$\mathbb{P}(B_1) = S_1-2S_2+3S_3-4S_4+...+(-1)^{n-1}n S_n$

Question

Let $A_1,A_2,\dots,A_n$ be events in a probability space $(\Omega, \mathcal{F},\mathbb{P})$, \begin{aligned} S_0 &= 1, \\ S_1 &= \sum_{i=1}^n \mathbb{P}(A_i), \\ S_2 &= \sum_{1 \leq k_1 < k_2 \leq n} \mathbb{P}(A_{k_1} \cap A_{k_2}),\\ \dots \\ S_r &= \sum_{\mathcal{J}_r}\mathbb{P}(A_{k_1} \cap A_{k_2} \cap \dots \cap A_{k_r}), \end{aligned} where the sum is taken over the unordered subsets $\mathcal{J}_r=[k_1,k_2,\dots,k_r]$ of $\{0,1,2,\dots,n\}$. Let $B_m$, $m \geq 0$, be the event in which exactly $m$ events from $A_1,A_2, \dots, A_n$ occur. Then

1. $\mathbb{P}(B_0) = 1-S_1+S_2-S_3+...+(-1)^n S_n$
2. $\mathbb{P}(B_1) = S_1-2S_2+3S_3-4S_4+...+(-1)^{n-1}n S_n$

I believe these results should follow from inclusion-exclusion principle. For the first part the proof that I found is for sets, not probabilities (https://nptel.ac.in/content/storage2/courses/111104026/lecture17.pdf). There is one question on mathstack (Inclusion Exclusion Probability Proof Using a Partition of the Space) where someone answered by working with sets, but not probabilities. Not sure how the proofs should look like in this case.

I couldn't find proofs for these equations anywhere so maybe there's someone who might have some sources where I could find the proofs? Some help of proving them ourselves would be appreciated as well.

Answer 1

For $[k]\in\mathcal I_m$, let $A([k]):=A_{k_1}\cap\cdots\cap A_{k_m}$. Then by the inclusion-exclusion principle:\begin{align*}\mathbb P(B_m)+\cdots+\mathbb P(B_n)&=\mathbb P\!\left(\bigcup_{[k]\in\mathcal I_m}A([k])\right)\\[.4em]&=\sum_{\emptyset\subsetneq J\subseteq\mathcal I_m}(-1)^{|J|+1}\,\mathbb P\!\left(\bigcap_{[k]\in J}A([k])\right)\\[.4em]&=\sum_{\emptyset\subsetneq J\subseteq\mathcal I_m}(-1)^{|J|+1}\,\mathbb P\!\left(A\Biggl(\bigcup_{[k]\in J}[k]\Biggr)\!\right)\\[.4em]&=\sum_{p=m}^n\sum_{[\ell]\in\mathcal I_p}\left[\sum_{\substack{\emptyset\subsetneq J\subseteq\mathcal I_m\\\bigcup_{[k]\in J}[k]=[\ell]}}(-1)^{|J|+1}\right]\mathbb P\bigl(A([\ell])\bigr).\end{align*} By relabelling, the sum between brackets depends only on $m$ and $p$ (not on $[\ell]$) and equals (again by the inclusion-exclusion principle): \begin{align*} \sum_{\substack{\emptyset\subsetneq J\subseteq\mathcal I_m\\\bigcup_{[k]\in J}[k]=[\ell]}}(-1)^{|J|+1} &=-{\sum_{K\subseteq[\ell]}(-1)^{|K|}\sum_{\substack{\emptyset\subsetneq J\subseteq\mathcal I_m\\\bigcup_{[k]\in J}[k]\subseteq[\ell]\setminus K}}}(-1)^{|J|}\\[.4em] &=-{\sum_{K\subseteq[\ell]}}(-1)^{|K|}\left[(1-1)^{\binom{p-|K|}m}-1\right]\\ &=(1-1)^p-\sum_{r=p-m+1}^p(-1)^r\sum_{\substack{K\subseteq[\ell]\\|K|=r}}1\\ &=0^p-\sum_{r=0}^{m-1}(-1)^{p-r}\binom pr\\[.4em] &=0^p+(-1)^{m+p}\binom{p-1}{m-1}. \end{align*} Therefore \begin{align} \mathbb P(B_m)+\cdots+\mathbb P(B_n)&=\sum_{p=m}^n\left[0^p+(-1)^{m+p}\binom{p-1}{m-1}\right]\sum_{\ell\in\mathcal I_p}\mathbb P\bigl(A([\ell])\bigr)\notag\\[.4em]&=\sum_{p=m}^n\left[0^p+(-1)^{m+p}\binom{p-1}{m-1}\right]S_p.\tag{$E_m$}\end{align} We finally get $\mathbb P(B_m)$ with $(E_m)-(E_{m+1})$: \begin{align*}\mathbb P(B_m)&=\left(0^m+\binom{m-1}{m-1}\right)S_m+\sum_{p=m+1}^n(-1)^{m+p}\left[\binom{p-1}{m-1}+\binom{p-1}m\right]S_p\\[.4em]&=S_m+\sum_{p=m+1}^n(-1)^{m+p}\binom pmS_p\\[.4em]&=\sum_{p=m}^n(-1)^{m+p}\binom pmS_p.\end{align*}