I would like your help to understand which assumptions are sufficient to get a desired conclusion about independence between random variables.
General setting:
Let $\mathcal{T}\equiv \{1,…,T\}$. Consider the random variables $A_t, B_t, Y_t$ $\forall t\in \mathcal{T}$.
Assume that $\{A_t\}_t$ are mutually independent across $t$. Moreover, $Y_t\equiv 3*A_t+B_t$ $\forall t \in \mathcal{T}$.
Assumptions I have doubts about:
A1: $\{B_t\}$ are mutually independent across t. Moreover, $\forall t \in \mathcal{T}$, $B_t$ is independent of $(A_t \text{ }\forall t\in \mathcal{T})$.
A2: $\{B_t\}$ are mutually independent across t. Moreover, $\forall t \in \mathcal{T}$, $B_t$ is independent of $A_t$.
Claim I am interested in:
Under A1 $$F_{Y_t}(\cdot | A_t \text{ }\forall t \in \mathcal{T})=F_{Y_t}(\cdot | A_t) \hspace{1cm} \forall t \in \mathcal{T}$$
where $F_{Y_t}(\cdot | A_t \text{ }\forall t \in \mathcal{T})$ is the probability distribution of $Y_t$ conditional on $A_t \text{ }\forall t \in \mathcal{T}$ and $F_{Y_t}(\cdot | A_t)$ is the probability distribution of $Y_t$ conditional on $A_t$.
Question: Is A2 sufficient for the claim above?
Best Answer
The answer to your question is "no". Under A2 you will have $\ \mathbb{E}\left(Y_t\left\vert A_s\, \forall s\in \mathcal T\right.\right)= 3A_t +\mathbb{E}\left(B_t\left\vert A_s\, \forall s\in \mathcal T\right.\right)\ $, but $\ \mathbb{E}\left(Y_t\left\vert A_t\right.\right)= 3A_t +\mathbb{E}\left(B_t\right)\ $, and these would have to be equal for your claim to be true. For a case where they're different, take $\ A_t\ $ to be identically distributed with positive variance $\ \sigma^2\ $, and $\ B_t=A_{t+1}\ $. Then A2 is satisfied, but the variance of $\ 3A_t +\mathbb{E}\left(B_t\left\vert A_s\, \forall s\in \mathcal T\right.\right)=3A_t + A_{t+1}\ $ is $\ 10\sigma^2\ $, whereas that of $\ 3A_t +\mathbb{E}\left(B_t\right)=3A_t+\mathbb{E}\left(A_{t+1}\right)\ $ is only $\ 9\sigma^2\ $.