If $X(\omega)$ is a Skorokhod representation, then is $X(g(\omega))$ of the same?

Question

So, I recently (re-)discovered that random variables learned in elementary probability such as the exponentially distributed random variable $X$ with cdf $F_X(x) = 1-e^{- \lambda x}$ can be explicitly represented as

$$X(\omega) := \frac{1}{\lambda} \ln(\frac{1}{1-\omega})$$

where the probability space is $((0,1), \mathscr B(0,1), \mu)$ where $\mu$ is Lebesgue measure.

This can be derived with the formula

$$X(\omega) = \sup\{x \in \mathbb{R}: F_X(x) < \omega\}$$

This is apparently called Skorokhod representation (so-called in David Williams' Probability with Martingales).

Now, apparently, $X(1-\omega)$ is not only also exponential but also does it have the same parameter $\lambda$ as $X(\omega)$, i.e. they have the same distribution.

Similarly, for $X \sim Be(p)$, I have found the following to be $\sim Be(p)$:

1. $1_{(0,1-p)}(\omega)$

2. $1_{(0,1-p)}(1-\omega) = 1_{(p,1)}(\omega)$

3. $1_{(0,1-p)}(1-(\omega+\frac{p}{2})) = 1_{(p,1)}(\omega+\frac{p}{2}) = 1_{(\frac{p}{2},1-\frac{p}{2})}(\omega)$

4. $1_{(0,1-p)}(1-(\omega+\varepsilon)) = 1_{(p,1)}(\omega+\varepsilon) = 1_{(\varepsilon,1-p+\varepsilon)}(\omega)$ for $0< \varepsilon < p < 1$

So, for what conditions on $g(\omega)$ are sufficient or necessary s.t. $X(g(\omega))$ has the same distribution as $X(\omega)$? I guess we need $g(\omega) : (0,1) \to (0,1)$.

Also does $g(\omega) = 1-\omega$ work for any X? It seems so:

$$X(1 - \omega) := \sup\{x \in \mathbb R | F_X(x) < 1 - \omega\}$$

$$ = \sup\{x \in \mathbb R | \omega \le 1 - F_X(x)\}$$

$$ = \sup\{x \in \mathbb R | \omega < 1 - F_X(x)\}$$

$$ = \sup\{x \in \mathbb R | \omega \ge F_X(x)\}$$

$$ = \sup\{x \in \mathbb R | \omega > F_X(x)\} := X(\omega)$$

QED?

Basically $\omega$ has the same probability of being in $[F_X(x),1)$ as $(0,F_X(x)]$ because both intervals have the same Lebesgue measure.

Answer 1

$X(g(\omega))$ has the same distribution as $x$ if $g:(0,1)\to (0,1)$ is what is called a measure preserving function. This means $m(g^{-1}A))=m(A)$ for all Borel sets $A$ (where $m$ is the Lebesgue measure on $(0,1)$.

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BCLC edit:

Proposition: If $g:(0,1)\to (0,1)$ is a measure preserving function, i.e. $Leb(g \in B) = Leb(B) \forall B \in \mathscr B(\mathbb R)$ then $F_{X(g)} = F_X$ or equivalently $\mathcal L_X(B)=\mathcal L_{X(g)}(B)$.

Pf:

$$P(X(g(\omega)) \in B)$$

$$ = P(g(\omega) \in X^{-1}(B))$$

$$ = P(\omega \in g^{-1}(X^{-1}(B)))$$

$$ = P(\omega \in X^{-1}(B)) \tag{*}$$

$$ = P(X(\omega) \in B)$$

QED

$(*)$ Here we make use that $P$ is Leb on $((0,1), \mathscr B(0,1))$.