Probability of Independent Events: showing abstractly that two events are independent

Feb 2013
2
0
Ohio
I haven't had a proofs class, so proofs always seem so complex to me. Here's the question: If A,B, and C are mutually independent, show that the following pairs of events are independent: A' and (B and C') (I understood the other ones). Also show that A', B', and C' are mutually independent.


Here's as far as I could get for the first part:

P(A') and P(B and C') = P(A') P(B) P(C')
1-p(a) and p(b) and (1-p(c))= p(a' and c') and p(b)
p(b)-p(a)p(b)(1-p(c))=(paUc)' and p(b)=
p(b)-p(b)p(c)-p(a)p(b)+p(a)p(b)p(c)=


For the second part I proved that each pair of a,b,c was independent, but I can't get very far with all three mutually independent. I know that p(a' and b' and c')= p(a')p(b')p(c') and then I multiplied out the first part taking the compliment of each as (1-p(x)) and multiplying the three.

I would really appreciate and help with this topic a lot! Thanks!
 
Aug 2006
22,432
8,611
Re: Probability of Independent Events: showing abstractly that two events are indepen

I haven't had a proofs class, so proofs always seem so complex to me. Here's the question: If A,B, and C are mutually independent, show that the following pairs of events are independent: A' and (B and C') (I understood the other ones). Also show that A', B', and C' are mutually independent.

There are major problems with definitions in this area of probability.
Here is the standard quote:
"If \(\displaystyle A_1,\cdots,A_n\) are mutually independent events then every event determined by a subcollection of these events is independent of every event determined by a subcollection of the remaining events, Jim Pitman.

If we use that definition, it seems to solve this question.
Have you proved that: If \(\displaystyle A~\&~B\) are independent events then \(\displaystyle A~\&~B'\) are independent events?
Here is the prove of that.
We know that \(\displaystyle \mathcal{P}(A)=\mathcal{P}(A\cap B)+\mathcal{P}(A\cap B')\).

So
\(\displaystyle \begin{align*} \mathcal{P}(A) \mathcal{P}(B^{\,\prime}) &= \mathcal{P}(A)[1- \mathcal{P}(B)]\\ &= \mathcal{P}(A)- \mathcal{P}(A) \mathcal{P}(B) \\ &=\mathcal{P}(A)- \mathcal{P}(A\cap B) \\ &=\mathcal{P}(A\cap B^{\,\prime}) \end{align*}\).

Now you should be able to finish.
 
Dec 2012
1,145
502
Athens, OH, USA
Re: Probability of Independent Events: showing abstractly that two events are indepen

Plato's response basically answers your question, but I thought I'd expand on it a "little". Here's my expansion:

MHFprobabilityindependencea.png

MHFprobabilityindependenceb.png

MHFprobabilityindependencec.png

If you bother to read the 3rd page, you'll see that Plato really completely answered the question.
 
Feb 2013
2
0
Ohio
Re: Probability of Independent Events: showing abstractly that two events are indepen

I'm sorry, I'm not sure I understand what you're talking about. The proof you gave makes sense @Plato, but I don't understand how it fits into the proof I've been given to work. Thanks for the input, though. I really appreciate your quick response!

-Lily
 
Dec 2012
1,145
502
Athens, OH, USA
Re: Probability of Independent Events: showing abstractly that two events are indepen

To prove A', B' and C' are mutually independent, you have to show:
1. The probability of the intersection of any two of them is the product of the probabilities.
2. The probability of the intersection of all three is the the product of the probabilities.

1 is covered in 3 b) of the 1st page of my previous response
2 is covered in 4 c) of the 2nd page.

Finally, the theorem on page 3 says in particular, if A, B, C, D, etc are mutually independent, then A', B', C', D' are mutually independent.