Proof concerning the union of a finite collection of events

[Portishead]

Homework Statement

Prove that[/B]
$$P(\cup_{i=1}^n E_i) \geq \max_i P(E_i) (1)$$ for n≥1

Homework Equations

I know that $$P(\cup_{i=1}^n E_i) \leq \sum_{i=1}^n P(E_i)$$.

The Attempt at a Solution

I know when n=1, trivially $$P(E_1) \geq \max_1 P(E_1) =P(E_1)$$. So I was hoping I could use induction to show that (1) is true for n≥1. But I don't know if this is the right approach...

 

[RUber]

I think if you can show it for n=2, then for the n+1 case, you can lump all the previous events into one, always having 2 events, like:
$Q= \bigcup_{i=1}^n E_i$ then show that $P(Q \cup E_{n+1}) \geq max(P(Q), P(E_{n+1}))$.
The key will be to how the probability of the union is defined.

 

[Portishead]

When n=2, my conjecture is that $\max_{1,2} P(E_i) \leq P(\cup_{i=1}^2 E_i) \leq P(E_1)+ P(E_2)$ by intuition, but I don't know how to write this formally...

My induction hypothesis is as follows, right?

Then, for all positive integers n, $\max_{i\in N} P(E_i) \leq P(\cup_{i=1}^n E_i)$

 

[RUber]

For *edit* mutually exclusive *edit* events, $P(\cup_{i=1}^2 E_i ) = P(E_1)+P(E_2)$.
Generally, $P(\cup_{i=1}^2 E_i ) = P(E_1)+P(E_2)- P(E_1 \cap E_2)$.
If you are allowed to start with this definition, there aren't too many additional steps to make the proof.
If you are required to derive this, then please tell me what you are allowed to assume.

 

[Portishead]

Thanks for the hint! So when n=2,
$\max_{1,2} P(E_i) \leq P(\cup_{i=1}^2 E_i)$
$=P(E_1)+P(E_2)-P(E_1\cap E_2)$
$\leq P(E_1) + P(E_2)$
Then, assume for when n=k , $\max_{i\in N} P(E_i) \leq P(\cup_{i=1}^k E_i)$
Define $Q= \bigcup_{i=1}^k E_i$ and let n=k+1.
Thus, $\max_{i\in N} P(E_i) \leq P(\cup_{i=1}^{k+1} E_i)$
$=P(Q \cup P(E_{k+1}))$
$=P(Q)+P(E_{k+1}) - P(Q \cap E_{k+1})$
$\leq \sum_{i=1}^{k+1} P(E_i)$

did i commit a logical fallacy?

 

[RUber]

I don't think you have shown that the max of the two probabilities is less then the union yet.
Your claim is that
$max(P(E_1),P(E_2)) \leq P(E_1 \cup E_2)$
You showed that the right side is bounded by the sum, but not that the probability of the union is not less than the max probability of a single event.
For that, you probably need to show that the $P(E_1 \cap E_2)$ term is bounded in some way.

 

[Portishead]

All I can think of is that:
1) $0 \leq P(E_1 \cap E_2) \leq 1$, by definition...

 

[RUber]

Try a different upper bound.

 

[RUber]

The question is what is the probability that both events occur.
If $P(E_1) = 0$ and $P(E_2) = 1$, what is $P(E_1 \cap E_2)$?
What can you learn from this?

 

[Portishead]

The question is what is the probability that both events occur.
If $P(E_1) = 0$ and $P(E_2) = 1$, what is $P(E_1 \cap E_2)$?
What can you learn from this?

It's intuitively zero and E1 and E2 are mutually exclusive/disjoint events?
If either $P(E_1)$ or $P(E_2)$ were the maximum, intuitively wouldn't that be less than or equal to the sum of both probabilities since $P(E_i) \geq 0$? I just don't know how to formally show my conjecture.
Is it necessarily true that $max (P(E_1), P(E_2)) \geq P(E_1 \cap E_2)$?

 

[RUber]

If one has a probability of zero, then the probability of both happening cannot be more than zero.
The intersection, or probability that both events occur cannot be more the the lowest probability.

 

[RUber]

Remember, you are trying to show that $P(E_1)+P(E_2) - P(E_1 \cap E_2) \geq \max(P(E_1),P(E,2))$.

 

[RUber]

Also, if you have 2 events, you can simply call one P_max and one P_min. This might make it more clear.

 

[Ray Vickson]

Homework Statement

Prove that[/B]
$$P(\cup_{i=1}^n E_i) \geq \max_i P(E_i) (1)$$ for n≥1

Homework Equations

I know that $$P(\cup_{i=1}^n E_i) \leq \sum_{i=1}^n P(E_i)$$.

The Attempt at a Solution

I know when n=1, trivially $$P(E_1) \geq \max_1 P(E_1) =P(E_1)$$. So I was hoping I could use induction to show that (1) is true for n≥1. But I don't know if this is the right approach...

It might help to just re-write the result: if $F = \cup_{i=1}^n E_i$ then you are being asked to show that
$$P(E_1 \leq P(F), \, P(E_2) \leq P(F), \, \ldots, \, P(E_n) \leq P(F)$$
This statement is logically equivalent to the one involving $\max_i P(E_i)$. (Why?)

 

[Portishead]

I'm using these lecture notes as a guide, just so you can know what assumptions were made for this problem:
http://ocw.jhsph.edu/courses/MethodsInBiostatisticsI/PDFs/lecture1.pdf
http://ocw.jhsph.edu/courses/MethodsInBiostatisticsI/PDFs/lecture2.pdf

It might help to just re-write the result: if $F = \cup_{i=1}^n E_i$ then you are being asked to show that
$$P(E_1 \leq P(F), \, P(E_2) \leq P(F), \, \ldots, \, P(E_n) \leq P(F)$$
This statement is logically equivalent to the one involving $\max_i P(E_i)$. (Why?)

I have no idea to be honest.

 

[Portishead]

I posted this post by mistake, please delete this post.