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This is taken from Pollard, A User's Guide to Measure Theoretic Probability problem 1.1.

Let $A_1, A_N$ be events in some probability space. Denote $\cap_{i \in J}A_i$ as $A_J$ for some set $J \subseteq [N]$. Also, denote the sum of the probabilities of the ways that $k$ events occur together as $S_k = \sum_{|J| = k}\mathbb P \{A_J\}$.

Show that the probability of exactly $m$ of the events $A_i$ occurring is given by:

$${m\choose m}S_m - {m+1\choose m} S_{m+1} ... +(-1)^{N-m} {N \choose m} S_N$$

What I've tried

I know that this is a generalization of the inclusion-exclusion rule which gives us the probability that none of the events occur. The inclusion-exclusion rule is easier to derive since it's clear what each term is adding and subtracting. Here I don't have strong intuition on what each of the terms represent.

The binomial coefficients remind me of the stars and bars result: the number of ways to distribute $m$ points into $k$ groups. So the first coefficient is the number of ways to distribute $m$ points into $1$ group, the second is the number of ways to distribute $m$ points into $2$ groups and so on until $N - m +1$ groups. This doesn't seem to be the right interpretation since it does not match up with the $S_k$ terms.

Any hints or intuition would be appreciated.

RobPratt
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dmh
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    You write "the sum of the probabilities of the ways that exactly $k$ events can occur", but I don't think you mean "exactly" here (otherwise the answer would simply be $S_m$); $A_J$ must be the probability that all the events $A_j\colon j\in J$ occur, and possibly others as well. – Greg Martin Feb 22 '21 at 04:08
  • Yes that's right, I'll correct it. Thanks! – dmh Feb 22 '21 at 04:09
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    Also, it's better to just write $(-1)^{N-m}$ (which automatically accounts for the parity of $N-m$) than to make up a new letter $a$ for that. – Greg Martin Feb 22 '21 at 04:11
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    Related: https://math.stackexchange.com/questions/1808129/generalised-inclusion-exclusion-principle – Mike Earnest Feb 22 '21 at 05:10

1 Answers1

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Let $\mathbb1_B$ be the indicator function of an event $B$, so that $\mathbb1_B$ is a random variable that equals $1$ if the event $B$ occurs and $0$ otherwise; note that the expectation of $\mathbb1_B$ equals the probability that $B$ occurs. Let $E_m$ denote the event that exactly $m$ of the $A_j$ occur. Your desired formula would follow from $$ \mathbb1_{E_m} = {m\choose m}\sum_{|J| = m}\mathbb 1_{A_J} - {m+1\choose m} \sum_{|J| = m+1} \mathbb1_{A_J} + \cdots + (-1)^{N-m} {N \choose m} \sum_{|J| = N}\mathbb 1_{A_J} $$ by taking expectations of both sides.

The point is that this equality of functions can be checked separately on every possible combination of the events $A_j$. In particular, if you let $K$ denote the exact set of events that occur, then you can verify that the two sides of the above identity are equal one $K$ at a time. The argument will depend only on $\#K$, and will essentially be "if $\#K = m+k$, then how many terms in the first sum equal $1$, how many terms in the second sum, etc.".

Greg Martin
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  • Thanks for the insight on how only the cardinality of $K$ matters. I eventually found the proof of the statement in my post: Corollary 5.2 on p. 184 of Martin Aigner's book A Course in Enumeration. The proof is more involved than I expected, I'll need to spend some extra time to carefully go through it. – dmh Feb 22 '21 at 19:30
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    Also check this answer for a proof that is much easier (for me) to follow: https://math.stackexchange.com/a/362516/352034 – dmh Feb 22 '21 at 19:45