# 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 ?

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Try using characteristic functions. –  Eckhard Mar 26 '13 at 9:56
this is actually false... You need $(X,Y)$ to be jointly normal. you only stated marginal distribution –  Lost1 Feb 1 '14 at 2:04

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

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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 $\frac 12$, 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, $E[XY] = E[X^2Z] = E[X^2]E[X] = 1 \cdot 0 = 0$, 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 $Y$.

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