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 '16 at 0:46

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 '20 at 0:29

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