joint distribution probability density function formula of multidimensional normal distribution random variables ：
   $$f\left(\overrightarrow{N}\right)=\frac{1}{{\left(2\pi \right)}^{\frac{n}{2}}{\left|\Sigma \right|}^{\frac{1}{2}}}{e}^{-\frac{1}{2}}{\left(\overrightarrow{N}-\mu \right)}^T{\Sigma }^{-1}\left(\overrightarrow{N}-\mu \right)$$
   among
   $$\begin{array}{rl}& \overrightarrow{N}yesmanyindividualalong\; withmachinechangeThe\; amountGroupbecomeOfalong\; withmachineArrowThe\; amount,nThat\text{'}s\; ok1Column\\ & \Sigma yesalong\; withmachineArrowThe\; amountOfAssociationFangBadMomentfront\\ & \mu yes\overrightarrow{N}OfperiodatArrowThe\; amount,nThat\text{'}s\; ok1Column\end{array}$$
   In two-dimensional random variables , Covariance matrix is a two-dimensional matrix , It is not difficult to calculate its inverse ,
   The formula of covariance matrix :
   $$\begin{gathered} \Sigma =\mathrm{E}\left[\left(\mathbf{N}-\mathrm{E}[\mathbf{N}]\right){\left(\mathbf{N}-\mathrm{E}[\mathbf{N}]\right)}^{\mathrm{T}}\right] \\ N=[X,Y{]}^T \end{gathered}$$
   You can follow the formula
   $A^{-1}=\frac{A^{\ast }}{|A|}$, among $A^{\ast }$ yes A Of Adjoint matrix , The adjoint matrix is Cofactor matrix The transpose matrix of , Each element of cofactor matrix is the original matrix $A$ Corresponding to the element Algebraic cofactor , The algebraic cofactor of an element is the determinant of the matrix obtained by removing the row and column of the element . This passage involves many concepts of linear algebra . You can look at the following formula to deduce step by step .
   $$\begin{array}{rl}A& =\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ A^{-1}& =\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\end{array}$$
   So there is , A two-dimensional A random variable ( That is to say 2 That's ok 1 Column Random vectors ) The probability density function of $f(x,y)$:
   $$\begin{array}{rl}f\left(x,y\right)& =\frac{1}{2\pi {\sigma }_1{\sigma }_2\sqrt{1-{\rho }^2}}{e}^q\\ Itsin:& \\ q& =\frac{-1}{2\left(1-{\rho }^2\right)}\left[\frac{{\left(x-{\mu }_1\right)}^2}{{\sigma }_1^2}-2\rho \frac{\left(x-{\mu }_1\right)\left(y-{\mu }_2\right)}{{\sigma }_1{\sigma }_2}+\frac{{\left(y-{\mu }_2\right)}^2}{{\sigma }_2^2}\right]\\ \rho & =\frac{Cov\left(X,Y\right)}{{\sigma }_1{\sigma }_2}\\ & {\sigma }_1,{\sigma }_{2}branchotheryesX,YOfmarkaccurateBad,{\mu }_1,{\mu }_{2}branchotheryesX,YOfperiodat\end{array}$$
   and :
   $$\begin{array}{rl}f\left(x\right)& ={\frac{1}{\sqrt{2\pi }{\sigma }_1}e}^{-\frac{{\left(x-{\mu }_1\right)}^2}{2{\sigma }_1^2}}\\ & \\ f\left(y\right)& ={\frac{1}{\sqrt{2\pi }{\sigma }_2}e}^{-\frac{{\left(y-{\mu }_2\right)}^2}{2{\sigma }_2^2}}\\ & \\ f(x)f(y)& ={\frac{1}{2\pi {\sigma }_1{\sigma }_2}e}^{\frac{{\left(x-{\mu }_1\right)}^2}{{\sigma }_1^2}+\frac{{\left(y-{\mu }_2\right)}^2}{{\sigma }_2^2}}\end{array}$$
   It can be seen that , Yes $\rho =0\Rightarrow f(x,y)=f(x)f(y)$, That is, irrelevant can deduce independent .
   Sum up 1,2, The left side of a proposition can be deduced from the right , The right side can also push out the left , So the proposition holds .