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The following convention is sometimes encountered in the text (here example from HMM statement, $\{X_n\}_{n \geq 1}$ is a Markov process):

$$X_n \mid (X_{n-1} = x_{n-1}) \sim f(x_n \mid x_{n-1})$$

It is clear, what it means for the "plain" case (Why are probability distributions denoted with a tilde?), but what about the meaning of full formal notation when we have the conditional part $\mid\ (X_{n-1} = x_{n-1})$?

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  • $\begingroup$ It's describing the conditional distribution (rather than the unconditional distribution). $\endgroup$
    – Matthew Gunn
    Commented Dec 17, 2015 at 9:05

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This might be translated to English as:

The distribution of $X_n$, conditional on $X_{n-1}$ taking the value $x_{n-1}$, is described by the probability density function $f(x_n \mid x_{n-1})$.

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  • $\begingroup$ Thanks, could you please also write it in another notation, as I believe the original is just an abbreviation, or is the $X|Y$ the most used one? (If my guess is right, it should be something with p()) $\endgroup$
    – Roman Susi
    Commented Dec 17, 2015 at 9:03
  • $\begingroup$ $X\mid Y$ is standard. For example, $P( X=x \mid Y =y)$, often written $P(X\mid Y)$ is the probability that $X$ takes the value $x$ given $Y$ takes the value $y$. $E[X \mid Y=y]$ is the expectation of $X$ conditional on $Y$ takes value $y$. $X \mid Y $ in general refers to $X$ conditional on $Y$. $\endgroup$
    – Matthew Gunn
    Commented Dec 17, 2015 at 9:07
  • $\begingroup$ Ok. Those others are of course very familiar, but the version without any "functors" around was confusing. $\endgroup$
    – Roman Susi
    Commented Dec 17, 2015 at 9:11
  • $\begingroup$ @RomanSusi No worries. There's a lot of notation around the use of probability. We've all had our confused "uhhh, so what is this?" moments! $\endgroup$
    – Matthew Gunn
    Commented Dec 17, 2015 at 9:14
  • $\begingroup$ Even better would be to translate it to English as: "The distribution of $X_n$, conditional on $X_{n-1}$ taking the value $x_{n-1}$, is described by the probability density function $f(\ \mid x_{n-1})$." $\endgroup$
    – Did
    Commented Dec 27, 2015 at 18:05

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