Is $g(X|Y=y)$ equivalent to $g(X)|Y=y$?

Question

Let $X$ and $Y$ be random variables with marginal pdfs $f_X(x)$ and $f_Y(y)$, respectively, and joint pdf $f_{X,Y}(x,y)$. Then for all $y$ such that $f_Y(y) \neq 0$ define the function $$f_{X|Y=y}(x) = \frac{f_{X,Y}(x,y)}{f_Y(y)}. \hspace{1cm} (*)$$

So for each fixed value of $y$, $f_{X|Y=y}(x)$ is a function of $x$. Now let $X|Y=y$ denote the random variable with the above pdf and let $g$ denote some arbitrary function. My question (which may seem rather trivial) concerns notation:

$$ \text{Are the random variables }\; g(X|Y=y) \;\text{ and }\; g(X)|Y=y \; \text{equivalent?} $$

(Here, $g(X)|Y=y$ is defined in an analogous way to $(*)$: Let $Z = g(X)$. Then $(g(X)|Y=y) = (Z|Y=y)$ is the random variable with pdf $\frac{f_{Z,Y}(z,y)}{f_Y(y)},$ where $f_{Z,Y}(z,y)$ denotes the joint distribution of $Z$ and $Y$.) Assuming the random variables are equivalent, is it because the two notations are equivalent? Or can we prove that these two random variables are equivalent? (Perhaps I am overthinking this...)

Answer 1

Okay, you have vector $(X,Y)$ with joint pdf (so that marginals has pdf too, due to Fubini). Then you can find conditional distribution of $X$ given $Y$, that is $\mathbb P(X \in \cdot | Y)$. Let $\mu(\cdot | Y)$ denote the conditional distribution $X|Y$ (that is $\mu(A|Y) = \mathbb P(X \in A | Y)$, one can actually show that it is absolutelly continuous so your density $f_{X|Y}$ exists). Now, take any $g$ which which is an diffeomorphism (so that for absolutelly continuous random variable $Z$, $g \circ Z$ is still absolutelly continuous). Denote $Z_y$ the random variable with $\mu(\cdot | y)$ distribution. You ask, whether $g \circ Z_y$ has the distribution $\nu(\cdot | y)$ where $\nu(A|Y) = \mathbb P(g(X) \in A | Y)$.

We have $\mathbb P( g(Z_y) \in A) = \mathbb P(Z_y \in g^{-1}(A)) = \mu(g^{-1}(A) | y)$.

Moreover $\nu(A|Y) = \mathbb P(g(X) \in A|Y) = \mathbb P(X \in g^{-1}(A) | Y) = \mu(g^{-1}(A)|Y)$, so $\nu(A|y) = \mu(g^{-1}(A)|y)$

That is, $g(Z_y)$ has $\nu(\cdot | y)$ distribution, which by definition is the distrubution of $g(X)$ given $Y=y$

Answer 2

There is no such thing as "conditional random variable" in probability theory. You can say "conditional PDF of $g(X)$ given $Y=y$", but not "conditional random variable $g(X|Y=y)$". Both $g(X|Y=y)$ and $g(X)|Y=y$ are not random variables, and if written as is, without context, are an abuse of notation.