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Sometimes I see the expression "two random variables are defined on the same probability space $(\Omega,F,P)$", and I'm curious what it means to be in the same probability space. Does it mean the same as being in the same sample space $(\Omega)$?

For example, let's say I have an object flying along the x-axis, and I pin one point as the x-coordinate, however, the flying object can be anywhere on the y-axis and on the z-axis. So two probability distributions are defined as [$g(y) dy$] and [$h(z) dz$].

And

$\int_{-\infty}^\infty g(y)\,dy = 1$

$\int_{-\infty}^\infty h(z)\,dz = 1$

Are the random variables $Y$ and $Z$ defined on the same probability space $(\Omega,F,P)$ ? (It seems like they are in a different sample space..)

x.projekt
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Naomi
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    Yes they are functions from the same sample space $\Omega$. So $X:\Omega\rightarrow\mathbb{R}$ and $Y:\Omega\rightarrow\mathbb{R}$. For any given problem it is usually assumed the random variables $X$ and $Y$ are on the same space (unless otherwise stated). The only reason to state it is to be formal. – Michael Dec 28 '21 at 06:47
  • The sentence you quoted says “on” but then you changed the word to “in”. – littleO Dec 28 '21 at 06:47
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    See https://math.stackexchange.com/questions/1578323/random-variables-defined-on-the-same-probability-space-with-different-distributi – Amaan M Dec 28 '21 at 06:54

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When it says $X, Y, Z$ are "on the same probability space" it means they are all part of the same probability experiment. So indeed there is just one sample space $\Omega$ and the random variables are all functions from that sample space to the real numbers: \begin{align} X:\Omega\rightarrow \mathbb{R}\\ Y:\Omega\rightarrow\mathbb{R}\\ Z:\Omega\rightarrow\mathbb{R} \end{align} So the particular outcome $\omega \in \Omega$ in the experiment determines the values $X(\omega)$, $Y(\omega)$ and $Z(\omega)$. It means that it makes sense to ask about the probability that $X+e^Y\leq Z$, and to define new random variables such as $W=X+Y+Z$ (otherwise these things would not make sense). Note that if $X, Y, Z$ are on the same probability space, they are not necessarily distributed the same, and they are not necessarily independent.

Further, it only makes sense to say that $X$ and $Y$ are independent if they are on the same probability space. Otherwise, the probability $P[\{X\leq 2.3\} \cap \{Y\leq 1.2\}]$ would not make sense because $\{X\leq 2.3\} \cap \{Y\leq 1.2\}$ would not be a valid event.

For a given problem, it is usually assumed that all random variables are on the same probability space (unless otherwise stated). For example, suppose we flip 10 coins, with all outcomes equally likely, and define $X$ as the number of HEADS, $Y$ as the number of TAILS, and $$ Z = \left\{\begin{array}{c} 1 & \mbox{ if $X=Y$} \\ 0 & \mbox{ else} \end{array}\right.$$ So $X, Y, Z$ are all on the same probability space, but that is so obvious that it is usually not stated. Usually you would only mention that random variables are on the same probability space if you want to be formal, or if you want to derive some general result without giving the details of the probability experiment.

Michael
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    Thank you for the clear explanation! – Naomi Dec 29 '21 at 13:10
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    I have a question. In the last example, suppose we flip 10 coins, with all outcomes equally likely. If we define $X$ as the number of HEADS and $Y$ as the number of TAILS, they are perfectly correlated. $X$ is observed means that the experiment has been done, so $Y$ has already been determined. It looks like something went wrong. So, how to define two uncorrelated random variables on the same probability space, e.g., the probability space in this example? – WoodyMiao Jul 24 '23 at 11:00
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    Here the sample space is $$\Omega = {(f_1, ..., f_{10}): f_i \in {H,T}\quad \forall i \in {1, ..., 10}}$$ and all $2^{10}$ outcomes are equally likely. Let $\omega = (f_1, ..., f_{10})$ represent an outcome. For $i \in {1, ..., 10}$ and $\omega \in \Omega$ define $$V_i(\omega) = \left{\begin{array}{cc} 1 & \mbox{ if $f_i=H$} \ 0 & \mbox{else} \end{array}\right.$$ It can be shown that $V_1$ and $V_2$ are independent $Bernoulli(1/2)$ variables. And we can indeed define $X(\omega) = \sum_{i=1}^n V_i(\omega)$ as the number of heads and $Y(\omega)=10-X(\omega)$ as the number of tails. – Michael Jul 24 '23 at 14:49
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    In this case if we know the value of $X$, that determines the value of $Y$, but does not necessarily determine which outcome $\omega$ occurred and does not necessarily determine the values of other random variables on that space. For example if $X=1$ then $Y=9$ but we do not know the value of $V_1$. There are exactly 10 outcomes $\omega\in \Omega$ that satisfy $X(\omega)=1$. However, the outcome $\omega \in \Omega$ determines everything that occurs, meaning it determines the value of all random variables $X(\omega), Y(\omega), V_1(\omega), ..., V_n(\omega)$ on that probability space. – Michael Jul 24 '23 at 14:56