Probability: Intersection and Union of sets

Question

I want to show the following statement,

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Let $E_1,\dots , E_n \in \mathcal{F}$. Let, $$\mathcal{C}=\left\{F \in \mathcal{F}: F = \cap_{i=1}^{n}F_i, F_i \in \left\{E_i, E_i^c\right\}, i = 1, . . . , n\right\}$$ For all $1\leq i_1 < \dots < i_k\leq n$ , where $k = 2, \dots , n − 1$, show that $\cap_{i=1}^{k}E_{i_j} = \cup\left\{F ∈ \mathcal{C} : F \subset \cap_{j=1}^{k}E_{i_j}\right\}$.

I don't see how to visualize the following set $\cup\left\{F ∈ \mathcal{C} : F \subset \cap_{j=1}^{k}E_{i_j}\right\}$.

Any ideas or leads?

Thanks :)

Answer 1

You can think of the $E_1,\dots,E_n$ as some "basic"-events. Then $E_1^c$ means $E_1$ doesn't happen, $E_1 \cap E_2$ means $E_1$ and $E_2$ happen; $E_1 \cup E_2$ means $E_1$ or $E_2$ happen.

Hence, $\mathcal{C}$ is the collection of events, where you have an oppinion on every basic-evenent $E_1, \dots, E_n$ -- whether it happend or not. Whereas in a set of the type $\cap_{i=1}^{k_j}E_{i_j}$ you know that the $E_{i_j}$ happend, but there may be some basic-event, where you don't have know.

The set $\left\{F ∈ \mathcal{C} : F \subset \cap_{i=1}^{k_j}E_{i_j}\right\}$ is the collection of events, where you know about every basic-event (they are in $\mathcal{C}$) and which are compatible with $\cap_{i=1}^{k_j}E_{i_j}$ (this is what the inclusion says).

The equality $$\cap_{i=1}^{k_j}E_{i_j} = \cup\left\{F ∈ \mathcal{C} : F \subset \cap_{i=1}^{k_j}E_{i_j}\right\}$$ basically says that you can write the event $\cap_{i=1}^{k_j}E_{i_j}$ as the union of the "compatible" basic-events. It's just an elementary set-theory/logic fact, where you don't need any measure- or probabilty theory for the proof. The direction $\supset$ should be clear. For the reverse inequality, I give you an example, which should make clear how to prove this:

Let $n=3$. Then $E_1 \cap E_3$ is not in $\mathcal{C}$. But $E_1 \cap E_2 \cap E_3$ and $E_1 \cap E_2^c \cap E_3$ are both in $\mathcal{C}$ and you have $E_1 \cap E_3 = (E_1 \cap E_2 \cap E_3) \cup (E_1 \cap E_2^c \cap E_3)$, which gives you the equality in that example.

The full proof is exactly the same idea, but with somewhat more notation. Feel free to ask if you get stuck there :)

Edit: The crucial point is the following. Let $\ell_1,\dots,\ell_p$ the remaing indezies, i.e. $\{1,\dots,n\}$ is the disjoint union of $\{i_1,\dots, i_k\}$ and $\{\ell_1,\dots,\ell_p\}$. Then $$ \cap_{j=1}^{k}E_{i_j} = \cup\left\{\bigcap_{j=1}^{k}E_{i_j} \cap \bigcap_{m=1}^p F_{\ell_m}: F_{\ell_m} \in \{ E_{\ell_m}, E_{\ell_m}^c \} \right\} $$ and $$\left\{\bigcap_{j=1}^{k}E_{i_j} \cap \bigcap_{m=1}^p F_{\ell_m}: F_{\ell_m} \in \{ E_{\ell_m}, E_{\ell_m}^c \} \right\} \subset \left\{F \in \mathcal{C} : F \subset \cap_{i=1}^{k_j}E_{i_j} \right\} $$

Answer 2

I will give an answer based on @user940347 idea.

Let $I=\left\{1,\dots,n\right\}$, $I_k=\left\{ i_1,\dots,i_k\right\}$ and $\tilde{I}_k=I\setminus I_k$. We have the following, \begin{align*} \cap_{j=1}^kE_{i_j}&=\cap_{i\in I_k}E_i\\ &= \bigcup\left(\cap_{i\in I_k}E_i\cap_{j\in\tilde{I}_k} \tilde{F}_j\right) \quad \text{where $\tilde{F}_j\in\left\{E_j,E_j^c\right\}$} \end{align*}

We have $\left(\cap_{i\in I_k}E_i\cap_{j\in\tilde{I}_k}\tilde{F}_j\right)\in\mathcal{C}$ by definition of $\mathcal{C}$ and of course $\left(\cap_{i\in I_k}E_i\cap_{j\in\tilde{I}_k}\tilde{F}_j\right)\in \cap_{i\in I_k}E_i$ by definition of intersection.

Therefore the $\subset$ is proved.