# What does $\Pr(X=a,Y=b)$ really mean?

Let's say that $X$ and $Y$ are random variables that take values in sets $\mathcal{A}$ and $\mathcal{B}$, respectively, then my question is: what does this probability really mean: $$\Pr(X=a, Y=b)$$

## My thoughts so far

Probabilities are functions that map from the space of events, into the space $[0,1]$. Specifically, this is the probability function prototype: $$\Pr: \text{space of events} \rightarrow [0,1]$$

An event is essentially a set of outcomes in $\Omega$. So the space of events is essentially the power set of the space of outcomes $\mathbb{P}(\Omega)$. So we can rewrite the probability definition: $$\Pr: \mathbb{P}(\Omega) \rightarrow [0,1]$$

Random variables $X$ and $Y$ are functions with the prototype $X:\Omega \rightarrow \mathcal{A}$, $Y:\Omega \rightarrow \mathcal{B}$, where $\mathcal{A}$ and $\mathcal{B}$ are some sets.

For any $a \in \mathcal{A}$, if we say $\Pr(X=a)$, clearly $X=a$ is not an event (it is literally a function name being equated against an element of a set, which is wrong in this context). So what we really mean is this: $$\Pr(\{\omega : \omega \in \Omega, X(\omega) = a\})$$

And that makes sense, because $\{\omega : \omega \in \Omega, X(\omega) = a\}$ is an event (a set of outcomes).

But $\Pr(X=a)$ is a fairly popular notation abuse because it makes writing easier. Some do further abuse by stating $\Pr(a)$ which in my view too abusive because it is potentially ambiguous.

Now let's look at my question $\Pr(X=a,Y=b)$. I am not sure what it really means, because I have two possibilities.

Possibility 1: this possibility makes sense to me (but I need your confirmation) since $\{\omega : \omega \in \Omega, X(\omega)=a, Y(\omega)=b\}$ is an event (set of outcomes). $$\Pr(\{\omega : \omega \in \Omega, X(\omega)=a, Y(\omega)=b\})$$

Possibility 2: this possibility makes no perfect sense to me since $\{(\omega_i, \omega_j) : (\omega_i,\omega_j) \in \Omega^2, X(\omega_i)=a, Y(\omega_j)=b\}$ is not a set of outcomes to me... Or is it? This causes me to be doubtful about my understanding. $$\Pr(\{(\omega_i, \omega_j) : (\omega_i,\omega_j) \in \Omega^2, X(\omega_i)=a, Y(\omega_j)=b\})$$

Possibility 3: perhaps you know of a better possibility that I didn't think about? If so kindly share.

• its actually closer to possibility 2. note in possibility 1 you are restricting the probability to the set of events such that $X(\omega) = a$ and $Y(\omega)=b$ (the same $\omega$). Feb 6, 2017 at 20:13

Possibility 1 is correct, as long as $\Omega$ is defined correctly. For example, you could have $\Omega = \mathcal A \times \mathcal B$, with $X(\omega) = X((\omega_1, \omega_2)) = \omega_1$ and $Y(\omega) = \omega_2$. Then \begin{align} \Pr( X = a, Y = b ) &= \Pr( \{ \omega \in \Omega \mid X(\omega)=a, Y(\omega)=b \}) \\&= \Pr( \{ \omega \in \Omega \mid \omega_1 = a, \omega_2 = b \}) \\&= \Pr( \{ (a, b) \} ) \end{align} in this simple setup. In a more complicated situation with other random variables, $\omega$ would contain more information, and the event $X=a,Y=b$ would correspond to more than just the one element of $\Omega$. Of course, usually we don't actually define it explicitly. The formal version of this is known as a product measure.

You could set up the situation so that $\Omega$ is some common base which you map to $\mathcal A$ and $\mathcal B$ separately, in which case the notation of possibility 2 would be basically correct – except that you have to keep track somehow that $X$ is a function on the first coordinate and $Y$ on the second, and $\Pr$ would be a measure on a sample space $\Omega^2$. This amounts to basically the same thing, but the notation is worse.

As an aside, the set of possible events is generally not the full power set of $\Omega$, since that leads to lots of difficulties. You might be interested in reading up on measure-theoretic probability.

• For the 1st part, I am a bit confused about $X(\omega) = X((\omega_1,\omega_2)) = \omega_1$. I am not seeing why you define the random variable to return the 1st component of the outcomes space which is $\omega_1$ in this case. Did you mean $X(\omega) = X((\omega_1,\omega_2)) = X_{\text{caveman}}(\omega_1)$? Feb 6, 2017 at 22:58
• $\Omega$ is a set of pairs, whose elements I'm writing as $\omega = (\omega_1, \omega_2)$; $X : \Omega \to \mathbb R$ is a function that gives the first component of the pair. This is one natural way to define a state space that allows for two real-valued random variables, though you can do it in any number of ways if you're so inclined. If you like, $X_\text{caveman}$ can be some other mapping, but you don't need to introduce it if you don't want to. Feb 6, 2017 at 23:02
• Thing is, I originally defined $X : \Omega \rightarrow \mathcal{A}$. You seem to have changed that into $X : \Omega \rightarrow \mathbb{R}$, which I don't understand why. Was this intentional to show me something? Or was it because of unrelated reasons? This makes me feel that maybe I am missing something that you wanted to say. Feb 6, 2017 at 23:22
• Sorry, I just missed that you had defined $\mathcal A$ and $\mathcal B$. Changed. Feb 6, 2017 at 23:23

It is possibility 1. Too see it better you can denote new bivariate r.v. $Z = (X,Y), Z:\Omega \rightarrow R^2$ and then

$$P(Z = (a,b)) = P(Z^{-1}(a,b)) = P(\{\omega \in \Omega\ |\ Z(\omega) = (a,b) \})$$