# Homework Help: Statistical Physics

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1. Aug 14, 2015

### rangatudugala

• New member warned about posting with no effort shown
How to prove this "if p(a)=p(b)=p then p(ab) ≤ p^2

2. Relevant equations

3. The attempt at a solution

2. Aug 14, 2015

### RUber

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

3. Aug 14, 2015

### 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.

4. Aug 14, 2015

### rangatudugala

Probability of a = probability b = p

5. Aug 14, 2015

### 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.

6. Aug 14, 2015

### rangatudugala

No Ruber

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

7. Aug 14, 2015

### RUber

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

8. Aug 14, 2015

### rangatudugala

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

9. Aug 14, 2015

### rangatudugala

10. Aug 14, 2015

### 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.

11. Aug 14, 2015

### 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 ?

12. Aug 14, 2015

### RUber

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

13. Aug 14, 2015

### 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?

14. Aug 14, 2015

### rangatudugala

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

15. Aug 14, 2015

### rangatudugala

hmm if p ≤ p^2

then p= 1 kw i dont think in that way

16. Aug 14, 2015

### 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.

17. Aug 14, 2015

### 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?

18. Aug 14, 2015

### rangatudugala

null set

19. Aug 14, 2015

### 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.

20. Aug 14, 2015

### rangatudugala

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

21. Aug 14, 2015

### RUber

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.

22. Aug 14, 2015

### rangatudugala

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 ?

23. Aug 14, 2015

### 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?

24. Aug 14, 2015

### rangatudugala

oh dear you confused me...

Last edited by a moderator: Aug 15, 2015
25. Aug 14, 2015

### Ray Vickson

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 $P(a \cap b) > p^2$).