3
$\begingroup$

Is there any way using the usual (Kolmogorov) axioms of probability to describe/model the following situation :

A value $v \in \mathbb{R}$ has an equal probability of being measured anywhere in the interval $(-\infty, +\infty)$. What we want to express is that :

  1. The probability of getting some value, in other words, some $v \in (-\infty, \infty)$, is 1.

  2. If there are two intervals $A$ and $B$, and $|A|=k\ |B|$, then the probability of $v$ being in $A$ is $k$ times the probability of $v$ being in $B$.

When attempting to model it with probability axioms, since the interval of all possible values for $v$ is infinite, each finite interval in $\mathbb{R}$ will have probability $0$. Countably adding the probability of finite intervals to get the entire $\mathbb{R}$ will then still yield $0$, which is not compatible with the axiom that $P(\mathbb{R})=1$. Also, there seems to be no way to express 2 above.

$\endgroup$
  • 1
    $\begingroup$ Your argument is right. There is no such probability measure. $\endgroup$ – André Nicolas Dec 31 '15 at 22:04
1
$\begingroup$

Your argument is solid. In fact, given any infinite set $X$ for which there is a countably-infinite partition $\mathscr P$ into sets of equal cardinality, there is no uniform probability measure on $X$ making all $\mathscr P$-sets measurable.

$\endgroup$
  • $\begingroup$ So then how does one model this situation, which can arise as a fairly natural notion - for example, if the value being measured is the position of a particle, then the two conditions essentially say "The particle can be found anywhere, will definitely be found somewhere, and with a chance proportional to the length of your detection device" ? $\endgroup$ – user9875 Dec 31 '15 at 23:15
  • $\begingroup$ One instead uses a measure, like the Lebesgue measure. It acts like a probability measure, but with the space having infinite measure. $\endgroup$ – Cameron Buie Dec 31 '15 at 23:33
0
$\begingroup$

Depending upon the source, there are two sets of axioms that are sometimes called "Kolmogorov's axioms."

  1. One set of axioms requires that probability be finitely additive, i.e., for any finite collection of disjoint events $A_1, \ldots, A_n$, the probability of the union $P(\bigcup_{i \leq n} A_i)$ is equal to the sum $\sum_{i \leq n} P(A_i)$.

  2. Often times, however, Kolmogorov's axioms are said to require countable additivity, i.e., for any countable collection of disjoint events $\{A_n: n \in \mathbb{N}\}$, the probability of the union $P(\bigcup_{n \in \mathbb{N}} A_n)$ is equal to the sum $\sum_{n \in \mathbb{N}} P(A_i)$.

The latter condition is obviously strictly stronger than the former.

Your two conditions are possible to fulfil if only finite additivity is required of a probability measure, but they are impossible to satisfy if a probability measure is required to be countably additive.

To see that they cannot be satisfied for a countably additive measure, note that the intervals $\{[z,z+1): z \in \mathbb{Z}\}$ cover the real number line. By countable additivity, at least one of those intervals must have positive probability if the entire real line has positive probability. Now if any interval $A$ has positive probability $r$, then the interval $[0,1]$ has probability $1/|A|$ by your second condition. By finite additivity (which is entailed by countable additivity), it follows that the intervals $[0,1), [1,2), \ldots [M, M+1)$ have probability greater than $1$, where $M$ is the ceiling of $(|A|)+1$. That's obviously impossible.

If probability is only required to be finitely additive, however, it is possible to define a measure that satisfies your two conditions. The reason is that one can assign every interval of finite length probability zero, and so your second condition is immediately satisfied. A good source to consult is Savage's Foundation of Statistics, which contains a discussion of the importance of finitely additive probability measures.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.