Given a r.v. $X$, can we choose a r.v. $Y$ independent of $X$ and with the same distribution? The question originates in the following problem: Given that $\phi$ is a characteristic function, show that so is $| \phi |^2$.
The solution given by a lecturer was as follows: Suppose $\phi$ is the characteristic function of a random variable $X$. Choose another random variable $Y$ which is independent of $X$ and has the same distribution. Then $| \phi |^2$ is the characteristic function of $X-Y$.
I agree with the last part once $Y$ is chosen, but it does not seem completely obvious to me that you can necessarily choose such a $Y$. $Y$ has to be defined on the same probability space as $X$ in order for $X-Y$ to be defined. So my question is whether it is so that given any random variable $X$ defined on the space $( \Omega, \mathcal{F}, P)$, we may choose another random variable $Y$ defined on $( \Omega, \mathcal{F}, P)$ such that $X$ and $Y$ are independent and have the same distribution? If so, how can we show that?
 A: In fact, we are not forced to define $Y$ over the same probability space. Instead, we can define a new probability space $(\Omega', \mathcal{F}', P')$, define $X'$ on $(\Omega', \mathcal{F}', P')$ so that it has the same distribution as $X$ and then define $Y$ on $(\Omega', \mathcal{F}', P')$ so that it is independent of $X'$ and has the same distribution as $X'$.
We end up in the desired situation where we have two independent random variables on the same probability space and with the same distribution described by the characteristic function $\phi$. This enables us to define $X-Y$ as required.
Now, in this particular case, we may choose to define our new probability space as
$$
(\Omega', \mathcal{F}', P')=(\Omega\times\Omega,\mathcal{F}\otimes\mathcal{F}, P\times P)
$$
where $\Omega'=\Omega\times\Omega$ is the Cartesian product, $\mathcal{F'}=\mathcal{F}\otimes\mathcal{F}$ is the product $\sigma$-algebra on $\Omega'$ and $P'=P\times P$ is the$^1$ product measure. Next, for $(\omega_1, \omega_2)\in\Omega'$ we define
$$
X'((\omega_1, \omega_2)) := X(\omega_1)\\
Y((\omega_1, \omega_2)) := X(\omega_2).
$$
These are easily seen to be measurable functions on $\Omega'$. Moreover, by definition of product measure, we have $P'(A\times B)=P(A)P(B)$ for any $A,B\in\mathcal{F}$. This can be used to show that $X'$ and $Y$ are independent. Finally, they both have the same distribution as $X$ and therefore their characteristic function is $\phi$.

$^1$ Product measure is unique here because probability measures are finite.
