Prove/Disprove: p(a∩b) ≤ q^2 with a,b Independant

[rangatudugala]

Prove or disprove the following statement:

If p(a)=p(b)=q then p(a∩b)≤q2

We know nothing know about event a , b.

The Attempt at a Solution

I tried this but don't know correct or not
Can some one help me
let a, b are independent event
0<q<1
then p(a∩b) = p(a) p(b) = q*q = q^2

[/B]

 

[DEvens]

Possibly reading this discussion of Bayes' theorem will help.

https://en.wikipedia.org/wiki/Bayes'_theorem

 

[RUber]

Did you give up on the other thread? https://www.physicsforums.com/threads/statistical-physics.827663/
I suggest you disprove the statement.
If a = b, then P(a) = P(b) =q, so your condition is met.
What is P(a∩b)?
Is P(a∩b) ≤ q² for all possible values of q?

 

[rangatudugala]

Did you give up on the other thread? https://www.physicsforums.com/threads/statistical-physics.827663/
I suggest you disprove the statement.
If a = b, then P(a) = P(b) =q, so your condition is met.
What is P(a∩b)?
Is P(a∩b) ≤ q² for all possible values of q?

previous thread I used confusing notation (It was my 1st attempt )

 

[rangatudugala]

previous thread I used confusing notation (It was my 1st attempt )

I tried but no idea feel something missing
let a, b are independent event
0<q<1
then p(a∩b) = p(a) p(b) = q*q = q^2

 

[RUber]

You won't prove that the statement is true or untrue by assuming that a and b are independent.
If a and b are independent, then the statement is true.
If a and b are not independent, you still don't know.
You either have to prove that
for all a and b such that p(a) = p(b) = q the statement is true
or
there exists at least one pair of a and b such that p(a) = p(b) = q AND the statement is false.
Let a = b and q = .5.
What do you find out?

 

[rangatudugala]

You won't prove that the statement is true or untrue by assuming that a and b are independent.
If a and b are independent, then the statement is true.
If a and b are not independent, you still don't know.
You either have to prove that
for all a and b such that p(a) = p(b) = q the statement is true
or
there exists at least one pair of a and b such that p(a) = p(b) = q AND the statement is false.
Let a = b and q = .5.
What do you find out?

Okay..
then p(a∩b) = p(a) p(b) =0.5*0.5 = 0.25

 

[rangatudugala]

Okay..
then p(a∩b) = p(a) p(b) =0.5*0.5 = 0.25

is there any other answer ??

 

[haruspex]

Okay..
then p(a∩b) = p(a) p(b) =0.5*0.5 = 0.25

No, that only works if they are independent. As RUber says, you need to think about two events that have some dependency. What's the most dependent you can make two events? (When two events have some dependency, knowing the outcome of one gives you information about the likely outcome of the other.)

 

[rangatudugala]

p(a∩b) = p(a) + p(b) - p(a∪b)Then how to get p(a∪b)??

 

[RUber]

Okay..
then p(a∩b) = p(a) p(b) =0.5*0.5 = 0.25

What is p(a ∩ a)?

 

[rangatudugala]

What is p(a ∩ a)?

What is p(a ∩ a)?

p(a ∩ a) = 0.5
:?

 

[rangatudugala]

p(a ∩ a) = 0.5
:?

isnt it?

 

[RUber]

Right. Is that less than or equal to q^2?
Is p(a)=p(a) =q?

 

[rangatudugala]

Right. Is that less than or equal to q^2?
Is p(a)=p(a) =q?

its greater than q^2

 

[rangatudugala]

its greater than q^2

p(a)=p(a) =q yes

!

 

[RUber]

So then...?

 

[rangatudugala]

So then...?

1/ let a, b are independent event
0<q<1
then p(a∩b) = p(a) p(b) = q*q = q^2

2/ a, b not independent event
0<q<1

then p(a∩a) = q >= q^2but why we going to calculate p(a∩a)??

question was p(a∩b) =< q^2 know i confused again.. Am I missed step ?

 

[RUber]

Okay, then let a and b be two separate events, that are highly correlated.
Let a = Is a student
Let b = Goes to school
Essentially, these are the same thing as saying a = b.

The reason we are doing this is because nothing in the statement you are trying to prove or disprove says we can't. If the statement is true, it must hold for all choices of a and b, including a = b. So if it doesn't hold for a = b, then it is disproved.
To disprove the statement, you don't have to show that it is always wrong, just that it could be wrong.
The fact is that the statement is true only when a and b are either independent or less likely to occur together than on their own.

So, like I have been saying. Use a = b and 0<q<1. Show that the example fits the assumptions for your statement (if ... ), then show that the outcome is false (then...) . This is how you disprove an if, then statement.

 

[rangatudugala]

Okay, then let a and b be two separate events, that are highly correlated.
Let a = Is a student
Let b = Goes to school
Essentially, these are the same thing as saying a = b.

The reason we are doing this is because nothing in the statement you are trying to prove or disprove says we can't. If the statement is true, it must hold for all choices of a and b, including a = b. So if it doesn't hold for a = b, then it is disproved.
To disprove the statement, you don't have to show that it is always wrong, just that it could be wrong.
The fact is that the statement is true only when a and b are either independent or less likely to occur together than on their own.

So, like I have been saying. Use a = b and 0<q<1. Show that the example fits the assumptions for your statement (if ... ), then show that the outcome is false (then...) . This is how you disprove an if, then statement.

Big Thanks Sir...
Now I got the Point
we don't know anything about event a & b so we can do both prove & disprove using "if statements.

Thank you sir again.
Big sorry if I bothered to you so much.. :)

 

[RUber]

You can only disprove the statement as written.
You can show that in some cases, p(a) = p(b)=q and p(a ∩ b) ≤ q^2, but you can also show that in some cases p(a) = p(b)=q and p(a ∩ b) > q^2.
That means that the statement as a whole is proved false by counterexample.