$Y$ is outcome and $A = (A_1, A_2, \dots, A_p)$ are absolute abundances. Let $S, V \subset {1, 2, \dots, p}, S \cap V = \emptyset$. Let $A_S$ be the joint of all $A_j$, $j \in S$ and $A_V$ the joint of all $A_{j^\prime}$, $j^\prime \in V$. $||\cdot||_1$ is the $L_1$ norm, i.e. total sum of abundances. Consider three null hypotheses:
$Y \perp (A_V, ||A||_1) | \frac{A_S}{||A_S||_1}$
$Y \perp A_V | \frac{A_S}{||A_S||_1}$
$Y \perp \frac{A_V}{||A||_1} | \frac{A_S}{||A_S||_1}$
Comments:
$1 \Rightarrow 2$, $1 \Rightarrow 3$. $2 \not\Rightarrow 3$, $3 \not\Rightarrow 2$.
It is not difficult to imagine distributions where 1 holds, and consequently, both 2 and 3. For this, simply imagine the generative model where $Y$ is sampled purely based on $\frac{A_S}{||A_S||_1}$ (i.e., your example).
Only 3 is testable with observed data. Based on 3, implied hypotheses can also be raised. For example, $Y \perp \frac{A_V}{||A_V||_1} | \frac{A_S}{||A_S||_1}$, or $Y \perp \frac{A_j}{||A||_1} | \frac{A_S}{||A_S||_1}$ for a specific $j$.
Our proposal is the special case where $V = {j}$ and $S = {-j}$.
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.