Let $X,A,Y_1,\dots,Y_n$ be random variables defined on some probability space $(\Omega,\mathcal{F},P)$. Suppose that $Y_1,\dots,Y_n$ are conditionally independent and identically distributed given the values of $A$ and $X$, i.e., $$ f_{Y_1,\dots,Y_n\mid A,X}(y_1,\dots,y_n\mid a,x) = \prod_{i=1}^n f_{Y_i\mid A,X}(y_i\mid a,x) $$ where $f_{Y_i\mid A,X}$ is the conditional density function of $Y_i$ given $A$ and $X$.
Suppose we also know the joint density function $f_{A,X}(a,x)$. Then our overall joint density function is fully specified as $$ f_{Y_1,\dots,Y_n,A,X}(y_1,\dots,y_n,a,x)=f_{Y_1,\dots,Y_n\mid A,X}(y_1,\dots,y_n\mid a,x) f_{A,X}(a,x). $$
We can calculate \begin{align*} f_{X\mid Y_1,\dots,Y_n}(x\mid y_1,\dots, y_n) &= \frac{ f_X(x) f_{Y_1,\dots,Y_n \mid X}(y_1,\dots,y_n\mid x) }{ \int f_X(x) f_{Y_1,\dots,Y_n \mid X}(y_1,\dots,y_n\mid x) dx } \\ &= \frac{ f_X(x) \int f_{Y_1,\dots,Y_n,A \mid X}(y_1,\dots,y_n,a\mid x) da }{ \int f_X(x) f_{Y_1,\dots,Y_n \mid X}(y_1,\dots,y_n\mid x) dx } \\ &= \frac{ f_X(x) \int f_{Y_1,\dots,Y_n \mid A, X}(y_1,\dots,y_n\mid a, x) f_{A\mid X}(a\mid x) da }{ \int f_X(x) f_{Y_1,\dots,Y_n \mid X}(y_1,\dots,y_n\mid x) dx }. \end{align*}
Now fix $y_1,\dots,y_n$. The denominator above is a constant depending on $y_1,\dots,y_n$. So $$ f_{X\mid Y_1,\dots,Y_n}(x\mid y_1,\dots,y_n) \propto f_X(x) \int f_{Y_1,\dots,Y_n \mid A, X}(y_1,\dots,y_n\mid a, x) f_{A\mid X}(a\mid x) da, $$ i.e., we are taking a "weighted average" with weights $f_{A\mid X}(a\mid x)$.
Now my question concerns how "uncertainty" in the density $f_{A\mid X}$ gets transferred to the density $f_{X\mid Y_1,\dots,Y_n}$. Specifically, if $\text{Var}[A \mid x]$ is increased for each $x$, does this imply that $\text{Var}[X\mid y_1,\dots,y_n]$ increases?
If we look at a special case where $Y_i\mid a,x$ is normal with mean $a$ and variance $x$, the answer is in the affirmative (see here). Does the result hold in general?