If $X\sim\mathbb{P}_{X}=g_{\ast}(\mathbb{P}_{Y})$ can we find $Y$ such that $X=g(Y)$ at least almost surely?

Question

Good morning,

It is the first time I write here on StackExchange, therefore any suggestion to improve my question is really appreciated.

It is known that the distribution $\mathbb{P}_{X}$ of the univariate random variable $$X=g(Y)$$ on the probability space $(\Omega,\mathcal{F},\mathbb{P})$, where $g: \mathbb{R}\rightarrow \mathbb{R}$ is a measurable function (with respect to Borel $\sigma$-algebras) is the push-forward of the distribution $\mathbb{P}_{Y}$ of the univariate random variable $Y$ through $g$, or (following Wikipedia notation) $$\mathbb{P}_{X}=g_{\ast}(\mathbb{P}_{Y}).$$ The equality $X=g(Y)$ holds "omega by omega".

1. Suppose that $g$ is not invertible (for instance, an increasing cadlag function). Is the opposite true? More precisely, if we know that $X$ is an univariate random variable with distribution $\mathbb{P}_{X}=g_{\ast}(\mathbb{P}_{Y})$, can we always construct a random variable $Y$ on the same probability space with distribution $\mathbb{P}_{Y}$ such that $X=g(Y)$ holds at least almost surely?

2. If the answer to the previous point is affirmative, can you suggest a textbook from which I can quote the result?

Many thanks in advance for your help.

Answer 1

If we cannot extend the space, then "no":

Let $(\Omega, \mathcal{F}, P)$ be a probability space that has only one outcome. Specfically, assume $\Omega = \{blue\}$. Define $X:\Omega\rightarrow\mathbb{R}$ by $X(blue)=0$.

Define the measurable function $g:\mathbb{R}\rightarrow\mathbb{R}$ by $g(x)=0$ for all $x \in \mathbb{R}$.

Let $Y$ be Bernoulli-1/2 on some other probability space. Then $X$ has the same distribution as $g(Y)$, but there is no Bernoulli 1/2 variable $Y$ on $(\Omega, \mathcal{F}, P)$ since that space has only one outcome $blue$.

If we can extend the space, then "yes":

Let $(\Omega, \mathcal{F}, P)$ and $(\Omega', \mathcal{F}', P')$ be two probability spaces.

-Let $X:\Omega\rightarrow\mathbb{R}$ be a random variable on the first space.

-Let $Y':\Omega'\rightarrow\mathbb{R}$ be a random variable on the second space.

-Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a Borel measurable function.

Define $X':\Omega'\rightarrow\mathbb{R}$ by $X'=g(Y')$. Suppose $X$ and $X'$ have the same distribution.

Suppose that, on the first probability space, there exists a uniformly distributed random variable $U:\Omega\rightarrow [0,1]$ that is independent of $X$. If the original space $(\Omega, \mathcal{F}, P)$ does not have such an independent random variable $U$, we can always extend the space using standard product space concepts.

Theorem 5.10 in Kallenberg's Foundations of Modern Probabilty ensures that since $(X', Y')$ lie on the second space, and since $X$ has the same distribution as $X'$, there is a random variable $Y$ on the first space such that $(X,Y)$ has the same joint distribution as $(X',Y')$. (Kallenberg constructs $Y$ as a Borel measurable function of $X$ and $U$).

Define the Borel measurable function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ by $$ f(u,v) = g(v)-u \quad \forall (u,v) \in \mathbb{R}^2$$ Define the random variables $D:\Omega\rightarrow\mathbb{R}$ and $D':\Omega'\rightarrow\mathbb{R}$ by \begin{align} D &= f(X, Y)\\ D'&= f(X',Y') \end{align} Since $(X,Y)$ and $(X',Y')$ have the same distribution, the random variables $D$ and $D'$ have the same distribution. However, $D'=0$ surely. It follows that $P[D=0]=1$. That is, $P[g(Y)=X]=1$. So $g(Y)=X$ almost surely.