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

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 $$X_i$$.

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.
• independence of $y_i$ and $y_j$ doesn't mean they're independent conditioned on some other variable, say $x_i,x_j$. Apr 10 '20 at 7:44