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I just want to confirm whether I am understanding this correctly: The requirements of a probability measure require the probabilities of pairwise disjoint sets to sum to one. Non-disjoint probabilities can sum to greater than one, then (I'm not sure if we would ever care about this, though).

An example I am thinking of is if we had 10 coin flips. The probability of the first coin flip being heads is $\frac{1}{2}$, the probability of the second coin flip being heads is $\frac{1}{2}$, the probability of the third coin flip being heads is $\frac{1}{2}$, and so on...

Daniel Fischer
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majmun
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2 Answers2

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Yes, the sum can be greater than $1$. (You're also right that we never really care about this - at least, as far as I know . . .)

An even simpler example:

  • What's the probability that $8=8$?

  • What's the probability that $1\not=17$?

  • What's the sum of these probabilities?

Noah Schweber
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  • Thank you. I will accept your answer once possible. Out of curiosity though, can a valid probability triple be defined where such probabilities are defined? – majmun Nov 24 '15 at 04:58
  • @majmun I'm not sure what you mean. Are you asking whether there are contexts in which probabilities greater than 1 are meaningful? (If so, my answer is "I'm pretty sure there aren't, but I could be wrong - stranger things have happened.") – Noah Schweber Nov 24 '15 at 05:00
  • I guess what I'm asking is that normally we say something like What is the probability of 8 occurring, or X=8 or something, correct? In that sense, I am not sure how valid your "simpler" example is, since I am not sure whether there is some probability triple where it would make sense to ask "does $8=8$, does $1 \not= 17$. (make sense as in being possible, not that anyone would ever want to ask it). – majmun Nov 24 '15 at 05:05
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    "$8=8$" and "$1\not=17$" are just silly ways of describing the trivial event - if you prefer, write "$X=X$" for each. – Noah Schweber Nov 24 '15 at 05:08
  • Ah, sorry for not getting that. Thank you. – majmun Nov 24 '15 at 05:10
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The requirements of a probability measure require the probabilities of pairwise disjoint sets to sum to one.

No. Given pairwise disjoint events $A_1, A_2, ...$ a probability space $(\Omega, \mathscr F, \mathbb P)$, we have by a Kolmogorov axiom:

$$P(\bigcup_n A_n) = \sum_n P(A_n)$$

Consider $(\Omega, \mathscr F, \mathbb P) = ([0,1], \mathscr B([0,1]), \lambda)$ and $A_1 = (0.4,0.6), A_2 = (0.7,0.8), A_m = \emptyset$ for $m \ge 3$

Then

$$P(\bigcup_n A_n) = \sum_n P(A_n) = 0.3 \ne 1$$

However, if we have pairwise disjoint $A_n$ s.t. $P(\bigcup_n A_n) = 1$, then

$$P(\bigcup_n A_n) = \sum_n P(A_n) = 1$$

Non-disjoint probabilities can be greater than 1 eg

$$\sum_{n=1}^{\infty} P([0,1/n]) = \sum_{n=1}^{\infty} 1/n = \infty$$

Consider the Borel-Cantelli Lemmas.

If we have independent events $A_1, A_2, ...$, then

$$\sum_n P(A_n) = \infty \to P(\limsup A_n) = 1$$

$$\sum_n P(A_n) < \infty \to P(\limsup A_n) = 0$$

As to why we would care? Look at cases 2,3,4,5 here.

We can have $1 = \sum_n P(A_n)$, $1 < \sum_n P(A_n) < \infty$ or $\sum_n P(A_n) = \infty$

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