Let me first introduce the definition of Gaussian I came across
A one dimensional random variable $\Gamma$ is Gaussian iff it has the characteristic function
$$\mathbb{E}[e^{i \xi \Gamma}]=e^{im\xi-\frac{1}{2}\sigma^2 \xi^2}$$
for some real numbers $m \in \mathbb{R}$ and $\sigma \geq 0$.
Now the question I have is, if $X,Y$ are two real-valued random variables which are Gaussian, and $Cov(X,Y)=0$, i.e. they are uncorrelated, how do I use the above definition to see that they are independent?
Thanks a lot in advance!
Best Answer
Note that, we require the two random variables to be jointly Gaussian for the result to hold (it is possible that two random variables are Gaussian but not jointly Gaussian). Also, I shall work with density functions instead of characteristic function for simplicity (you can deduce the density from characteristic function, and vice versa). For notation purpose, I denote the random vector by $\left(X_1,X_2\right)$.
The joint probability density function of a bivariate normal distribution is given by $$f_{X_1,X_2}\left(x_1,x_2\right)=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}} \times \exp{\left[-\frac{\left\{\left(\frac{x-\mu_1}{\sigma_1}\right)^2+\left(\frac{y-\mu_2}{\sigma_2}\right)^2-2\rho\left(\frac{x-\mu_1}{\sigma_1}\right)\left(\frac{y-\mu_2}{\sigma_2}\right)\right\}}{2\left(1-\rho^2\right)}\right]}$$ Now, $Cov(X_1,X_2)=0 \Leftrightarrow \rho=0$. Putting $\rho=0$ the joint density function reduces to \begin{align*} f_{X_1,X_2}\left(x_1,x_2\right)&=\frac{1}{2\pi\sigma_1\sigma_2} \times \exp{\left[-\frac{\left\{\left(\frac{x-\mu_1}{\sigma_1}\right)^2+\left(\frac{y-\mu_2}{\sigma_2}\right)^2\right\}}{2\left(1-\rho^2\right)}\right]}\\ &=\frac{1}{\sigma_1\sqrt{2\pi}}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu_1}{\sigma_1}\right)^2\right\} \times \frac{1}{\sigma_2\sqrt{2\pi}}\exp\left\{-\frac{1}{2}\left(\frac{y-\mu_2}{\sigma_2}\right)^2\right\} \end{align*}
Since, we can factor out the joint density function in the form $g(x) \times h(y)$ (which automatically are the respective marginal densities), we conclude that $X_1$ and $X_2$ are independent.