A statistics conjecture: so simple. but i donno how to do.

[tea217]

Conjecture: Let A, B, and C by any three events, with P(A) > 0, P(B) > 0, and P(A&B) > 0.
If P(C|A) > P(C) and P(C|B) > P(C), then P(C|A&B) > P(C).

Prove it either it's true OR give a counterexample~! (A&B) means "A intersect B"

(Is me again, the guy with another stats question. These are the questions i donno how to do.)

I'm like 99% sure this conjecture is true man. I tried to find counterexamples, but could'nt.
But i tried so hard proving it too, but couldn't. Life is so hard~

 

[KLin]

if you think life is hard because you don't get statistics, you're in for a rude awakening.

 

[tea217]

I don't even want to take stats... But..i had to ~`
SO many assignments. so little time.

 

[Zenmervolt]

The statement can be false if P(A&amp;B)<P(A) and P(A&amp;B)<P(B)

ZV

 

[tea217]

So are you saying this conjecture is false?? not can be false.

Given the question... If is false,. can you find me an example?? (Like some dice rolling, or coin tossing, etc.)
B/C i don't think it can be false,..

 

[TuxDave]

I'm sure it's false... coming up with a counter example.

 

[TuxDave]

Bah... can't get a story-book counterexample, but just draw a venn diagram with A and B intersecting and C existing everywhere in A and B except where A &amp; B intersect. Now go make up some numbers and QED.

 

[Gunslinger08]

False when all elements that are in both A&amp;B are also in C.

 

[glugglug]

Its false.

In particular, your conjecture fails whenever C < A n B

(< = is a subset of, couldn't find real subset symbol)

 

[tea217]

Hmm.. But you guys say is False,..when C is a subset of A n B (A n B means A intersect B)
Consider this example:

* * * ----> 1 2 3
* * * ----> 4 5 6
* * * ----> 7 8 9
We got a bunch of stars,.and lets just number the stars like that for convenience of refering to them

If A = stars 123456 and B = 456789. then P(A) = 6/9 and P(B) = 6/9
You guys said C is a subset of A n B fails the conjecture. So let's decide C to be just star 5. so P(C) = 1/9

Conjecture:
Let A, B, and C by any three events, with P(A) > 0, P(B) > 0, and P(A n B) > 0.
If P(C|A) > P(C) and P(C|B) > P(C), then P(C|A n B) > P(C).

So P(C|A) = 1/6 > P(C) = 1/9
And P(C|B) = 1/6 > P(C) = 1/9
BUT P(C|A n B) = 1/3 > P(C) = 1/9
My conjecture is still true when C is a subset of A n B.

Did i calculated my P(C|A n B) incorrectly? If so, what's the right answer

 

[glugglug]

Oops I got it backwards. Its A n B is a subset of C when it fails.

So if C in your example is instead {4,5,6}, P(C|A) = 2/3, P(C|B) = 2/3, P(C) = 1/3 and P(C|A n B) = 2/3

 

[tea217]

i don't get how you find P(C|A n B) = 2/3 ??
I thought P(A n B) = 3/9 = 1/3 ...and... P(A n B n C) = 3/9 = 1/3
so P(C|A n B) = P(A n B n C) / P(A n B) = 1

But let's assume you got P(C|A n B) = 2/3 is correct
still.. is greater P(C) = 1/3 , so my conjecture still is true.

 

[TuxDave]

Here's your counter example.

You roll a 9 sided dice.

P(A) = You rolled either a 1,2,3,4
P(B) = You rolled either a 4,5,6,7
P(C) = You rolled a 1,2,3,5,6,7

P(C) = 6/9=2/3
P(C|A) = 3/4
P(C|B) = 3/4

P(C|A&amp;B) = 0

 

[tea217]

Just a little confused here "You roll an 8,9 sided dice"
is it 8 or 9??

I see that P(C) = 6/9, which implies the dice is 9sided.
so the dice is 9sided??

But yes, your example look very right. All b/c P(A&amp;B&amp;C) = 0.


THank you TuxDave.
Before you said "just draw a venn diagram with A and B intersecting and C existing everywhere in A and B except where A &amp; B intersect"....i couldn't really get it.
But the example makes it so clear now....!!!

Appreciated..

 

[zerocool1]

what does '|' mean can't remember the notation

 

[chuckywang]

Originally posted by: zerocool1
what does '|' mean can't remember the notation

it means "given that"

it's used for conditional probability.

so P(A|B) is probability of A given that B is true.

 

[TuxDave]

Originally posted by: tea217
Just a little confused here "You roll an 8,9 sided dice"
is it 8 or 9??

I see that P(C) = 6/9, which implies the dice is 9sided.
so the dice is 9sided??

But yes, your example look very right. All b/c P(A&amp;B&amp;C) = 0.


THank you TuxDave.
Before you said "just draw a venn diagram with A and B intersecting and C existing everywhere in A and B except where A &amp; B intersect"....i couldn't really get it.
But the example makes it so clear now....!!!

Appreciated..

No prob... sorry, about the confusion, I meant 9 sided dice. You got it right.

 

[glugglug]

Originally posted by: tea217
i don't get how you find P(C|A n B) = 2/3 ??
I thought P(A n B) = 3/9 = 1/3 ...and... P(A n B n C) = 3/9 = 1/3
so P(C|A n B) = P(A n B n C) / P(A n B) = 1

But let's assume you got P(C|A n B) = 2/3 is correct
still.. is greater P(C) = 1/3 , so my conjecture still is true.

Wow, I keep goofing up.

'|' is given that? Was thinking Union, with the &amp; symbol being used for intersection.

 

[tea217]

Thank you for all your help guys. I guess this thread ends here.

But in my other thread...http://forums.anandtech.com/me...418498&amp;STARTPAGE=1
is my other stats question..

 

[Howard]

Originally posted by: glugglug

Originally posted by: tea217
i don't get how you find P(C|A n B) = 2/3 ??
I thought P(A n B) = 3/9 = 1/3 ...and... P(A n B n C) = 3/9 = 1/3
so P(C|A n B) = P(A n B n C) / P(A n B) = 1

But let's assume you got P(C|A n B) = 2/3 is correct
still.. is greater P(C) = 1/3 , so my conjecture still is true.

Wow, I keep goofing up.

'|' is given that? Was thinking Union, with the &amp; symbol being used for intersection.

I've always expressed domain and range like this: {x E R| x > 0 } or something like that.

 

[Skiguy411]

Originally posted by: TuxDave
Now go make up some numbers and QED.

It has been shown! Woo! Gotta love that latin......