$f(\{Y \vert X = x\})$ vs $f(Y) \vert X = x$

Question

This is probably a question with a simple answer, but the reason is likely to be more technical.

Suppose $Y$ and $X$ are random variables and $f$ is a function of single variable (it is not to mean probability density, neither marginal nor conditional) without any assumptions regarding the type of the function or dependence of the random variables, is there a difference between $f(Y) \vert X = x$ and $f(\{Y \vert X = x\})$?

Basically does it matter whether the conditional information is supplied before or after applying the function?

EDIT:

Due to notational confusion I am adding @Xi'an's better formulation here. Why is the probability distribution of the random variable $f(Y)$ conditional on the event $X=x$ is formally identical to the probability distribution of the random variable $f(Z)$ when $Z$ is distributed from the conditional distribution $p(y|X=x)$?

Answer 1

The conditional probability distribution of the random variable $W=f(Y)$ conditional on the event $X=x$ is identical to the conditional probability distribution of the random variable $f(Z)$ when $Z$ is distributed from the conditional distribution $p(y|X=x)$.

To wit, by the (annoyingly named) "law of the unconscious statistician" \begin{align} \mathbb P^Y(f(Y)\in A)=\mathbb P^W(W\in A) &=\mathbb E^X[\mathbb E^{W|X}\{\mathbb I_A(W)|X \}] \end{align} and \begin{align} \mathbb P^Y(f(Y)\in A) &= \mathbb E^X[\mathbb P(f(Y)\in A|X)]\\ &=\mathbb E^X[\mathbb E^{Y|X}\{\mathbb I_A(f(Y))|X \}]\\ &=\mathbb E^X[\mathbb E^{Z|X}\{\mathbb I_A(f(Z))|X \}] \end{align} where the last step is just a change of notation (see note below).

As an illustration, take $X$ to be a Poisson $\mathcal P(\lambda)$ random variable and $Y$ conditionally on $X$ to be a Binomial $\mathcal B(X,p)$ random variable (making $Y$ marginally a Poisson $\mathcal P(p\lambda)$ random variable). And take $f(y)=\mathbb I_0(Y)$. Then

• conditional on $X=x$, $f(Y)$ is a Bernoulli random variable with probability $(1-p)^x$ as the transform by $f$ of a Binomial $\mathcal B(x,p)$ random variable
• $W=f(Y)$ is a Bernoulli random variable with conditional distribution a Bernoulli distribution with probability$$\mathbb P(W=1|X=x)=\mathbb P(Y=0|X=x)=(1-p)^x$$

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Note that the issue with the notation in the question is that $Y|X$ is not a standard notation, if one means "the random variable with distribution equal to the conditional distribution of $Y$ given $X$" because $Y|X$ does not exist as such: $Y$ has both a marginal distribution and a conditional distribution, conditional on another random variable such as $X$. In other words, generating $Y$ from its marginal or (i) generating $X$ from its marginal then (ii) $Y$ from its conditional given the realisation of $X$ produces the same random variable. (This is why I conditioned upon $X=x$ in this answer.)

Answer 2

Strictly speaking, without any context about what $f(\cdot)$ denotes, I would say there is a difference from a notation standpoint.

If $Y$ is a random variable, then $f(Y)$ is also a random variable, and so the notation $f(Y)|X$ would read to me as:

"The random variable $f(Y)$ given/conditioned on the random variable $X$".

On the other hand, many could easily interpet $f(Y|X)$ as meaning:

"The density function of $Y$ conditioned on the random variable $X$"

So if $f(\cdot)$ is just a generic function, and not the density function of $Y$, you should avoid putting the condition within brackets to make this clear.