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In Bayesian statistics, parameters are said to be random variables while data are said to be nonrandom. Yet if we look at the Bayesian updating formula $$ p(\theta|y)=\frac{p(\theta)p(y|\theta)}{p(y)}, $$ we find probability (density or mass) conditioned on the data as well as the conditional and unconditional probability (density or mass) of the data itself.

How does it make sense to consider probability (density or mass) conditioned on a constant or probability (density or mass) of a constant?

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    $\begingroup$ No no no, data is random, otherwise it would not be associated with a probability distribution. $\endgroup$
    – Xi'an
    Commented Jun 9, 2021 at 7:31
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    $\begingroup$ The Bayesian approach conditions [in a probabilistic sense] upon the data, which is completely different from assuming the data is non-random. The posterior distribution on the parameter is a conditional distribution, given the data, which only makes sense if the data itself is endowed with a probability distribution, ie, is the realisation of a random variable $\endgroup$
    – Xi'an
    Commented Jun 9, 2021 at 7:39
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    $\begingroup$ In probability theory, the conditional distribution $p(\theta|y)$ treats $y$ as given or fixed and $\theta$ as random or varying. This does not mean that $y$ is not random or more accurately that $y$ is not the realisation of a random variable. $\endgroup$
    – Xi'an
    Commented Jun 9, 2021 at 7:56
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    $\begingroup$ There is no difference between Bayesian and frequentist take on randomness of the data. The cherry on pie from the Bayesian approach is to turn the parameter random as well, or more accurately the realisation of a random variable. I never encountered the statement that "data is not random". The difference is rather that frequentist procedures are evaluated based on their frequency properties, ie by averaging over all possible realisations, instead of conditional on the actual realisation, as the Bayesian approach does. $\endgroup$
    – Xi'an
    Commented Jun 9, 2021 at 8:06
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    $\begingroup$ As part of formal undergrad economics training, I encountered something akin to this pedagogical simplification in 2nd year econometrics, using baby Wooldridge. I recall raising my hand during the lecture to ask why in a linear regression setting he was writing $p(y | x , \beta)$, and why the fixed parameter $\beta$ was appearing on the right hand side of a conditioning statement along with other conditioning variables $x$. The instructor corrected this to $p(y | x ; \beta)$. $\endgroup$
    – microhaus
    Commented Jun 9, 2021 at 14:25

4 Answers 4

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The Bayesian approach to (parametric) statistical inference starts from a statistical model, ie a family of parametrised distributions, $$X\sim F_\theta,\qquad\theta\in\Theta$$ and it introduces a supplementary probability distribution on the parameter $$\theta\sim\pi(\theta)$$ The posterior distribution on $\theta$ is thus defined as the conditional distribution of $\theta$ conditional on $X=x$, the observed data. This construction clearly relies on the assumption that the data is a realisation of a random variable with a well-defined distribution. It would otherwise be impossible to define a conditional distribution like the posterior, since there would be no random variable to condition upon.

The possible confusion may stem from the fact that a difference between Bayesian and frequentist approaches is that frequentist procedures are evaluated and compared based on their frequency properties, ie by averaging over all possible realisations, instead of conditional on the actual realisation, as the Bayesian approach does. For instance, the frequentist risk of a procedure $\delta$ for a loss function $L(\theta,d)$ is $$R(\theta,\delta) = \mathbb E_\theta[L(\theta,\delta(X))]$$ while the Bayesian posterior loss of a procedure $\delta$ for the prior $\pi$ is $$\rho(\delta(x),\pi) = \mathbb E^\pi[L(\theta,\delta(x))|X=x]$$

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  • $\begingroup$ Could you re-phrase that, for clarity? "… data… can… be conditioned on…" seems too strange. $\endgroup$ Commented Jun 12, 2021 at 22:42
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Maybe the confusion comes from the short hand $p(\theta|y)$ which actually means $p(\theta|Y=y)$, the random variable $Y$ interpreted as generating the data takes the fixed value $y$, fixed after actually having observed the data? So the data are random in the sense of having a distribution as long as they're uncertain, i.e., not fully observed, and then they become fixed by observation. (Nothing particularly Bayesian about this, though.)

Reading a comment on the original question, "To a subjective Bayesian, nothing is random" - nothing is really/objectively random (to a subjective Bayesian at least), however it can be random in the sense of being modelled by a random variable. So another source of confusion may be mixing up the use of the term "random" in a "philosophical" manner (referring to something that is "truly random", in the sense of having randomness as intrinsic property), and in a mathematical/technical manner, referring to something that appears as random variable in a probability model.

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    $\begingroup$ This is helpful. For the second paragraph, aleatory (regarding intrinsic properties) and epistemic (regarding our knowledge) are two relevant terms. $\endgroup$ Commented Jun 10, 2021 at 11:02
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Be very careful with the statement you choose. "nonrandom" is very different from "observed".

In Bayesian statistics everything is a random variable, the only difference between these random variables is some are observed and some are hidden.

For example in your case $y$ is an observed random variable and $\theta$ is a hidden random variable, your goal is to estimate the posterior distribution of $\theta$ conditioned on the observed $y$.

That says in Bayesian mindset we shouldn't reat $y$ like a constant as in the traditional sense, instead, it's an instance, or reliazation, of an random variable. (The observed values of the variables are also called "evidence" in most of the Bayesian statistics literatures.)

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  • $\begingroup$ Thanks. By this time I have learned it is simply a misleading language used incautiously by some. $\endgroup$ Commented Jun 23, 2021 at 19:06
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To be concrete, consider the simple case of throwing a dice. Every face has a probability to be thrown. The outcome of all throws is non-random (it is a fixed pattern determined by throwing the dice a lot of times).
On this pattern, you can apply a new chance of appearing. If you throw with two dices, a new pattern will emerge. This is because different dices will produce different outcomes (only in the case of perfect dices, the chance distribution of each one is the same as for the others). The chance that the dices used in separate throws is the same is very small. But there is a chance. And this chance is measured by applying the chance to the non-random chance distributions of a dice (for every dice this is a different distribution, though they are all quite alike).

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    $\begingroup$ Thank you. The outcome of all throws is non-random. Do you mean the probability distribution (chance distribution for Bayesians) of the random variable arising from the experiment is fixed? On this pattern, you can apply a new chance of appearing. What do you mean here? I also find it hard to follow the rest of the answer. You are probably using nontechnical language to make the content easier to grasp, but for me this introduces a threshold as nontechnical language is hard to parse exactly. Would it be possible to formulate this in the established technical language? $\endgroup$ Commented Jun 10, 2021 at 10:01
  • $\begingroup$ @RichardHardy Hi there. Sorry for reacting a bit late. I only saw it now... What I mean by the pattern is (in the case of the dices) is the pattern each dice will produce (after throwing the dice many times). This pattern will be different for real dices (not for idealized ones though). If you want to know the chances of different patterns to occur (of dices with small variations) you have to apply your chances to the different patterns. Then to every (non-random) pattern of chances you attribute a chance. In technical language? Let me pause for reflection for a second (or a bit more...). $\endgroup$ Commented Jun 10, 2021 at 11:16

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