Suppose $X$ and $Y$ are defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Is possible that $X$ and $Y$ have different CDF? I know the answer is yes, I don't understand why.
I don't understand the link between the measure $\mathbb{P}$ and the CDFs $P_X(X \le x), P_Y(Y \le y) $
Intuitively what doest it mean for two RVs to be defined on the same probability space?
Could you provide an example?
If $X:\Omega\to\mathbb R$ is measurable in the sense that $X^{-1}(B)\in\mathcal F$ for every element $B$ of Borel $\sigma$-algebra $\mathcal B(\mathbb R)$ then its CDF $F_X$ is defined by:
$$F_X(x):=\mathbb P(\{\omega\in\Omega\mid X(\omega)\leq x\})$$
The RHS is abbreviated by $\mathbb P(\{X\leq x\})$ or (a bit shorter) by $\mathbb P(X\leq x)$.
The expression $P_X(X\leq x)$ that appears in your answer is wrong.
The random variable $X$ induces a probability measure $P_X$ on measurable space $(\mathbb R,\mathcal B(\mathbb R))$ by:$$P_X(B)=\mathbb P(\{\omega\in\Omega\mid X(\omega)\in B\})=\mathbb P(X\in B)$$ where the utmost RHS is again an abbreviation.
Applying that on $B=(-\infty,x]$ we arrive at: $$F_X(x):=\mathbb P(\{\omega\in\Omega\mid X(\omega)\leq x\})=\mathbb P(X\leq x)=P_X((-\infty,x])$$
So the argument of $P_X$ is a subset of $\mathbb R$ and not necessarily a subset of $\Omega$ as the (wrong) notation $P_X(X\leq x)$ suggests.
If $Y:\Omega\to\mathbb R$ is another measurable function then it will have likewise its CDF and there is no reason at all to think that this function will coincide with the CDF of $X$.
E.g. define $X$ by $\omega\mapsto 0$ and define $Y$ by $\omega\mapsto1$.
Then it can be deduced that $F_X(x)=1_{[0,\infty)}(x)\neq1_{[1,\infty)}(x)=F_Y(x)$.
If $X$ and $Y$ are defined on the same probability space then expressions like $X+Y$ make sense and are new random variables on that probability space.
Let $\Omega=\{0\}$, with the only possible probability measure, namely $P(\Omega)=1$, $P(\emptyset)=0$. Define $X,Y:\Omega\to\mathbb R$ by $X=0$, $Y=2$.
What is the probability that $X>1$? What is the probability that $Y>1$? Do $X$ and $Y$ have the same CDF?
To start with an example, imagine a fair dice with numbers from $1$ to $6$. These numbers are the outcomes and are collected in $\Omega$, i.e. $\Omega = \{1,2,3,4,5,6\}$ in this case.
As a probability space you want to choose one, such that you can "say something" about all the events you are interested in. A event is a set of outcomes. In the upper case interesting events are of course the singletons with the outcomes after rolling the dice, i.e. $$\{\{1\},\{2\},\{3\},\{4\},\{5\},\{6\}\} \subset \mathcal{F}.$$ But it might for example be also interesting only to know if the outcome is even or odd, because you are not really interested in its exact value. Therefore we want to include as many sets as possible into $\mathcal{F}$. Thats easy in this case, because one can think of $\mathcal{P}(\Omega)$.
But as you might know this choice is not always that easy, because there is one more step to make - defining the measure $\mathbb{P}$. And for this $\mathcal{P}(\Omega)$ is sometimes just too large. Because of this we are fine with only taking a subset $\mathcal{F} \subset \mathcal{P}(\Omega)$, such that we can define a probability measure $\mathbb{P}$. Back to our example this would be the uniform distribution over $\Omega$, i.e. $$\mathbb{P}(\{i\}) = \frac{1}{6} \ \ \ \forall 1\leq i \leq 6.$$
Now we are ready to define random variables:
As a first one imagine $X(\omega) = 1_{\{\omega \ is \ odd\}}$. What this one tells you is easy to explain: You throw a dice and $X$ tells you whether the outcome is odd or not.
As a second one we imagine a random variable which tells you more than just if the outcome was odd, namely $Y(\omega) = \omega$. So if you throw your dice $Y$ tells you the value of the outcome and you can decide for your self whether the outcome was odd or not.
And of course the distribution functions $F_X$ and $F_Y$ are entirely different: $$F_X(x) = \cases{0 \mathrm{\ \ for \ }x<0 \\ \frac{1}{2} \mathrm{\ \ for \ }0 \leq x<1 \\1 \mathrm{\ \ for \ }x\geq 1}$$ and $$F_Y(y) = \cases{0 \mathrm{\ \ for \ }x<1 \\ \frac{1}{6} \mathrm{\ \ for \ }1 \leq x<2\\ \frac{2}{6} \mathrm{\ \ for \ }2 \leq x<3\\ \frac{3}{6} \mathrm{\ \ for \ }3 \leq x<4\\ \frac{4}{6} \mathrm{\ \ for \ }4 \leq x<5\\ \frac{5}{6} \mathrm{\ \ for \ }5 \leq x<6 \\1 \mathrm{\ \ for \ }x\geq 6}$$
I hope this example helps a bit. Feel free to ask and correct or improve if something is not precise or correct enough.
TL;DR I think the source of your confusion is seeing $X$ and $Y$ as being both the identity random variable in $(\Omega, \mathscr{F},\mathbb P)$.
I'm going to give an example of explicit exponential and uniform distributions in the same probability space in $(\Omega, \mathscr{F},\mathbb P)$.
Consider a random variable $X$ in $((0,1), \mathcal{B}(0,1), Leb)$ given by
$$X(\omega):=\frac{1}{\lambda} \ln \frac{1}{1-\omega}, \lambda > 0$$
It has cdf $F_X(x) = P(X \le x) = (1-e^{-\lambda x})1_{(0,\infty)}$, which we know to be the cdf of an exponentially distributed random variable. (*)
Actually,
$$X(1-\omega):=\frac{1}{\lambda} \ln \frac{1}{\omega}, \lambda > 0$$
also has cdf $F_X(x) = P(X \le x) = (1-e^{-\lambda x})1_{(0,\infty)}$.
Are all cdfs in this mysterious probability space exponential? No!
Now consider the identity random variable $U$ in $((0,1), \mathcal{B}(0,1), Leb)$:
$$U(\omega):=\omega$$
It has cdf $F_U(u) = P(U \le u) = u1_{(0,1)}+1_{(1,\infty)}$, which we know to be the cdf of a uniformly distributed random variable.
The above $X$ and $U$ have different CDFs under the same probability space. The aforementioned explicit representations of the exponential and uniform distributions in this probability space are called Skorokhod representations in $((0,1), \mathcal{B}(0,1), Leb)$.
Now consider the identity random variable $U$ in $(\mathbb R, \mathscr B(\mathbb R), (1-e^{-\lambda \{u\}})1_{(0,\infty)})$
No surprise that $U$ is exponential by definition: $F_U(u) = P(U \le u) = (1-e^{-\lambda \{u\}})1_{(0,\infty)}$.
Now you're wondering: Aha! So every random variable here is exponential right? Well no, for any distribution you can think of, say, uniform, Bernoulli, etc, all have a place here and their Skorokhod representaion is given by:
$$Y(\omega) = \sup(y \in \mathbb{R}: F(y) < \omega)$$
Try for yourself to see for yourself that
$$Y(\omega) = \sup(y \in \mathbb{R}: y1_{(0,1)}+1_{(1,\infty)} < \omega)$$
has uniform distribution in $(\mathbb R, \mathscr B(\mathbb R), (1-e^{-\lambda \{u\}})1_{(0,\infty)})$, i.e. $$P(Y \le y) := P(\sup(y \in \mathbb{R}: y1_{(0,1)}+1_{(1,\infty)} < \omega) \le y) = y1_{(0,1)}+1_{(1,\infty)}$$
Also try to see for yourself that $X(\omega)$ above no longer has exponential distribution in this probability space. (**)
Conclusion: I think the source of your confusion is seeing $X$ and $Y$ as being both the identity random variable in $(\Omega, \mathscr{F},\mathbb P)$. If you were to see them explicitly, you would know that they definitely don't necessarily have the same distribution.
What $\mathbb P$ does is tell you the probabilities of $\omega$'s. So you know how likely the sample point $$0.5 \in \Omega = (0,1)$$ is but not directly how likely the random variable $X$ is equal to a number in its domain such as the real number $$X(0.5) = \frac{1}{\lambda} \ln \frac{1}{1-0.5} \in \mathbb R$$ is. (*) Of course, the probability that $X$ is the real number $X(0.5)$ is
dependent on the probability that the sample point $0.5$ because $0.5=1-e^{-\lambda X(0.5)}$
not expected to be the same in another probability space, assuming of course that $X$ is in the new probability space, because it now depends on the probability of the sample point/s $X^{-1}(X(0.5))$ aka $X \in \{X(0.5)\}$. (**)
Pf of (*):
Two steps in computing $P(X \le x)$:
Find all $\omega \in \Omega = (0,1)$ s.t. $X(\omega) \le x$
Compute the probability of all those $\omega$'s.
For $x \le 0$, $P(X\leq x) = P(X \in \emptyset^{\mathbb R}) = P(\emptyset^{\Omega}) = 0$
For $x > 0$, $X(\omega) \le x$
Step 1:
$$ \iff \frac1{\lambda}\ln(\frac{1}{1-\omega}) \le x$$
$$ \iff \omega \le \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$
$$ \iff \omega \in (0,1) \cap (-\infty,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$
$$ \iff \omega \in (0,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$
Step 2:
$$Leb(\omega | \omega \in (0,\frac{e^{\lambda x} - 1}{e^{\lambda x}}))$$
$$= Leb((0,\frac{e^{\lambda x} - 1}{e^{\lambda x}}))$$
$$= \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$
QED
Pf of (**):
Actually $X \notin (\mathbb R, \mathscr B(\mathbb R), (1-e^{-\lambda \{u\}})1_{(0,\infty)})$ because we need $\frac{1}{1-\omega} > 0 \iff \omega < 1$.
QED
Same for $X(1-\omega)$ where we need $\frac{1}{\omega} > 0 \iff \omega > 0$.
But we can can further try to show $X$ is not exponential in $((-\infty,1), \mathscr B((-\infty,1)), (1-e^{-\lambda \{u\}})1_{(0,\infty)})$
Pf:
For $x \le 0$, $P(X\leq x) = P(X \in \emptyset^{\mathbb R}) = P(\emptyset^{\Omega}) = 0$
For $x > 0$, $X(\omega) \le x$
Step 1:
$$ \iff \frac1{\lambda}\ln(\frac{1}{1-\omega}) \le x$$
$$ \iff \omega \le \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$
$$ \iff \omega \in (-\infty,1) \cap (-\infty,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$
$$ \iff \omega \in (-\infty,\min\{1,\frac{e^{\lambda x} - 1}{e^{\lambda x}}\})$$
Step 2:
$$P(\omega | X(\omega) \le x)$$
$$ = P(\omega | \omega \in (-\infty,\min\{1,\frac{e^{\lambda x} - 1}{e^{\lambda x}}\}))$$
$$= \int_{-\infty}^{\min\{1,\frac{e^{\lambda x} - 1}{e^{\lambda x}})\} d((1-e^{-\lambda \{u\}})1_{(0,\infty)})$$
$$\int_{-\infty}^{\min\{1,\frac{e^{\lambda x} - 1}{e^{\lambda x}})\} \lambda e^{-\lambda u}$$
$$1-e^{-\lambda \min\{1,1-e^{-\lambda t}\}}$$
Doesn't look exponential to me.
QED
Perhaps it would be helpful to understand what these definitions try to capture. "A random variable is a real-valued function defined on a probability space that is measurable" though precise is perhaps thrust little prematurely to you.
Please see my answer elsewhere for a related question. It gives simple illustrations in a natural setting without any symbols or technical terms.
Here is the link:
https://mathoverflow.net/questions/250500/why-do-we-need-random-variables/252491#252491
