# Conditional likelihood, conditional independence and joint independence

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:

$$L(D;\theta) = \prod_{i=1}^n p_i(x_i, y_i;\theta) \tag{1}$$

$$p_i$$ - 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:

$$L(D;\theta) = \prod_{i=1}^n p_i(y_i | x_i;\theta) p_i(x_i;\theta) \tag{2}$$

Conditional probability can always be formulated by definition 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 random vectors are independent, i.e. $$(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j), \forall j \neq i$$

then below conditional random variables are also independent? $$(Y_i | X_i) \mathrel{\unicode{x2AEB}} (Y_j | X_j), \forall j \neq i$$

NB I only make assumption of independence, the sequence of random vectors could have different distributions.

• Your notation and setting obscure the basic simplicity of the answer: any function of $(x_i,y_i)$ is independent of any function of $(x_j,y_j)$ for $i\ne j.$
– whuber
Commented Apr 13 at 20:43
• @whuber thanks a lot, that provided me a lot of clarity! Commented Apr 14 at 20:55
• You say "Given the data [are] generated independently..." which is not true, but also is not a necessary condition for expressing the 3rd display. This is rather the definition of conditional probability. Commented Jul 30 at 20:21

After submitting to correct some of your very confused notations especially your incorrect concept of independence of probabilities about observed different data samples instead of the correct concept of independence of different random vectors (also see @whuber's question commented under your own latest confused answer), the linked post in the comment hasn't completely clarified this question for you.

Please bear in mind that here $$(X_i,Y_i)$$ is a single (atomic) random vector while your concerned question is not about the usual conditional independence of random vectors which should be expressed as $$(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j) | Z, \forall i \neq j$$ following your notation. Yet your real question is about independence of the split conditional random variables $$Y_i|X_i \mathrel{\unicode{x2AEB}} Y_j|X_j, \forall i \neq j$$ which is not directly provable by plugging to any of the referenced theorem which basically says if $$X \mathrel{\unicode{x2AEB}} Y$$ then $$g(X) \mathrel{\unicode{x2AEB}} h(Y), \forall g, h$$. Before trying that out indirectly we can first use a simple example to feel this subtle kind of 'split-conditional' pairwise independence.

Suppose your DGP follows an additional simple deterministic rule $$Y_i=X_i, \forall i$$, while all the random vectors are still pairwise independent $$(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j), \forall i \neq j$$. In this case apparently the conditional random variable $$Y_i|X_i=X_i$$ trivially, and note the $$X$$-component projection of each vector $$Z_i=(X_i,Y_i)$$ is a function of random vectors. Then we can apply above referenced theorem to rigorously prove $$X_i \mathrel{\unicode{x2AEB}} |X_j, \forall i \neq j$$, so we have $$Y_i|X_i \mathrel{\unicode{x2AEB}} Y_j|X_j, \forall i \neq j$$.

Now to be able to rigorously prove the general case, let $$l_i$$ be above projection function onto the $$X$$ component of each random vector $$Z_i=(X_i,Y_i)$$. Then a generic conditional random variable $$Y_i|X_i \sim p_i(X_i)=p_i(l_i(Z_i))$$ which is still a (composite) function of the random vector $$Z_i$$ (here $$p_i$$ is a general distribution function assuming each conditional random variable $$Y_i|X_i$$ is well-defined), therefore we now can prove the same independence conclusion in the most general case only assuming independence of the sequence of random vectors.

Finally back to the common regression cases as you've already understood a fixed identical conditional distribution $$p_{Y|X}$$ is assumed implicitly for each conditional rv $$Y_i|X_i$$, we can still assert they're independent even if their conditional distribution functions are the same as an extreme edge case because any value $$y_i$$ of a conditional rv cannot give any clue about another conditional rv's value $$y_j$$ assuming IID data for regression.

• keeping notations aside, as @whuber pointed above, functions of independent random vectors are also independent. The joint distribution or "splitted" conditional distribution is a function of the said random vectors. Hence they ought to be independent as well. Commented Jul 28 at 17:42