Let $X_1, \ldots, X_n$ be marginally Gaussian distributed random variables that are uncorrelated. Does it imply that they are independent?

  • 3
    $\begingroup$ There's an example of dependent uncorrelated random normals here. It's possible to use the same basic idea to generate others. SIlverfish's answer here gives another. $\endgroup$
    – Glen_b
    Sep 9, 2016 at 0:46

1 Answer 1


If $X = (X_1,\ldots, X_n)$ are jointly normal, too, then yes. Otherwise, no.

In this case $\Sigma = \text{diag}(\sigma_1^2,\ldots, \sigma_2^2)$ and $\mu = (\mu_1,\ldots,\mu_n)'$

\begin{align*}f_X(x) &= (2\pi)^{-\frac{n}{2}}|\Sigma|^{-\frac{1}{2}}\exp\left[-\frac{1}{2}(x-\mu)'\Sigma^{-1}(x-\mu) \right] \\ &= (2\pi)^{-\frac{n}{2}} (\sigma_1^2\cdots\sigma_2^2)^{-\frac{1}{2}}\exp\left[-\frac{1}{2}\sum_{i=1}^n\frac{(x_i-\mu_i^2)^2}{\sigma_i^2} \right] \\ &= \prod_{i=1}^n \left[\frac{1}{\sqrt{2\pi\sigma_i^2}} \exp\left(-\frac{(x_i-\mu_i)^2}{2\sigma_i^2} \right)\right] \\ &= \prod_{i=1}^n f_{X_i}(x_i). \end{align*}

For an example of two dependent $X_1$ and $X_2$ that are uncorrelated, but dependent, check out the example here. You can take $n=2$. Define $X_1 \sim \text{Normal}(0,1)$, $W$ is $1$ or $-1$ with probability $.5$ and independent from $X_1$. Then define $X_2 = WX_1$.

The $X$s are un-correlated because \begin{align*} \text{Cov}(X_1,X_2) &= \text{Cov}(X_1,W X_1)\\ &= E[X_1^2W] \\ &= E[X_1^2]E[W] \\ &=0 \end{align*}

But they are very dependent.

  • 1
    $\begingroup$ No problem, sir. $\endgroup$ May 6, 2020 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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