Hierarchical models and conditional independence

Question

Suppose that we have a hierarchical model given by (this is Example 4.4.5 of Berger and Casella(2002)) \begin{align*} X\mid Y&\sim\text{binomial}(Y,p),\\ Y\mid\Lambda&\sim\text{Poisson}(\Lambda),\\ \Lambda&\sim\text{exponential}(\beta). \end{align*} Is it possible to show that $X$ is conditionally independent of $\Lambda$ given $Y$? Or is this an implicit assumption that is made in this hierarchical model?

Intuitively, $\Lambda$ affects $X$ only through $Y$ and if $Y$ is given, the value of $\Lambda$ should have no effect on the value of $X$. However, I am trying to come up with a rigorous argument. We would have to show that $$ f(x,\lambda\mid y) =f(x\mid y)f(\lambda\mid y) $$ or, equivalently, $$ f(x\mid\lambda,y) =f(x\mid y). $$

We know how $f(x\mid y)$ looks like and we can use Bayes' theorem to calculate $f(\lambda\mid y)=f(y\mid\lambda)f(\lambda)/f(y)$. We have that $$ f(x,\lambda\mid y) =\frac{f(x,\lambda, y)}{f(y)} =\frac{f(x\mid\lambda, y)f(y\mid\lambda)f(\lambda)}{f(y)} $$ but I am not sure how to proceed with $f(x\mid\lambda,y)$. How can we obtain an expression of $f(x\mid\lambda,y)$?

Answer 1

Given that $X|Y \sim Binomial(Y,p)$, the distribution of $X$ depends only on $Y$ and $p$. The conditional posterior distribution of $X$ given $Y$ is proportional to the binomial likelihood times the prior for $p$

The expression for $f(X|y,\lambda)$ is $$f(X|y,\lambda)=f(X|y)=\binom{y}{x}p^x(1-p)^{y-x}.$$