My question regards whether it's possible to know whether two Gaussian random variables are independent when we know only that their respective laws are governed by conditional distributions of jointly distributed Gaussians.

Specifically, consider the following situation. Suppose we know that $(Y_1, Y_2, Y_3)$ are jointly multivariate gaussian: i.e. $\left[\begin{array}{l} Y_1\\Y_2\\Y_3\end{array}\right]\sim N\left(\left[\begin{array}{l}\mu_1\\\mu_2\\\mu_3\end{array}\right],\left[\begin{array}{l}\Sigma_{11} &\Sigma_{12}&\Sigma_{13}\\\Sigma_{21}&\Sigma_{22}&\Sigma_{23}\\\Sigma_{31}&\Sigma_{32}&\Sigma_{33}\end{array}\right]\right)$

Now, suppose we condition $Y_2$ on $Y_1$ and $Y_3$ respectively so we have from standard Gaussian conditioning that

\begin{equation} p(y_2|y_1) = N\left(\mu_{2|1}, \Sigma_{2|1}\right) \end{equation} where \begin{equation} \mu_{2|1}=\mu_2+\Sigma_{21}\Sigma_{11}^{-1}(y_1-\mu_1) \end{equation} and \begin{equation} \Sigma_{2|1} = \Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}. \end{equation}

Likewise, \begin{equation} p(y_2|y_3) = N\left(\mu_{2|3}, \Sigma_{2|3}\right) \end{equation} where \begin{equation} \mu_{2|3}=\mu_2+\Sigma_{31}\Sigma_{33}^{-1}(y_3-\mu_3) \end{equation} and \begin{equation} \Sigma_{2|3} = \Sigma_{22}-\Sigma_{23}\Sigma_{33}^{-1}\Sigma_{32}. \end{equation}

Now suppose we know $Y_{*}$ and $Y_{**}$ are such that $Y_{*}\sim N\left(\mu_{2|1}, \Sigma_{2|1}\right)$ and $Y_{**}\sim N\left(\mu_{2|3}, \Sigma_{2|3}\right)$.

Can we say anything about the independence of $Y_{*}$ and $Y_{**}$ through this distributional knowledge alone?


No. Given two Gaussian random variables with specified means and variances, one cannot assume that they are jointly Gaussian, including as a special case, independent Gaussian random variables. In your example, $Y_{*}$ and $Y_{**}$ cannot be assumed to be independent, or jointly Gaussian.

| cite | improve this answer | |

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.