# Normal distribution $\rho_{X,Y} = 0 \rightarrow X \bot Y$

Assume $X \sim \mathcal N(\mu_1, \sigma_1^2)$ and $Y \sim \mathcal N(\mu_2, \sigma_2^2)$. If $\rho_{X,Y} = 0$ then $X \bot Y$.

Can someone give a hint why this is true ?

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Try using characteristic functions. –  Eckhard Mar 26 at 9:56
The joint distribution function of $X$ and $Y$ is
$$f_{X,Y}(x,y)=\frac 1 {2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}exp\left(-\frac{\begin{bmatrix}x-\mu_1 y-\mu_2\end{bmatrix}C^{-1}\begin{bmatrix}x-\mu_1 \\ y-\mu_2\end{bmatrix}}{2}\right)$$
where $C=\begin{bmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{bmatrix}$.
Put $\rho=0$ and $f_{X,Y}(x,y)$ factorises itself into the product of pdfs of two normal distributions, $\mathfrak N(\mu_1,\sigma_1^2)$ and $\mathfrak N(\mu_2,\sigma_2^2)$, which means $X$ and $Y$ are independent.