Does $0$ correlation imply independence for marginally normal distributions? 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 ?
 A: what you said is true only when $(X,Y)$ is known to be jointly Gaussian. What you said in the question suggests that you only know $X$ and $Y$ are marginally Gaussian, for which correlation being $0$ is not sufficient!
Consider this construction:
take $X$ to be $N(0,1)$ distribution. Take $Y=X$ if $|X|\leq c$, $Y=-X$, if $|X|>c$.
where $c>0$ is chosen such that 
$E X^2 1_{\{X\leq c\}}=E X^2 1_{\{X> c\}}$
Convince yourself $Y$ is also distributed as $N(0,1)$!
Now note this:
$\rho = EXY -EXEY =EX^21_{\{X\leq c\}}- E X^2 1_{\{X> c\}} -EXEY =0$ but $X$ and $Y$ clearly are not independent
Moral of the story: you always need the joint distribution to be Gaussian! Knowing the marginals is not enough
A: Here is a standard very-well-known example.
Let $X \sim N(0,1)$, $Z$ a discrete random variable taking on values $+1$ and $-1$ with equal probability (a Rademacher distribution) $\frac 12$, and define $Y = XZ$.  Then,
\begin{align}
P\{Y \leq y\} &= P\{XZ \leq y\}\\
&= P\{X \leq y, Z = +1\} + P\{X \geq -y, Z = -1\}\\
&= P\{X \leq y\}P\{Z = +1\} + P\{X \geq -y\}P\{Z = -1\}, 
&\scriptstyle{\text{by independence of} ~ X ~\text{and}~ Z}\\
&= \Phi(y)\cdot \frac 12 + \Phi(y)\cdot \frac 12
&\scriptstyle{\text{sketch the CDF if this step is not obvious}}\\
&= \Phi(y)
\end{align}
showing that $Y \sim N(0,1)$ also. Note that $E[X]=E[Y]=0$. Also, $E[Z]=0$.
But, $E[XY] = E[X^2Z] = E[X^2]E[Z] = 1 \cdot 0 = 0$,
and so we get that $X$ and $Y$
are uncorrelated random variables. However, $X$ and $Y$ are
very much dependent random variables. Consider that conditioned
on $X = a$, $Y$ is a discrete random variable
that takes on values $+a$ and $-a$ with equal probability.
Had $X$ and $Y$ been independent (as you want them to be),
the conditional distribution
of $Y$ would have continued serenely to be $N(0,1)$, secure in the
knowledge that $Y$ is independent of $X$ and so knowledge that $X$ has
value $a$ cannot affect the conditional distribution of $Y$.
