Is the conditional likelihood of a sample of $(X,Y)$ a conditional joint distribution of all $Y$ on all $X$'s?

Question

Given random variables from a random sample $(X_1,Y_1),\dots,(X_n,Y_n)$, the conditional likelihood of observing $y_i |x_i$ (for all $i$) given parameters $\theta_1,\dots,\theta_n$ is usually written as something along the lines of $L(y|x;\theta_1,\dots,\theta_n)$. Is it correct to view the conditional likelihood of the entire sample as $f(y_1,\dots,y_n|x_1,\dots,x_n,\theta_1,\dots,\theta_n)$, where $f$ is a conditional probability density function? If $Y_i$ and $Y_j$ ($i \neq j$) is conditionally independent given $X_1,\dots,X_n$, then I assume the joint distribution can be broken down to equal $\Pi f(y_i|x_1,\dots,x_n,\theta_1,\dots,\theta_n)$? Most notations I see don't explicitly denote conditioning of $Y_i$ on all $X$'s, only on $X_i$.

Answer 1

As you point out, $f(y_1, ..., y_n | x_1, ..., x_n; \theta_1, ..., \theta_n)$ is valid for the most general case.

When $y_j$ are independent, $$ f(y_1, ..., y_n | x_1, ..., x_n; \theta_1, ..., \theta_n) = \Pi f(y_i| x_1, ..., x_n; \theta_1, ..., \theta_n). $$ You say you "assume it". You can easily prove it by iteratively applying the following: $$ f(y_1, ..., y_n | x_1, ..., x_n; \theta_1, ..., \theta_n) \\= f(y_1| x_1, ..., x_n; \theta_1, ..., \theta_n) f(y_2, ..., y_n | y_1, x_1, ..., x_n; \theta_1, ..., \theta_n) \\= f(y_1| x_1, ..., x_n; \theta_1, ..., \theta_n) f(y_2, ..., y_n | x_1, ..., x_n; \theta_1, ..., \theta_n). $$ The first step is the law of conditional probability. The last step follows from the independence of $y_j$ on $y_i$, $i \neq j$.

In many problems of the literature, I guess that $y_i$ only depends on $x_i$. Hence the notation that you mention.