I was wondering if it might be possible to generate 2 correlated $Beta$ random variables?

In other words, I want to generate two Beta random variables which can be said to have come from two Beta distributions whose correlation is known to be "$\rho =$ some number"?

In R, I tried a simple linear transformation approach to achieve this:

X1 = rbeta(1e4, 5, 1) ; X2 = rbeta(1e4, 6, 1) ;  X3 = rbeta(1e4, 7, 1) ; a = -.5

Y1 = X1 + (a*X2) ; Y2 = X2 + (a*X3) ## Y1 and Y2 are meant to be correlated

cor(Y1, Y2)

plot(Y1~Y2, col = densCols(Y1, Y2) ) ; abline(lm(Y1~Y2), lty = 3, lwd = 2)

But the linear transformation approach above is particularly unprincipled. If you change a to .5, then the meant-to-be $Beta$ random variables go beyond 1:

enter image description here

P.S. There is apparently a principled way to do this as shown HERE.

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    $\begingroup$ There is nothing preventing Y1 or Y2 from being outside $(0,1)$ so they cannot be beta distributed? $\endgroup$ – Christoph Hanck Jun 13 '17 at 15:17
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    $\begingroup$ Exactly how do you want the variables to be correlated? There are myriad ways to construct them, so we need further restrictions. A general method is to use a copula; another method would be to select a suitable Dirichlet distribution. $\endgroup$ – whuber Jun 13 '17 at 15:53
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    $\begingroup$ The interesting paper at arxiv.org/pdf/1406.5881.pdf develops a bivariate distribution with beta marginals, that admits of all correlations in $[-1,1]$. They do not give an simulation algorithm but that shouldn't be to difficult. $\endgroup$ – kjetil b halvorsen Jun 13 '17 at 16:08
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    $\begingroup$ Proportions don't necessarily have Beta distributions. In fact, for a fixed denominator that would be a mathematical impossibility: the proportion is discrete while a Beta variable is continuous. You could immediately adopt the solution I posted yesterday about generating correlated Binomial variates, because those provide nice models of the counts that you really need for the t-test: you can't (validly) do a t-test on proportions alone. $\endgroup$ – whuber Jun 13 '17 at 16:54
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    $\begingroup$ Indeed, the method I proposed for generating correlated binomials seems to fit your situation perfectly, because it constructs them as sums of correlated binary ("dichotomously scored") random variables. Thus, all you have to do is decide how much correlation you want among the pre- and post-responses, item by item, and you're on your way. $\endgroup$ – whuber Jun 13 '17 at 17:41

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