Statistical Physics: Proving "if p(a)=p(b)=p then p(ab) ≤ p^2

[rangatudugala]

How to prove this "if p(a)=p(b)=p then p(ab) ≤ p^2

Homework Equations

The Attempt at a Solution

 

[RUber]

Do you know anything about p, a, or b?

 

[RUber]

Please provide some information about what you have already tried or methods you are familiar with so we can point you in the right direction--i.e. fill in the template.

 

[rangatudugala]

Probability of a = probability b = p

 

[RUber]

What does p(ab) mean? Both a and b happen? What if a is b? then p(a and b) = p(a and a) = p.

 

[rangatudugala]

No Ruber

if p(a) = p(b) = p (let say some value) then prove or disprove p(a ∩ b) ≤ p^2

 

[RUber]

So do you know anything about a and b? Assume a = b, then p(a ∩ b) = p(a) = p ≥ p^2 .

 

[rangatudugala]

nothing mention disjoint or not i need to prove this is right or not

 

[rangatudugala]

let say 'a' ≠ b

 

[RUber]

Clearly if a and b were disjoint, the probability of a and b happening together is zero which will surely be less than p.
The key here would be if a and b were independent. If they are, then you might have something to prove...otherwise, you just have 0 ≤ p(a ∩ b) ≤ p.

 

[rangatudugala]

yes i got the point if and be disjoint then p(ab) =0

let say they are not disjoint then how to prove that ?

 

[RUber]

There is nothing to prove unless you know they are independent.

 

[RUber]

If the events a and b are independent, then, by the definition of independence, p(a ∩ b) = p(a) p(b) = p^2.
If they are not independent, then like I said before, they can be anywhere from disjoint to completely coincident, i.e. 0≤p(a ∩ b)≤p.
Is p ≤ p^2?

 

[rangatudugala]

thing is no any hints (information) given in the question.. okay what if not ?

 

[rangatudugala]

hmm if p ≤ p^2

then p= 1 kw i don't think in that way

 

[RUber]

Think of a Venn diagram with two circles representing a and b, both the same size (p). What is the maximum size of the overlapping region?
If no other information is given in the question, then you can assume that anything is possible other than what you know to be true.
If you are to prove the statement, you need to show it holds true all the time. If you are to disprove it, you just need one counterexample.

 

[RUber]

So, if p can be any value between 0 and 1, you have to prove that the statement is true for all values of p, not just p=1.
I don't think you will be able to prove it to be true without more constraints or assumptions.
Can you prove that it is not true?

 

[rangatudugala]

null set

 

[RUber]

If p(a∩b) = 0, then p(a∩b) ≤ p^2.
That is not a good counterexample.
Similarly, if p = 0, then p(a∩b) ≤ p^2. So, that's no good.

 

[rangatudugala]

so you trying to explain that p(null set) = 0 so its not good example is it ?

 

[RUber]

so you trying to explain that p(null set) = 0 so its not good example is it ?

Right.

If p(a) = p(b) and the problem doesn't state that a is not b, then a = b should be your first example.
Look at post 7. Assume 0<p<1 to eliminate the option for p = p^2.

 

[rangatudugala]

Right.

If p(a) = p(b) and the problem doesn't state that a is not b, then a = b should be your first example.
Look at post 7. Assume 0<p<1 to eliminate the option for p = p^2.

okay i think i got the answer

so
1/ if a, b mutually exclusive then p(a∩b) =0

2/ if a,b independent then p(a∩b)= p(a)*p(b) =p^2

is it ?

 

[RUber]

Both of your statements 1/ and 2/ are true, but this is not a proof.
You don't know anything about a and b.
What if a and b are entirely coincident, i.e. if a then b?

 

[rangatudugala]

oh dear you confused me...

 

[Ray Vickson]

How to prove this "if p(a)=p(b)=p then p(ab) ≤ p^2

Homework Equations

The Attempt at a Solution

okay i think i got the answer

so
1/ if a, b mutually exclusive then p(a∩b) =0

2/ if a,b independent then p(a∩b)= p(a)*p(b) =p^2

is it ?

Please use different letters: use $P(a)$ and $P(b)$ for the probabilities of $a$ and $b$, but the letter $p$ for their value; that is, you should say $P(a) = P(b) = p$. That will avoid a lot of confusion.

Both of your examples obey $P(a \cap b) \leq p^2$. But: are you finished? No: you have not proved that $P(a \cap b) \leq p^2$ for all possible cases where $P(a) = P(b) = p$, nor have you discovered a counterexample (that is, an example where $P(a \cap b) > p^2$).

 

[rangatudugala]

Please use different letters: use $P(a)$ and $P(b)$ for the probabilities of $a$ and $b$, but the letter $p$ for their value; that is, you should say $P(a) = P(b) = p$. That will avoid a lot of confusion.

Both of your examples obey $P(a \cap b) \leq p^2$. But: are you finished? No: you have not proved that $P(a \cap b) \leq p^2$ for all possible cases where $P(a) = P(b) = p$, nor have you discovered a counterexample (that is, an example where $P(a \cap b) > p^2$).

okay I'm fail to prove that can you please tell me how to do that ?

 

[Ray Vickson]

okay I'm fail to prove that can you please tell me how to do that ?

No. We are not permitted to solve problems for students---we are allowed to give hints, but nothing more. (Anyway, to be honest, I cannot see how to deal with the problem right now!)

 

[RUber]

a=b satisfies the assumption for you conditional argument. Start there.

 

[rangatudugala]

No. We are not permitted to solve problems for students---we are allowed to give hints, but nothing more. (Anyway, to be honest, I cannot see how to deal with the problem right now!)

Big Thanks .. I also dnt kw that's why i posted it.. its okay.. atleast we tried kw.

 

[rangatudugala]

independent
P(a/b)= P(a) ---> 1
P(b/a)= P(b) ---> 2

also P(a/b) = P(a)* P(a ∩ b) / P(b) ---> 3
P(b/a) = P(b)* P(a ∩ b) / P(a) ---> 4

 

[RUber]

Where are you going with this?
Assuming independence will change the problem.
We already covered that.
You don't know anything about P(a|b), so it can be anywhere from 0 to 1.

 

[rangatudugala]

sorry .. i dnt kw homework to prove this .. there may be a mistake some where or i dnt know nothing

Thanks for your help.. Ill try next prob

 

[RUber]

I have already told you a few times how to prove it.
If it were true that " if P(a) =P(b) =p then P(a ∩ b) ≤ p²" then the relation would have to hold for all values of p and all a and b such that P(a) =P(b) =p.
You have shown that it is true for independent a and b, and for mutually exclusive a and b. I will also add in that it is true for certain p =1.
It is certainly not true for any a and b such that 0<p<1 and p(a|b) > p.

 

[RUber]

independent
also P(a/b) = P(a)* P(a ∩ b) / P(b) ---> 3
P(b/a) = P(b)* P(a ∩ b) / P(a) ---> 4

Are you sure about this?
Your notation P(a/b) means probability of a given b, right?

This implies that P(a/b) = P(a ∩ b). Shouldn't it just be P(a/b) = P(a ∩ b) / P(b)?

 

[rangatudugala]

Are you sure about this?
Your notation P(a/b) means probability of a given b, right?

This implies that P(a/b) = P(a ∩ b). Shouldn't it just be P(a/b) = P(a ∩ b) / P(b)?

P(a/b) means probability of a given b, right? yes its true

how come this possible ?P(a/b) = P(a ∩ b) / P(b)