2
$\begingroup$

I think the question (stated in the title) is self explanatory, but here is the complete question:

Consider the triple: $\langle S, \mathscr S, Pr \rangle$

Axiom 1: $Pr(S)=2$, $Pr(\emptyset)=0$

Axiom 2: $Pr(s)+Pr(s^c)=2$ for each $s\in \mathscr S$.

Basically, this is all I am asking: if we push the bound up (for the first axiom), and likewise tweak the second axiom will we have the same Probability Theory we have right now (perhaps with slight numeric differences, but consistent nonetheless)?

Further Explanation and Context:

Given a coin with even density, we know that the probability of landing $\{H\}$ or $\{T\}$ is $\frac{1}{2}$. Now, if we modify axiom 1, as suggested, $Pr\{H\}$ would be $1$, likewise for $\{T\}$; however, modifying the second axiom will, fortunately, help us reformulate conditional Probability $Pr\{H\}$ or $Pr\{T\}$ given we land either $\{H\}$ or $\{T\}$, and we get $Pr\{H\}=\frac{1}{2}$ and $Pr\{T\}=\frac12$ given we land either $\{H\}$ or $\{T\}$.

Gae. S. rightfully pointed out that for countable and uncountable sets the measure might stop being $\sigma-$finite.

This is the essence of my question: Most "physical" spaces are finite; will such a measure, axiomatized by what was stated, above be sufficient? My guess is that it will work out just fine, but I am not quite sure if I can prove this.

Note: I found these two questions to be similar to what I am asking, but I don't think they were helpful to me.

Probability Axioms

Why must the probability of an event be between 0 and 1?

$\endgroup$
12
  • 5
    $\begingroup$ I don't quite see how products of probability spaces would work. The product of $n$ spaces of measure $2$ has measure $2^n$, and I think the behaviour for infinite products is catastrophically different (i.e. I would guess it may stop being $\sigma$-finite). $\endgroup$
    – user239203
    Jan 1, 2021 at 10:50
  • $\begingroup$ @Gae.S. That was a good catch. I think you are right, but that would be for countable sample spaces though, right? What about strictly finite spaces, do you think it will work? $\endgroup$ Jan 1, 2021 at 11:14
  • 3
    $\begingroup$ Basically there are two useful values for the measure (probability) of the entire space one (probability) and infinite (measure theory). I have never thought about any other possibilities and I find it hard to see the point. $\endgroup$ Jan 2, 2021 at 18:13
  • 2
    $\begingroup$ I find it very confusing that you keep referring to the first and second axioms and never state them. There is no single ordering of the axioms of probability, and in fact sometimes they are stated as three or four separate axioms. $\endgroup$
    – Mars
    Jan 3, 2021 at 5:46
  • 1
    $\begingroup$ How is one supposed to interpret "$Pr(E) + Pr(E^c) = 2$", which literally translates as "the probability of either $E$ or not $E$ is 200%." $\endgroup$ Jan 3, 2021 at 5:49

2 Answers 2

2
$\begingroup$

Every statement in the usual probability theory of the space $(S, \mathscr{S}, \mathbb{P})$ can be changed into a statement in your alternative theory of the space $(S,\mathscr{S},Pr)$ by changing $\mathbb{P} \mapsto \frac{Pr}{2}$. Similarly every statement in your alternative theory becomes a statement of the usual theory by $Pr \mapsto 2\mathbb{P}$. In other words, they are equivalent and one is consistent if and only if the other is.


That said, changing $1$ to $2$, in my mind, just adds needless complication. When dealing with products, for instance, the usual theory holds that $$\mathbb{P}_{S_1\times S_2}(A_1\times A_2) = (\mathbb{P}_{S_1}\otimes \mathbb{P}_{S_2})(A_1\times A_2) = \mathbb{P}_{S_1}(A_1)\mathbb{P}_{S_2}(A_2)$$ which extends how we compute area as length times width. In your theory, the equivalent statement would be $$Pr_{S_1\times S_2}(A_1\times A_2) = \frac{1}{2} Pr_{S_1}(A_1)Pr_{S_2}(A_2)$$ and it becomes worse with higher powers $$Pr_{\prod_{i=1}^n S_i}\left(\prod_{i=1}^nA_i\right) = \frac{1}{2^{n-1}}\prod_{i=1}^nPr_{S_i}(A_i),$$ which directly complicates every statement involving independence, most likely with mixed powers of $2$.

No doubt there's a philosophical razor we could use here to reject your alternative theory as being more complicated with no worthwhile payoff.


To create an alternative theory that may result in an actual benefit, you would need to do something more drastic like taking as an axiom $\widetilde{Pr}(S) = x$ is a fixed variable and work in the ring $\mathbb{R}[x,x^{-1}]$ or some extension thereof. This might force the theory to keep track of another form of "dimension", but I'm skeptical that the payoff, if it ever materializes, would be worth the massive headache caused.

$\endgroup$
1
1
$\begingroup$

This was a bit too long for a comment, so I will post it as an answer.

I think that (at least the proofs of) various zero-one laws like Kolmogorov's will suffer from such a definition. Such laws say that certain events have probability $0$ or $1$, i.e., occur almost never, or almost surely. The proofs I know proceed by showing that the relevant events $E$ are independent of themselves, and thus satisfy $P(E) = P(E\cap E) = P(E)^2$, and this means they must have probability $1$ or probability $0$ (as these are the only roots of $p^2-p = 0$). Since we would thus be deducing that these events have probability $1$ "out of $2$," I feel like this theory you are proposing must be inconsistent with the "usual" probability theory.

$\endgroup$
3
  • $\begingroup$ Hi, thanks for that insight. I think "$Pr(S)=2$, $Pr(\emptyset)=0$, $Pr(s)+Pr(s^c)=2$ for each $s\in \mathscr S$" will induce their own set of laws much like "Zero-One" of Kolmogorov. I think, like you said, it will be inconsistent with theorems deduced from the usual probability theory, but self consistent. Is there any way you can show it's inconsistent with measure theory as a whole? If you can show that, then that would imply these axioms are self-inconsistent. Unfortunately I have no clue when it comes to that. $\endgroup$ Jan 3, 2021 at 8:39
  • 2
    $\begingroup$ @BertrandWittgenstein'sGhost: I do not see how one could show the axioms you laid out are inconsistent with measure theory, since you are essentially describing measure spaces of total measure $2$, which measure theory has no problems with. To me, the main thing that distinguishes probability theory from pure measure theory is the concept of independence. I think with the total measure $2$ assumption, you will have to see what you need to change about the definition of independence. For example, in usual probability, the total space is independent of itself, but no longer in this version. $\endgroup$
    – Alex Ortiz
    Jan 3, 2021 at 20:52
  • $\begingroup$ I see. I appreciate your time! $\endgroup$ Jan 4, 2021 at 2:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .