1
$\begingroup$

In the Probability section of the Inclusion–exclusion principle Wikipedia page the general formula is: $$ \mathbb {P} \left(\bigcup _{i=1}^{n}A_{i}\right)=\sum _{i=1}^{n}\mathbb {P} (A_{i})-\sum _{i<j}\mathbb {P} (A_{i}\cap A_{j})+\sum _{i<j<k}\mathbb {P} (A_{i}\cap A_{j}\cap A_{k})+\cdots +(-1)^{n-1}\sum _{i<...<n}\mathbb {P} \left(\bigcap _{i=1}^{n}A_{i}\right)$$

In my understanding, the last summation should be removed to give: $$ \mathbb {P} \left(\bigcup _{i=1}^{n}A_{i}\right)=\sum _{i=1}^{n}\mathbb {P} (A_{i})-\sum _{i<j}\mathbb {P} (A_{i}\cap A_{j})+\sum _{i<j<k}\mathbb {P} (A_{i}\cap A_{j}\cap A_{k})+\cdots +(-1)^{n-1}\mathbb {P} \left(\bigcap _{i=1}^{n}A_{i}\right)$$

since there is only 1 term left with the probability of the intersection of all sets from $1$ to $n$.

Anyone to give me a confirmation/refutation?

$\endgroup$
5
  • 2
    $\begingroup$ it is not written well, but the intent is clearly for there to be only one term in the sum so that it matches what you wrote. $\endgroup$ Commented Sep 15, 2020 at 20:37
  • $\begingroup$ @MatthewTowers if it's a sum with only one term, it seems pretty confusing to leave it, esp. with the i<...<n indices. $\endgroup$
    – Marc
    Commented Sep 15, 2020 at 20:43
  • $\begingroup$ I think they want to emphasize that the final term isn't a special case by giving it a sum just like the others. Nevertheless what they've written isn't correct notation $\endgroup$ Commented Sep 15, 2020 at 20:44
  • $\begingroup$ @MatthewTowers I see. So I probably shouldn't propose a fix in Wikipedia I guess :) Thank you for your clarification. $\endgroup$
    – Marc
    Commented Sep 15, 2020 at 20:46
  • $\begingroup$ I think it would be fine to propose a change, it's at best poor notation $\endgroup$ Commented Sep 15, 2020 at 20:47

1 Answer 1

3
$\begingroup$

Presumably the intent is to have a sum, for consistency with the other terms, but for the sum to have only one summand. Though the notation is bad. A better way would be to have the first sum being over $i_1 : 1 \leq i_1 \leq n$, the second over $i_1, i_2 : 1 \leq i_1 < i_2 \leq n$, and so on, with the last being $i_1,\ldots, i_n : 1 \leq i_1 < i_2 < \cdots < i_n \leq n$ so that it only has one term $i_1=1,i_2=2, \ldots$

$\endgroup$

You must log in to answer this question.

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