Given two probability spaces $\textstyle (\Omega_1,\mathcal{F}_1,P_1) , \textstyle (\Omega_2,\mathcal{F}_2,P_2)$ , X is a random variable defined on $(\Omega_1,\mathcal{F}_1,P_1)$, Y is a random variable defined on $(\Omega_2,\mathcal{F}_2,P_2)$. Is it possible X and Y have the same distribution?
if have, how to prove?
Sure it's possible. That $X$ and $Y$ has the same distribution means that their respective probability distributions (probability measures) $P_X:=P_1\circ X^{-1}$ and $P_Y:=P_2\circ Y^{-1}$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ are the same. To prove it, you have to show that $$ P_X(B)=P_1(X\in B)=P_2(Y\in B)=P_Y(B),\quad B\in\mathcal{B}(\mathbb{R}), $$ where $\mathcal{B}(\mathbb{R})$ is the Borel sigma-algebra on $\mathbb{R}$. Since $\mathcal{B}(\mathbb{R})$ is generated by the sets of the form $(-\infty,a]$ for $a\in\mathbb{R}$, it is actually enough to show that $$ P_X((-\infty,a])=P_1(X\leq a)=P_2(Y\leq a)=P_Y((-\infty,a]),\quad a\in\mathbb{R}. $$ Note that this corresponds to checking if their respective cumulative distribution functions $F_X$ and $F_Y$ are the same.
Other methods also apply: $X$ and $Y$ have the same distribution, i.e. $P_X=P_Y$ if
their characteristic functions agree, i.e. ${\rm E}[e^{itX}]={\rm E}[e^{itY}]$, $t\in\mathbb{R}$,
their moment-generating functions agree, i.e. ${\rm E}[e^{tX}]={\rm E}[e^{tY}]$, $t\in\mathbb{R}$,
their densities agree, i.e. $f_X(t)=f_Y(t)$, $t\in\mathbb{R}$, provided that both $X$ and $Y$ are absolutely continuous,
their probability mass functions agree, i.e. $p_X(t)=p_Y(t)$, $t\in\mathbb{R}$, provided that both $X$ and $Y$ are discrete variables,
under additional assumptions, that ${\rm E}[X^n]={\rm E}[Y^n]$ for all $n\in\mathbb{N}$.
Yes. When you say that two variables have the same distribution, what you mean is that they have the same distribution function (e.g., http://www.statlect.com/glossary/distribution_function.htm). Comparing two distribution functions is possible also when the two variables are defined on different spaces. You just need to check that the two functions take the same value at each point on the real line. Comparisons between variables defined on different spaces are not possible when using concepts such as almost sure equality or zero distance (in probability, in mean square, etc.). Note that you can check equality of two distribution functions also using transforms, such as the moment generating function or the characteristic function, that have a one-to-one relation with the distribution function.
Let's try this answer. We assume that $X : \Omega_1 \rightarrow \mathbb{R}$ and $Y : \Omega_2 \rightarrow \mathbb{R}$. The distribution of $X$ is given by $$F_{P_1,X}(x) = P_1(X \leq x),$$ for all $x \in \mathbb{R}$, and for $Y$, $$F_{P_2,Y}(y) = P_2(Y \leq y),$$ for all $y \in \mathbb{R}$. So, if $X$ and $Y$ have the same distribution, we have $$ F_{P_1,X}(x) = F_{P_2,Y}(x),$$ for all $x \in \mathbb{R}$. So, I would say that it's possible for them to have the same distribution.
