Let there be $n$ events given by $A_1,A_2,\ldots,A_n \subseteq \Omega$. $B_m$ denotes the event wherein at least $m$ events occur, and $C_m$ denotes the event wherein exactly $m$ events occur. I must show that $$ P(B_m) = \sum_{k=m}^n (-1)^{k-m}{k-1\choose k-m}S_k$$ $$ P(C_m) = \sum_{k=m}^n (-1)^{k-m}{k\choose m}S_k$$
Following is my argument and approach to computing $P(B_m)$ and $P(C_m)$. I have found $P(C_m)$, and I need to convert a double summation to a single summation to find the expression for $P(B_m)$. I would love it if someone could help me refine my argument wherever necessary, and let me know if it's correct. Thanks a lot!
We will call the intersection of $k$ sets as a $k$-intersection. $$S_k = \sum_{1\le i_1 < i_2 < \ldots < i_k\le n} P(A_{i_1}A_{i_2}\ldots A_{i_k})$$ Note that $P(A\cap B)$ is written as $P(AB)$ for convenience (and similarly for more intersections).
We wish to compute $P(C_k)$, given events $(A_i)_{i=1}^n$. We know that $S_m$ is an over-estimate for $P(C_m)$, since we need to subtract off some terms. $$P(C_m) = S_m - ??$$
Exactly which terms are these?
The extra terms are the $m+1,m+2,\ldots,n$ intersections. Given a $k$ intersection, say $A_{i_1}\ldots A_{i_k}$ with $k\ge m$ - how many $m$ intersections is it contained in? Exactly $k\choose m$, since this is the same as selecting $m$ elements out of $\{i_1,\ldots,i_k\}$.
Let us first get rid of the $m+1$ intersections: $$P(C_m) = S_m - {m+1 \choose m}S_{m+1} + ??$$ We realize that we have subtracted the $m+1$ intersections but by an argument similar to the one above - it is evident that these $m+1$ intersections all contain $m+2,\ldots$ intersections too. So, the expression we have now is an underestimate - since we ended up subtracting more than we should have. We correct this by adding the $m+2$ intersections, and we know that for a given $m+2$ intersection, there are exactly $m+2\choose m$ intersections of size $m$ that contain it: $$P(C_m) = S_m - {m+1 \choose m}S_{m+1} + {m+2\choose m}S_{m+2} - ??$$ Continuing in this fashion, we arrive at $$ P(C_m) = S_m - {m+1 \choose m}S_{m+1} + {m+2\choose m}S_{m+2} - \ldots = \sum_{k=m}^n (-1)^{k-m}{k\choose m}S_k$$ Now if $B_m$ denotes the event that at least $m$ of $A_1,\ldots,A_n$ occur, we have $$P(B_m) = \sum_{i=m}^n P(C_i) = \sum_{i=m}^n \sum_{k=i}^n (-1)^{k-i}{k\choose i}S_k$$
Thanks again!
