A question related to cardinality and probability

[baiyang11]

Dear all,

I have a question attached related to both probability and cardinality. Let me know if my formulation of the problem is non-rigorous or confusing. Any proof or suggestions are appreciated.Thank you all.
The question follows.Consider a set \(I\) consists of \(N\) incidents.
\[I=\{i_{1},i_{2},...,i_{k},...i_{N}\}\]
Each incident has a probability to happen, i.e. incident \(i_{k}\) happens with the probability \(r_{k}\). Without loss of generality, we assume \(r_{1}\geq r_{2}\geq ... \geq r_{k}\geq ... \geq r_{N}\)
Given a constant \(n<N\), we can have set \(I_{1}=\{i_{1},i_{2},...,i_{n}\}\). Apparently, \(|I_{1}|=n\) and \(I_{1}\subset I\).
Define a mapping \(I\to S\) with \(S=\{s_{1},s_{2},...,s_{k},...s_{N}\}\) subject to
\[
s_{k} = \left\{ \begin{array}{ccc}
1 &\mbox{ (Pr=$r_{k}$)} \\
0 &\mbox{ (Pr=$1-r_{k}$)} \\
\end{array} \right.
\]
Pick out the incidents with correspond \(s\) being 1 to form the set \(I_{2}\) , i.e.
\[I_{2}=\{i_{m_{1}},i_{m_{2}},...,i_{m_{M}}\} \quad \mbox{and} \quad s_{m_{k}}=1 \quad k=1,2,...,M \]
Apparently, \(|I_{2}|=M\) and \(I_{2}\subset I\). Note that there could be \(I_{2}\ne I_{1}\) and \(|I_{2}| \ne |I_{1}|\).

The question is,
If we have two set \(A\) and \(B\) with \(|A|=|B|=n\) and assume
\[ |A \cap I_{1}| \geq |B \cap I_{1}| \]
Is the following statement true?
\[ E(|A \cap I_{2}|) \geq E(|B \cap I_{2}|) \]
where \(E\) means expected value.
If this is true, how to prove it? If not, how to prove it’s not true?

 

[Siron]

I already looked a couple of times at this question and it still confuses me.
First of all the statement itself. I think it has to be stated as
$$\mathbb{E}(A \cap I_2) \geq \mathbb{E}(B \cap I_2)?$$

Second, is there no information about $A$ en $B$ at all? Perhaps, something like $A \subset I$ and $B \subset I$?

 

[baiyang11]

I already looked a couple of times at this question and it still confuses me.
First of all the statement itself. I think it has to be stated as
$$\mathbb{E}(A \cap I_2) \geq \mathbb{E}(B \cap I_2)?$$

Second, is there no information about $A$ en $B$ at all? Perhaps, something like $A \subset I$ and $B \subset I$?

Thank you for reply!
(1) Yes. Here the 'E' character means expected value. I just didn't know how to make it like your character when I wrote this question. Now I know, but I can't edit it.
(2) Yes. $A \subset I$ and $B \subset I$.
Since I can't edit the original post, let me repost the question here.

Consider a set $I$ consists of $N$ incidents.
$I=\{i_{1},i_{2},...,i_{k},...i_{N}\}$
Each incident has a probability to happen, i.e. incident $i_{k}$ happens with the probability $r_{k}$. Without loss of generality, we assume $r_{1}\geq r_{2}\geq ... \geq r_{k}\geq ... \geq r_{N}$
Given a constant $n<N$, we can have set $I_{1}=\{i_{1},i_{2},...,i_{n}\}$. Apparently, $|I_{1}|=n$ and $I_{1}\subset I$.
Define a mapping $I\to S$ with $S=\{s_{1},s_{2},...,s_{k},...s_{N}\}$
subject to
$s_{k} = \left\{ \begin{array}{ccc}
1 &\mbox{ (Pr=$r_{k}$)} \\
0 &\mbox{ (Pr=$1-r_{k}$)} \\
\end{array} \right.$
Pick out the incidents with correspond $s$ being 1 to form the set $I_{2}$ , i.e.

$I_{2}=\{i_{m_{1}},i_{m_{2}},...,i_{m_{M}}\} \quad \mbox{and} \quad s_{m_{k}}=1 \quad k=1,2,...,M$

Apparently, $|I_{2}|=M$ and $I_{2}\subset I$. Note that there could be $I_{2}\ne I_{1}$ and $|I_{2}| \ne |I_{1}|$.

The question is,
If we have two set $A$ and $B$ satisfying the following three assumptions:

(1)$A\subset I$ and $B\subset I$

(2)$|A|=|B|=n$

(3)$|A \cap I_{1}| \geq |B \cap I_{1}|$

Is the following statement true?

$\mathbb{E}(A \cap I_2) \geq \mathbb{E}(B \cap I_2)$

where $\mathbb{E}$ means expected value.
If this is true, how to prove it? If not, how to prove it’s not true?