If $(X,Y)$ is Gaussian and $\mathbb E[XY]=\mathbb E[X]\mathbb E[Y]$, then $X$ and $Y$ are independent Let $(X,Y)$ a Gaussian vector. I know that if $X$ and $Y$ are independent, then $\mathbb E[XY]=\mathbb E[X]\mathbb E[Y]$. But why is the converse true ? i.e. why $\mathbb E[XY]=\mathbb E[X]\mathbb E[Y]$ implies that $X$ and $Y$ are independent ?
 A: The joint characteristic function takes the form of  a product of the characteristic functions of the marginals. The is because the variance-covariance matrix is diagonal.  
You can also write down the joint density function explicitly and see that it is product of the marginals. Ref: Vol. II of Feller's book. 
A: Because a Gaussian joint distribution is determined by its mean vector and covariance matrix.
A: Here's a subtle point: It says $(X,Y)$ is Gaussian.
It could have said $X,Y$ are ("are", not "is", and the pair is not enclosed in parentheses) jointly Gaussian. Then it's not as subtle.
To call a pair of random variables "Gaussian" is to say that the two are jointly Gaussian.
It can happen that $X,Y$ are Gaussian and uncorrelated but not independent, but not if they're jointly Gaussian. An example is this: Let $X=\pm Y$ where $Y$ has a Gaussian distribution with expectation $0$ and variance $1,$ and the plus or minus sign is chosen indepenently of the value of $Z$ with each sign having probability $1/2.$ Then $X$ and $Y$ are uncorrelated but obviously not independent. They're not jointly Gaussian since $\Pr(X+Y=0) = 1/2,$ so $X+Y$ is not Gaussian. Being jointly Gaussian means being so distributed that every non-constant linear combination of them is Gaussian (in the singular case, this would include the possibility of some linear combinations having a degenerate Gaussian distribution, i.e. variance $0$).


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*Show that $(X,Y)$ cannot be Gaussian, i.e. $X,Y$ (without parentheses) cannot be jointly Gaussian, unless the joint density is of the form $$ f(x,y) \propto \exp\left( -\frac 1 2 ((x,y)-(\mu,\nu))' V^{-1} \left(\begin{array}{c} x-\mu \\ y-\nu \end{array}\right) \right) $$ for some $2\times2$ non-negative-definite symmetric matrix $V.$

*Show that $\operatorname{cov}(X,Y)$ is the off-diagonal entry of $V.$

*Show that if the off-diagonal entry of $V$ is $0,$ then the density factors as a function of $x$ times a function of $y.$
Steps 2 and 3 are easy; I am not sure at this moment how much work is involved in step 1.
