For random variable X on a probability space there is another probability space with random variable Y that has the same probability as its negative

Question

$\newcommand{\P}{{\mathbb{P}}} \newcommand{\A}{{\mathcal{A}}} \newcommand{\O}{{\Omega}}$ $\newcommand{\Pp}{{\mathbb{P}^\prime}} \newcommand{\Aa}{{\mathcal{A}^\prime}} \newcommand{\Oo}{{\Omega^\prime}}$ $\newcommand{\phi}{{\varphi}} \newcommand{\E}{{\mathbb{E}}}$ Let $X$ be a random variable on the probability space $ (\Omega,\A,\P)$. Show there is a probability space $ (\Oo,\Aa,\Pp)$ with a random variable $Y$ on it, such that $$\Pp_Y = \Pp_{-Y}$$ and for all $t\in\mathbb{R}$ $$\phi_Y(t) = \phi_X(t)\phi_X(-t).$$

Do the following:

1. Let $ (\Oo,\Aa,\Pp)$ be a probability space with independent random variables $X_1,X_2$ such that $\Pp_{X_1}=\Pp_{X_2}=\P_{X}$.
2. Show that $Y:=X_1-X_2$ proofs the statement above.

Proof

I solved this with the answer of @Did in a another question.

1 )

Using the product space $ (\Oo,\Aa,\Pp)$ given by $$\begin{aligned} \Oo &= \O_1\times\O_2 \\ \Aa &= \A\otimes\A \\ \Pp &= \P\otimes\P , \end{aligned}$$

let $X_i(t)=X(\omega_i)$ with $\omega=(\omega_1,\omega_2)\in\Oo$ and $t\in\mathbb{R}$, than $X_1,X_2$ are independent. With $\{X_i = t\}_n=\{\omega_n \mid (\omega_1,\omega_2)\in\{X_i=t\}\}$ we observe $$\begin{aligned} \Pp_{X_i}(t) &= \Pp(X_i=t) \\ &= \P(\{X_i=t\}_1)\P(\{X_i=t\}_2) \\ &= \P(X=t)\P(\O) \\ &= \P(X=t)\cdot 1 \\ &= \P_X(t). \end{aligned}$$

Thus $\Pp_{X_1}=\Pp_{X_2}=\P_X$.

2 )

Let $Y:=X_1-X_2$. With the uniqueness of the characteristic function and the previous result from 1 ) we get that

$$\phi_{X_1}=\phi_{X_2}=\phi_{X}.$$

Thus the characteristic function of $Y$ is given by

$$\begin{aligned} \phi_Y(t) &= \E(e^{it(X_1-X_2)}) \\ &= \E(e^{it X_1}e^{it(-X_2)}) \\ &= \E(e^{it X_1})\E(e^{i(-t)X_2}) \\ &= \phi_{X_1}(t)\phi_{X_2}(-t)\\ &= \phi_{X}(t)\phi_{X}(-t). \end{aligned}$$

We also observe that

$$\begin{aligned} \phi_{-Y}(t) &= \E(e^{-it(X_1-X_2)}) \\ &= \E(e^{it X_2}e^{it(-X_1)}) \\ &= \E(e^{it X_2})\E(e^{i(-t)X_1}) \\ &= \phi_{X_2}(t)\phi_{X_1}(-t)\\ &= \phi_X(t)\phi_X(-t)\\ &= \phi_Y(t), \end{aligned}$$

thus $\Pp_Y=\Pp_{-Y}$ with of the uniqueness of the characteristic function.

Questions

1. Did I miss something/ is there something I should add to the proof?
2. Is there a probability space $ (\Oo,\Aa,\Pp)$ that is not a product space $(\O^{T},\A^{\otimes T},\P^{\otimes T})$ for some $T$ such that the above conditions are fulfilled?
   • This question/answer, shows there exists another random variable on another probability space with the same distribution (not just the product space). My proof is based on the part with the product space. I first tried $\Omega_2=\Omega_1\cup E$ with $E\cap\Omega_1=\emptyset$ and $E\neq \emptyset$, but didn't find two independent random variables.