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 $X_i$.


1 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.

  • 1
    $\begingroup$ independence of $y_i$ and $y_j$ doesn't mean they're independent conditioned on some other variable, say $x_i,x_j$. $\endgroup$
    – gunes
    Apr 10, 2020 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.