My question is motivated by the question here: How to calculate probabilities for football match?. In that question, we have a situation where we have calculated (based on historical data), the probabilities of football teams winning in particular situations. So for example, Chelsea wins at home 50% of the time, and Liverpool wins an away game 20% of the time. So a fundamental question in that post is "Given these two probabilities, what is the probability that Chelsea win when playing Liverpool at thome?"

This is also my question, but I am looking for an answer in terms of rigorous probability theory. Both the answers to the other question don't explain things rigorously. In particular, my understanding is that we effectively have two different probability spaces, let's call them $(C, \mathscr{A_1},P_1)$ and $(L, \mathscr{A_2},P_2)$, representing the probability measures on Chelsea and Liverpool respectively. Here the space on $C$ represents Chelsea playing games at home, and the second space represents Liverpool playing away.

It seems reasonable to me that we may define $C = L = \mathbb{N}^2$, where we would interpret the elementary event $\{(i,j)\}$ as meaning that Chelsea (assuming we are in the first probability space) has scored $i$ goals, while the opponent has scored $j$ goals. Here the 'opponent' is not necessarily Liverpool. I think this means that the probability space models what happens 'on average' when the team plays against any opponent, but I'd like to know if this is actually a reasonable interpretation? I also think it seems reasonable to define the sigma algebra as the Borel sigma algebra on $\mathbb{R}^2$.

Then the only difference between the two spaces is the probability functions $P_1$ and $P_2$. Is there any meaningful way to assign a probability to the event that Chelsea beat Liverpool (in a home game for Chelsea)?

The problem I see s that $P_1$ and $P_2$ inherently encode information that we want to model. For example, the function $P_1$ models Chelsea playing an arbitrary opponent at home, while $P_2$ is a model for Liverpool playing an arbitrary opponent away. So while the underlying state-space and sigma algebras are equal mathematically, the probability function 'encodes' information to mean that the event $\{(1,2)\}$ in $C$ is not the same as $\{(1,2)\}$ in $L$, since in the former, it's the event that Chelsea scores one goal and another team score 2, while in the latter, it's the event that Liverpool score one goal and another team score 2. So while mathematically they are the same objects, the interpretation is different.

How do we reconcile this difference of interpretations? It seems to me that due to this issue of interpretation, the state spaces fundamentally look different, so I am not sure how to define the event that is "Chelsea beats Liverpool"?

  • $\begingroup$ I really like the fact that you found out yourself that the probability spaces are wrong to begin with. That's one step. Now take this from me, when you compare to random variables in whatever sense then both of them have to live in the same state space. So you have one state space (what is it?) and two random variables. $\endgroup$
    – Shashi
    Commented Nov 12, 2020 at 10:50
  • $\begingroup$ @masiewpao Since you sample points are $(i, j) \in \mathbb{N}^2$, you only need the discrete measure not the Borel measure on $\mathbb{R}^2$. $\endgroup$
    – balddraz
    Commented Nov 12, 2020 at 10:57
  • $\begingroup$ @Shashi This is what I'm confused by! I can't think of a way to build a model (from historical data at least) that uses the same state space for both teams. I was thinking there maybe needs to be an entirely different approach to constructing $P_1$ and $P_2$? $\endgroup$
    – masiewpao
    Commented Nov 12, 2020 at 11:09
  • $\begingroup$ @0XLR Ah yes of course, although I suppose we can just define the support to be $\mathbb{N}^2$ $\endgroup$
    – masiewpao
    Commented Nov 12, 2020 at 11:10

1 Answer 1


I extend the comment that I wrote here. I will not give an answer to your question, but this will help you to find an answer yourself (that maybe an answer to the original question does not even exist).

So to extend the comment that I wrote, if in some sense $X$ is the random variable that corresponds to the score of Chelsea and $Y$ the score of Liverpool, then you are basically after $$\mathbb P(X>Y),\ \ \ .... \text{right?!}$$ Since $\{X>Y\}$ is nothing else but $$\{\omega\in \Omega\ : \ X(\omega)>Y(\omega)\} $$ these two random variables must live in the same probability space.

So assume you got that right, you defined your probability space and all. If you know joint law and marginal law, then I hope you see that the statement $$\text{"Liverpool wins 20% at home"}$$ does not say much about the joint law. While the one that you are interested in requires the joint law. Do we have the joint law? As long as the answer is no, then you cannot answer the question. Why? See it as follows: consider a case where the Premier League is so unpopular that there are only three teams playing, Chelsea, Liverpool and Arsenal. Since you want to practice how to do this rigorously, I leave this exercise for you:

Exercise. Construct a probability space where Liverpool has probability $\frac 1 5$ to win in an arbitrary match, Chelsea has probability $\frac 4 5 $ to win an arbitrary match, and Chelsea wins from Liverpool with probability $x$ (you choose $x$ yourself from the infinite amount of choices).

Sorry, not every story has a happy ending.

PS: oh, also if you ever encounter occasions where the state space is countable, then taking the power set as the $\sigma$-algebra may not be a bad idea.

  • $\begingroup$ Sorry for the delay in accepting, it took a while for me to think about the problem but I think I see what you're saying now. Am I correct in thinking that the way we defined the 'models' (i.e. using historic data in that particular way) simply doesn't lend itself to using the joint law that we would need? I.e., we cannot meaningfully assign a probability given the info? $\endgroup$
    – masiewpao
    Commented Nov 13, 2020 at 12:45
  • 1
    $\begingroup$ Exactly!! I'm glad that I could help! $\endgroup$
    – Shashi
    Commented Nov 13, 2020 at 12:59

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