Let $(\Omega,\mathcal{F},\mathbb{P})$ be some probabilistic space and $A_1,\ldots,A_n\in \mathcal{F}$. Is it true that:

$$\sum\limits_{i=1}^{n} \mathbb{P}(A_i)^2 - \sum\limits_{1\le i<j\le n}\mathbb{P}(A_i \cap A_j)^2+\ldots + (-1)^{n+1} \mathbb{P}(A_1 \cap A_2 \cap \ldots\cap A_n)^2\ge 0?$$

When $A_1,\ldots,A_n$ are independent, then we can write left side of this inequality as:

$$1-\prod\limits_{i=1}^{n} (1-\mathbb{P}(A_i)^2)\ge 0.$$

What happens in the general case?


Let $X_r$, for $r=1,2,\ldots,s$, be independent random elements of $\Omega$. We have $\text{Prob}\left(X_r\in E\right)=\mathbb{P}(E)$ for all $E\in\mathcal{F}$. Let $T$ be the event that there exists $i=1,2,\ldots,n$ such that $X_r\in A_i$ for every $r=1,2,\ldots,s$. Clearly, $0 \leq \text{Prob}(T) \leq 1$.

Now, by the Principle of Inclusion and Exclusion, $$\displaystyle\text{Prob}(T)=\sum_{k=1}^n\,(-1)^{k-1}\,\sum_{1\leq i_1<i_2<\ldots<i_k\leq n}\,\text{Prob}\left(X_r\in\textstyle\bigcap_{j=1}^k\,A_{i_j}\text{ for all }r=1,2,\ldots,s\right)\,.$$ Since $X$ and $Y$ are independent, $$\displaystyle\text{Prob}(T)=\sum_{k=1}^n\,(-1)^{k-1}\,\sum_{1\leq i_1<i_2<\ldots<i_k\leq n}\,\prod_{r=1}^s\,\text{Prob}\left(X_r\in\textstyle\bigcap_{j=1}^k\,A_{i_j}\right)\,.$$ Thus, $$\displaystyle\text{Prob}(T)=\sum_{k=1}^n\,(-1)^{k-1}\,\sum_{1\leq i_1<i_2<\ldots<i_k\leq n}\,\prod_{r=1}^s\,\mathbb{P}\left(\textstyle\bigcap_{j=1}^k\,A_{i_j}\right)\,.$$ Consequently, $$\displaystyle\text{Prob}(T)=\sum_{k=1}^n\,(-1)^{k-1}\,\sum_{1\leq i_1<i_2<\ldots<i_k\leq n}\,\Bigg(\mathbb{P}\left(\textstyle\bigcap_{j=1}^k\,A_{i_j}\right)\Bigg)^s\,.$$ This proves that $$0 \leq \sum_{k=1}^n\,(-1)^{k-1}\,\sum_{1\leq i_1<i_2<\ldots<i_k\leq n}\,\Bigg(\mathbb{P}\left(\textstyle\bigcap_{j=1}^k\,A_{i_j}\right)\Bigg)^s \leq 1\,.$$

In fact, if we have $A_i^r\in\mathcal{F}$ for $i=1,2,\ldots,n$ and $r=1,2,\ldots,s$, then $$\displaystyle0\leq\sum_{k=1}^n\,(-1)^{k-1}\,\sum_{1\leq i_1<i_2<\ldots<i_k\leq n}\,\prod_{r=1}^s\,\mathbb{P}\left(\textstyle\bigcap_{j=1}^k\,A^r_{i_j}\right)\leq 1\,.$$

  • $\begingroup$ How are $X_r$ and Prob constructed? $\endgroup$ – simonzack Jul 10 '15 at 23:16
  • $\begingroup$ Simple, equip $\Omega^s$ with the product $\sigma$-algebra on $\Omega$, and define $X_r:\Omega^s\to\Omega$ to be the canonical projection onto the $r$-th coordinate. The probability function $\text{Prob}$ is inherited from the probability measure $\mathbb{P}$ on $\Omega$. $\endgroup$ – Batominovski Jul 10 '15 at 23:28
  • $\begingroup$ Thanks for the explanation! $\endgroup$ – simonzack Jul 11 '15 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.