I see this kind of notation often
$$ p_{\theta} (x|z, y) = f(x; z, y, \theta) $$
I understand the conditional prob noation on the left. What is the significance of the ; on the joint prob on the right?
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8$\begingroup$ My preference is to write $X|Y, Z; \theta, \phi$, to show conditional distributions, conditional on other random variables, and fixed, unknown parameters. $\endgroup$– tchakravartyCommented Sep 4, 2017 at 19:54
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$\begingroup$ You've got different guesses. Perhaps you could give one or two examples from published work and more context. $\endgroup$– Nick CoxCommented Sep 5, 2017 at 6:36
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1 Answer
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What this notation says is that
$$ f(x; z, y, \theta) $$
is a function of $x$ with "parameters" $z, y, \theta$. It is just a way to visually show that they are of different kind (e.g. data vs parameters, random vs fixed quantities). While the conditional notation has precise meaning, this notation is used informally, to improve readability of the formulas rather than to convey any specific meaning.
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$\begingroup$ $z, y$ would be unusual notation for parameters. With no other information and if obliged to guess I would go for @tchakrcarty's interpretation. $\endgroup$– Nick CoxCommented Sep 5, 2017 at 6:35
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$\begingroup$ Sorry; that should be @tchakravarty. $\endgroup$– Nick CoxCommented Sep 5, 2017 at 7:37
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$\begingroup$ @NickCox I have seen the notation $p(b_i \mid T_i, \delta_i, y_t; \theta)$ used recently and had similar questions to what the notation meant. This quantity is the posterior distribution for $b_i$, and $T_i, \delta_i$ and $y_i$ are data, and $\theta$ is a parameter. For this reason, I agree with the answer that the ";" is used (or can be used) to separate data from parameters. $\endgroup$– JLeeCommented Nov 20, 2019 at 15:23
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1$\begingroup$ @JLee I think what you say is fully consistent with my comment. Ending symbolism with $|$ data; parameters) is what tchakravarty was surmising and I was supporting. $\endgroup$– Nick CoxCommented Nov 20, 2019 at 16:03
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$\begingroup$ @NickCox Sorry, yes. I have just reread. I agree with tchakravarty's interpretation. $\endgroup$– JLeeCommented Nov 20, 2019 at 20:16