# Homework Help: Proof concerning the union of a finite collection of events

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1. Jun 24, 2015

1. The problem statement, all variables and given/known data
Prove that

$$P(\cup_{i=1}^n E_i) \geq \max_i P(E_i) (1)$$ for n≥1

2. Relevant equations
I know that $$P(\cup_{i=1}^n E_i) \leq \sum_{i=1}^n P(E_i)$$.
3. 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...

Last edited: Jun 24, 2015
2. Jun 24, 2015

### 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.

3. Jun 24, 2015

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, $\max_{i\in N} P(E_i) \leq P(\cup_{i=1}^n E_i)$

Last edited: Jun 24, 2015
4. Jun 24, 2015

### 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.

5. Jun 24, 2015

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?

6. Jun 24, 2015

### RUber

I don't think you have shown that the max of the two probabilities is less then the union yet.
$max(P(E_1),P(E_2)) \leq P(E_1 \cup E_2)$
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.

7. Jun 24, 2015

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

8. Jun 24, 2015

### RUber

Try a different upper bound.

9. Jun 24, 2015

### 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?

10. Jun 24, 2015

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 $max (P(E_1), P(E_2)) \geq P(E_1 \cap E_2)$?

Last edited: Jun 24, 2015
11. Jun 24, 2015

### 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.

12. Jun 24, 2015

### RUber

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

13. Jun 24, 2015

### RUber

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

14. Jun 24, 2015

### Ray Vickson

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?)

15. Jun 24, 2015

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
I have no idea to be honest.

16. Jun 24, 2015