ke a joint normal distribution with a correlation of 0.8. But it is not. The above are <em>not</em> interpreted as conditional entities; rather, they are factors in the joint density. Multiplying them gives, <span class="math inline">\(\textsf{normal}(x \mid 0.8y,\sigma)
\times \textsf{normal}(y \mid 0.8x,\sigma)\)</span>, which is what it is (you can work out the algebra) but it is not the joint distribution where the conditionals have regressions with slope 0.8.</p>
<p>With censoring and truncation, Stan 