Proof related to the expected value of cardinality

[baiyang11]

Consider [tex]N[/tex] random variables [tex]X_{n}[/tex] each following a Bernoulli distribution [tex]B(r_{n})[/tex] with [tex]1 \geq r_{1} \geq r_{2} \geq ... \geq r_{N} \geq 0[/tex]. If we make following assumptions of sets [tex]A[/tex] and [tex]B[/tex]:

(1) [tex]A \subset I [/tex] and [tex]B \subset I[/tex] with [tex]I=\{1,2,3,...,N\}[/tex]

(2) [tex]|A \cap I_{1}| \geq |B \cap I_{1}|[/tex] with [tex]I_{1}=\{1,2,3,...,n\}, n<N[/tex]

(3) [tex]|A|=|B|=n[/tex]

Do we have [tex]\mathbb{E}(\Sigma_{a\in A} X_{a}) \geq\mathbb{E}(\Sigma_{b\in B} X_{b})[/tex]?

To avoid confusion, [tex]\mathbb{E}[/tex] means expected value.Thanks!

 

[Siron]

This question looks really similar to your previous one so I hope I can be of more help this time. I'll give my thoughts about this question.

We have given $N$ random variables with a Bernoulli distribution. That means $\mathbb{P}[X_n = 1] = r_n$ and $\mathbb{P}[X_n = 0]=1-r_n$. Furthermore, $\mathbb{E}[X_i]=r_i$ for $i=1,\ldots,N$.

By condition $(2) |A \cap I_1| \geq |B \cap I_1| \neq 0 $ we know that $A$ has more elements in common with $I_1$. Since the random variables $X_1,\ldots,X_n$ have greater expected values then the random variables $X_j$ with $j \in I \backslash I_1$, the statement
$$\mathbb{E}\left(\sum_{a \in A} X_a \right) \geq \mathbb{E}\left(\sum_{b \in B} X_b \right)$$
probably holds. Of course this is not a formal proof but it should give some intuition.

The only thing we know about $n$ and $N$ is $n<N$ so it could be the case that $N>2n$. In this case we can suppose $A \cap I_1 = \emptyset$ and $B \cap I_1 = \emptyset$. Now it is possible to choose $A$ en $B$ such that the statement does not hold anymore.

To make my point more clear. Suppose $I = \{1,2,\ldots,7\}$ and $I_1 = \{1,2,3\}$, thus $n=3$ and $N = 7$. Set $A = \{5,6,7\}$ and $B=\{4,5,6\}$. Then we have
$$\mathbb{E}\left(\sum_{a \in A} X_a \right) = \mathbb{E}[X_5]+\mathbb{E}[X_6]+\mathbb{E}[X_7] = r_5+r_6+r_7$$
$$\mathbb{E}\left(\sum_{b \in B} X_b \right) = r_4+r_5+r_6$$.

Since we can say $r_4 > r_7$, we have in this case that the statement does not hold.

I hope I made my point clear though I did not give a formal explanation. But I wanted to give my thoughts because in my opinion this is an interesting question.

 

[baiyang11]

This question looks really similar to your previous one so I hope I can be of more help this time. I'll give my thoughts about this question.

We have given $N$ random variables with a Bernoulli distribution. That means $\mathbb{P}[X_n = 1] = r_n$ and $\mathbb{P}[X_n = 0]=1-r_n$. Furthermore, $\mathbb{E}[X_i]=r_i$ for $i=1,\ldots,N$.

By condition $(2) |A \cap I_1| \geq |B \cap I_1| \neq 0 $ we know that $A$ has more elements in common with $I_1$. Since the random variables $X_1,\ldots,X_n$ have greater expected values then the random variables $X_j$ with $j \in I \backslash I_1$, the statement
$$\mathbb{E}\left(\sum_{a \in A} X_a \right) \geq \mathbb{E}\left(\sum_{b \in B} X_b \right)$$
probably holds. Of course this is not a formal proof but it should give some intuition.

The only thing we know about $n$ and $N$ is $n<N$ so it could be the case that $N>2n$. In this case we can suppose $A \cap I_1 = \emptyset$ and $B \cap I_1 = \emptyset$. Now it is possible to choose $A$ en $B$ such that the statement does not hold anymore.

To make my point more clear. Suppose $I = \{1,2,\ldots,7\}$ and $I_1 = \{1,2,3\}$, thus $n=3$ and $N = 7$. Set $A = \{5,6,7\}$ and $B=\{4,5,6\}$. Then we have
$$\mathbb{E}\left(\sum_{a \in A} X_a \right) = \mathbb{E}[X_5]+\mathbb{E}[X_6]+\mathbb{E}[X_7] = r_5+r_6+r_7$$
$$\mathbb{E}\left(\sum_{b \in B} X_b \right) = r_4+r_5+r_6$$.

Since we can say $r_4 > r_7$, we have in this case that the statement does not hold.

I hope I made my point clear though I did not give a formal explanation. But I wanted to give my thoughts because in my opinion this is an interesting question.

Thank you for reply.
You provide a clear example when $A$ and $B$ have nothing in common with $I_{1}$ at all. But I think my original intention is to make $|A \cap I_{1}| \geq |B \cap I_{1}|>0$. I admit it is my mistake not to make this clear.
So if $|A \cap I_{1}| \geq |B \cap I_{1}|>0$, is the answer "YES" or "NO"?

It may be still No. Someone else gives me the following example.
Let $n \geq 2$, $N=n+1, A=I-\{1\}, B=I-\{2\}$ and $r_{1}>r_{2}$. Then
$$\mathbb{E}\left(\sum_{a \in A} X_a \right)=\sum_{i} r_{i}-r_{1} $$
$$\mathbb{E}\left(\sum_{b \in B} X_b \right)=\sum_{i} r_{i}-r_{2} $$
With $r_{1}>r_{2}$, we get $\mathbb{E}\left(\sum_{a \in A} X_a \right)<\mathbb{E}\left(\sum_{b \in B} X_b \right)$ to make the answer "NO".

In fact, by asking my original question, I what to abstract the following statement which seems to be true intuitively.

If I have better knowledge of what incidents have the top probabilities to happen (like I know set A rather than set B), is the knowledge trustable (A is still favored) when those incidents realize (the expected value thing)?

Now, inspired by the counterexample above, it seems that we can always manipulate different $r_{i}$ to get $B$ favored with $|A|=|B|$.

Now that under the original assumptions $\mathbb{E}\left(\sum_{a \in A} X_a \right) \geq \mathbb{E}\left(\sum_{b \in B} X_b \right)$ does not always hold, I am trying to see under what assumptions, we can always favor A, i.e. $\mathbb{E}\left(\sum_{a \in A} X_a \right) \geq \mathbb{E}\left(\sum_{b \in B} X_b \right)$. But I am a bit lost.

 

[Siron]

Thank you for reply.
You provide a clear example when $A$ and $B$ have nothing in common with $I_{1}$ at all. But I think my original intention is to make $|A \cap I_{1}| \geq |B \cap I_{1}|>0$. I admit it is my mistake not to make this clear.
So if $|A \cap I_{1}| \geq |B \cap I_{1}|>0$, is the answer "YES" or "NO"?

It may be still No. Someone else gives me the following example.
Let $n \geq 2$, $N=n+1, A=I-\{1\}, B=I-\{2\}$ and $r_{1}>r_{2}$. Then
$$\mathbb{E}\left(\sum_{a \in A} X_a \right)=\sum_{i} r_{i}-r_{1} $$
$$\mathbb{E}\left(\sum_{b \in B} X_b \right)=\sum_{i} r_{i}-r_{2} $$
With $r_{1}>r_{2}$, we get $\mathbb{E}\left(\sum_{a \in A} X_a \right)<\mathbb{E}\left(\sum_{b \in B} X_b \right)$ to make the answer "NO".

In fact, by asking my original question, I what to abstract the following statement which seems to be true intuitively.

If I have better knowledge of what incidents have the top probabilities to happen (like I know set A rather than set B), is the knowledge trustable (A is still favored) when those incidents realize (the expected value thing)?

Now, inspired by the counterexample above, it seems that we can always manipulate different $r_{i}$ to get $B$ favored with $|A|=|B|$.

Now that under the original assumptions $\mathbb{E}\left(\sum_{a \in A} X_a \right) \geq \mathbb{E}\left(\sum_{b \in B} X_b \right)$ does not always hold, I am trying to see under what assumptions, we can always favor A, i.e. $\mathbb{E}\left(\sum_{a \in A} X_a \right) \geq \mathbb{E}\left(\sum_{b \in B} X_b \right)$. But I am a bit lost.

Nice counterexample! I did not see it that way. Anyway your question is interesting but not something I can answer right away. I'll think about it but in the mean time I hope some other members can help you out.

Ps: where does this problem come from? I have had some probability theory courses but I've never discussed problems like this before.

 

[baiyang11]

Nice counterexample! I did not see it that way. Anyway your question is interesting but not something I can answer right away. I'll think about it but in the mean time I hope some other members can help you out.

Ps: where does this problem come from? I have had some probability theory courses but I've never discussed problems like this before.

I appreciate your continuous attention on this problem and I do hope attentions from some other members, maybe from various areas, because it seems to be interdisciplinary between set theory and probability theory.

This abstracted problem comes from a concrete problem in my major, which is electrical engineering (EE). We come up with a concrete method for a EE problem we are researching currently. Then I realize the essence of the concrete method is trying to answer the question, the italic one in my previous post. Then I looked at the EE problem and what we do in the concrete method, and come up with this mathematical problem, which I found interesting as well.