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Suppose $S$ is a sample space (the set of all outcomes $\omega_i$) for an experiment. A random variable $X$ is defined as a real-valued function which maps elements from the sample space to real numbers, i.e. $X:S\to \mathbb R$.

Discrete Random variable:

The definition of the conditional probability mass function of $X$ given $Y=y$ is $$\mathbb P(X=x|Y=y)=\frac{\mathbb P(X=x, Y=y)}{\mathbb{P}(Y=y)} .$$

Question: In lecture slides I have seen the notation, for example, that $X|(Y=y) \sim \text{Bin}(m, \lambda).$ What is the definition of $X|(Y=y)$? Is it a random variable itself with a restricted sample space? Maybe $X|(Y=y): \{\omega\in S: Y(\omega)=y \} \to \mathbb R$?

What would be the definition of $X|(Y=y)$ for $X$ and $Y$ being continuous random variables?

(Note: If it isn't a random variable, then how can we talk about it's distribution and expected value?)

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    $\begingroup$ It denotes the conditional distribution of $X$ given $Y$. However, $X\mid Y=y\sim \text{Bin}(m,\lambda)$ implies that $X$ is independent of $Y$ unless $m$ and/or $\lambda$ depend on $y$. $\endgroup$
    – user140541
    Mar 19, 2020 at 1:04
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    $\begingroup$ A (probability) distribution is just a function having certain properties. $\endgroup$
    – user140541
    Mar 19, 2020 at 1:12
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    $\begingroup$ No, $X\mid Y=y$ is not a random variable. I've seen similar questions here. Check, for example, this post. $\endgroup$
    – user140541
    Mar 19, 2020 at 1:21
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    $\begingroup$ @user523384 Technically speaking, no new object denoted by $X|Y=y$ is ever defined in the theory of conditional probability. The new object introduced is the conditional distribution: conditional on the event $\{Y=y\}$ one can define a (new and valid) e.g. mass function $P(X=x|Y=y)$ that obeys all the usual axioms of probability, for each fixed $y$. There simply are no “conditional random variables”, only random variables that when we have some extra knowledge, may turn out to have a more useful conditional distribution, conditional on some event, than their unconditional distribution. $\endgroup$ Mar 19, 2020 at 1:26
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    $\begingroup$ Technically, for a given $y$ one may define a probability space and a random variable $X_y$ living on that space s.t. the distribution of $X_y$ is the distribution implied by your notation. $\endgroup$
    – user140541
    Mar 19, 2020 at 1:30

1 Answer 1

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Summarising the very helpful comments from @Nap D. Lover and @d.k.o. - In the original theory of conditional probability, there is no such definition of a "conditional random variable."

Before addressing the notation, a thought about the "requirement" of a conditional random variable

  • The purpose of a conditional distribution, $\mathbb P(X=x|Y=y)$, is a way to "recalibrate" the probability assignment/distribution for $X$, given we received information about $Y$. (Which intuitively, could be the probability distribution of the temperature $X$ as $\mathbb P(X=x)$ vs. the probability distribution of the temperature $X$, given the humidity $Y$ was $y$, being $\mathbb P(X=x|Y=y)$). It is still a probability distribution designed for the random variable $X$, just "recalibrated" to better model the "true" probabilities for the given situation.

  • So I guess, in a way, a new random variable for a "conditional random variable" is not really necessary. While it is possible to define a random variable $X_y$ living on a new restricted sample space, maybe it moves away from the idea of this distribution being "rediagnosis" of what the probability distribution of $X$ should be, given the new "symptoms" ($Y=y$).

  • Hence it makes sense to only need Conditional distributions and Conditional expectation (The expected value of $X$, but weighted in a different way to account for the new information) etc, and not a new random variable itself.

The notation: So the interpretation of the notation can be left as what @d.k.o. said in the very first comment, $X|(Y=y) \sim \text{Bin}(m, \lambda)$ is just shorthand notation for saying "The distribution of $X$, conditioned on $Y=y$, is (from the definition in the question) $\text{Bin}(m, \lambda)$.

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    $\begingroup$ $X|(Y=y) \sim \text{Bin}(m, \lambda)$ is an abuse of notation, and you should avoid using it. $\endgroup$
    – kludg
    Mar 19, 2020 at 4:34
  • $\begingroup$ Hi @user523384: I wrote the "other" question on this topic, and I have thought about it some since then. I agree with this summary. But I think that defining $X_y$ (in your notation) tends to actually emphasize the conditional nature of its value -- the condition is right there in its name! Another thought: without getting to far into measure theory, there are definitely technical issues with defining this idea in general (i.e., there isn't going to be a single expansive definition that covers all the cases). But the "intuition" is sound. Conditioning really is a kind of composition $\endgroup$
    – nomen
    Sep 7, 2021 at 23:51
  • $\begingroup$ which is why our intuition was so strongly drawn to the idea of "picking a distribution at random" or "picking a density function at random" in this way. The best analogy I can come up with is as if we're doing naive set theory: It is correct on simple domains but incoherent on complicated enough domains (which is why we need measure theory in the first place). $\endgroup$
    – nomen
    Sep 7, 2021 at 23:53
  • $\begingroup$ Also, I found some resources relevant to actually doing this project in its potential generality on the other math se: mathoverflow.net/questions/20740/…. There is also a pretty famous paper that discusses the nature of the conditioning operator in terms of "disintegrations" (stat.yale.edu/~jtc5/papers/ConditioningAsDisintegration.pdf) $\endgroup$
    – nomen
    Sep 8, 2021 at 16:41

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