Relationship between two Normal Random Variables with covariance matrices that are inverse of each other Let $$x_1 \sim N(0, \Sigma_1)$$ $$x_2 \sim N(0, \Sigma_2)$$ $$\Sigma_1 = \Sigma_2^{-1}$$ 
That is each covariance matrix can be seen as the precision matrix (or inverse) of the other. 
Simple question: What is the relationship between $x_1$ and $x_2$?
Edit: I am looking for relationships that suggest a solution to the following question – If I can sample $x_1$ can I use this to efficiently sample $x_2$? I am happy if it ends up being an approximation to samples of $x_2$. (By efficiently I mean something faster than using Cholesky factors). 
 A: For simplicity of notation, consider 2-vectors of random variables and also that that the random variables comprising $\mathbf X_1$ are standard normal random variables.  Then, the assumption is that $\mathbf X_1$ and $\mathbf X_2$ are pairs of jointly normal random variables with covariance matrices
$$\Sigma_1 = \left[\begin{matrix}1 & \rho\\\rho & 1\end{matrix}\right]\quad \text{and}\quad\Sigma_2 = \Sigma_1^{-1} = \left[\begin{matrix}\frac{1}{1-\rho^2} & \frac{-\rho}{1-\rho^2}\\\frac{-\rho}{1-\rho^2} & \frac{1}{1-\rho^2}\end{matrix}\right] = \left[\begin{matrix}\sigma^2 & -\rho\sigma^2\\-\rho\sigma^2 & \sigma^2\end{matrix}\right]$$
respectively where $\sigma^2 = \frac{1}{1-\rho^2} \geq 1$.  So we see immediately that the random variables in $\mathbf X_2$ have larger variances than those in $\mathbf X_1$ and also the correlation coefficient between them has the same magnitude as the correlation coefficient between the random variables in $\mathbf X_1$ but the opposite sign!
If the random variables have different means and/or variances, the results are messier, and of course for vector random variables of larger dimensions, even simple observations such as those made above are more difficult.

Returning to the case of 2-vectors, my interpretation of what the OP calls "the key question" is that of constructing samples of $\mathbf X_2$ given that we have available to us some samples of $\mathbf X_1$.  That is, we are given $n$ pairs of real numbers (the $n$ known samples of $\mathbf X_1$) and asked what samples of $\mathbf X_2$ might look like. This is easy: from the results $\operatorname{cov}(X,-Y) = -\operatorname{cov}(X,Y),  \operatorname{var}(aX)=a^2\operatorname{var}(X)$, we have that if $\mathbf X_1 = (X_1,Y_1)$ is a pair of standard (jointly) normal random variables with covariance (and correlation coefficient) $\rho$), that is,  $\mathbf X_1 \sim N(0,\Sigma_1)$, then $(\sigma X_1, -\sigma Y_1)$ is a pair of zero-mean jointly normal random variables with variance $\sigma^2$ and covariance $-\rho\sigma^2$, that is $\mathbf X_1\sim N(0,\Sigma_2)$. Thus, if we have $n$ samples $(x_i,y_i), 1 \leq i \leq n,$ of $\mathbf X_1$, then we can take
$(\sigma x_i, -\sigma y_i), 1 \leq i \leq n$ to be $n$ samples of $\mathbf X_2 \sim N(0, \Sigma_2)$.
Hey, it is not possible to make a silk purse out of a sow's ear. Nobody said that samples of $\mathbf X_2$ have to be independent of the samples of $\mathbf X_1$, and if we are constructing samples of $\mathbf X_2$ from samples of $\mathbf X_1$ without any random or pseudorandom deux ex machina, then there has to be some connection between the samples, right? And if we are bringing in randomness, then just sample directly from $\mathbf X_2$ and be done with it. Why bring in samples of $\mathbf X_1$ into the picture at all?
