Consider a sequence of data samples generated from $n$ independent random vectors $(X_1, Y_1), (X_2,Y_2), (X_3,Y_3) ...$
$$D = (x_1,y_1), (x_2,y_2), (x_3,y_3) ...$$
Where $(X_i, Y_i)$ - is a random vector and $X_i$, $Y_i$ are either scaler or vector valued random vectors and $(x_i,y_i)$ are the data samples from these.
The likelihood function can be defined as the following:
$$ L(D) = \prod_{i=1}^n p_i(x_i, y_i) \tag{1} $$
$p_i$ - the probability distribution function of the $i$'th random vector.
We can take a product of the probabilities since these random vectors are independent.
$(1)$ can also be formulated as:
$$ L(D) = \prod_{i=1}^n p_i(y_i | x_i) * p_i(x_i) \tag{2} $$
Given the data is generated independently, conditional distribution can also be formulated as:
$$ P(y_1, y_2, ... y_n | x_1, x_2, ... x_n) = \frac{P(x_1, x_2 ... x_n, y_1, y_2 ... y_n)}{P(x_1, x_2 ... x_n)} \tag{3} $$
$$ P(y_1, y_2, ... y_n | x_1, x_2, ... x_n) = \frac{\prod_{i=1}^n p_i(x_i, y_i)}{\prod_{i=1}^n p_i(x_i)} \tag{4} $$
$$ P(y_1, y_2, ... y_n | x_1, x_2, ... x_n) = \prod_{i=1}^n p_i(y_i | x_i) \tag{5} $$
Equation $(3)$ to $(4)$ due to the independence of the data generating process.
Based on the above:
Can one say that if the data samples is generated independently i.e. $$p_i(x_i,y_i) \mathrel{\unicode{x2AEB}} p_j(x_j,y_j) \forall j \neq i $$
then the conditional distribution of data sample $$p_i(y_i | x_i) \mathrel{\unicode{x2AEB}} p_j(y_j | x_j) \forall j \neq i $$ ?
NB I only make assumption of independence, the sequence of random vectors could have different distributions.
