Assume that I have a posterior distribution $p(\theta_1, \theta_2|X)$ and I obtain an additional information in the form of a marginal density $q(\theta_1|Y)$ that is of the same type as $p(\theta_1|X)$. For instance, let $p$ be bivariate normal and $q$ univariate normal.
My question: is there any way how to incorporate the information carried by $q$ into $p(\theta_1, \theta_2|X)$? First, I though about merging the marginals $p(\theta_1|X)$ and $q(\theta_1|Y)$, but in my opinion this is not a correct way as it neglects the dependence between $\theta_1$ and $\theta_2$. So I'm lost and will be happy for any reasonable ideas.
I'd write:
$$\, p(\theta_1, \theta_2 \mid X,Y)=\frac{p(X,Y \mid \theta_1, \theta_2)\, p(\theta_1, \theta_2)}{p(X,Y)}= \tag{1}$$ $$=\frac{p(X \mid \theta_1, \theta_2) \, p(Y \mid \theta_1, \theta_2)\, p(\theta_1, \theta_2)}{p(X,Y)}= \tag{2}$$ $$=\frac{\, p(\theta_1, \theta_2 \mid X) \, p(Y \mid \theta_1, \theta_2) P(X)}{P(X,Y)}=\tag{3}$$ $$=\frac{\, p(\theta_1, \theta_2 \mid X)\, p(\theta_1, \theta_2 \mid Y) P(Y)P(X)}{P(X,Y)\, p(\theta_1, \theta_2)}=\tag{4}$$ $$=\frac{\, p(\theta_1, \theta_2 \mid X)\, p(\theta_1 \mid Y) \, p(\theta_2 \mid \theta_1)}{\, p(\theta_1, \theta_2)} \frac{ P(Y)P(X)}{P(X,Y)}=\tag{5}$$ $$=\frac{\, p(\theta_1, \theta_2 \mid X)\, p(\theta_1 \mid Y) }{\, p(\theta_1)} \tag{6}$$
This involves several assumptions or approximations:
For $(2)$ : $X,Y$ conditionally independent on the parameters (looks ok)
For $(5)$ : $p(\theta_2 \mid \theta_1, Y)=p(\theta_2 \mid \theta_1)$ (debatable, but it seems the only way to reflect the fact that $Y$ gives information only about $\theta_1$)
For $(6)$ : $X$ and $Y$ independent. This looks more questionable than $(2)$.