Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
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

2 Answers 2

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

share|improve this answer

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 $Y$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.