# Conditional probability of a joint independent distribution

• July 31st 2012, 01:08 PM
Salain
Conditional probability of a joint independent distribution
Hi :)

From the definition of independence I know that if A and B are independent $P(A\cap B)=P(A)*P(B)$.

But if I have a conditional probability of a joint probability, such as $P(A\cap B|R)$ and I make the same assumption that A and B are independent, can I now say that $P(A\cap B|R)=P(A|R)*P(B|R)$?

It seems correct to me, but I'm concerned there may be some weird interaction I'm not thinking of.

Thanks!
• July 31st 2012, 02:41 PM
Salain
Re: Conditional probability of a joint independent distribution
For what it's worth, here's as much of a proof as I can work out

$P(A\cap B|R)=\frac{P(A\cap B\cap R)}{P(R)}$ by one of the definitions of a conditional probability. And since by the same definition

$P(A|R)*P(B|R)=\frac{P(A\cap R)*P(B\cap R)}{P(R)*P(R)}$ The only way that

$P(A\cap B|R)=P(A|R)*P(B|R)$ under the assumption that A and B are independent is if

$P(A\cap B\cap R) \stackrel{?}{=}\frac{P(A\cap R)*P(B\cap R)}{P(R)}$ under the same assumption. And you can use the same identity in the opposite direction to reduce the right side to

$P(A\cap B\cap R) \stackrel{?}{=} P(A|R)*P(B\cap R)$ which I'm not sure about :)
• July 31st 2012, 03:33 PM
Salain
Re: Conditional probability of a joint independent distribution
Ok I believe I've got the rest of it, but please check my work since I'm definitely not an expert and it's important that I get the correct result!

Where I left off before,
$P(A\cap B\cap R) \stackrel{?}{=} P(A|R)*P(B\cap R)$ and since $P(A\cap R)=P(R)*P(B|R)$ we can further expand to

$P(A\cap B\cap R) \stackrel{?}{=} P(A|R)*P(B|R)*P(R)$ and finally from the chain rule of probability we know that

$P(A\cap B\cap R) = P(A|B\cap R)*P(B|R)*P(R)$ which by our assumption of independence between A and B reduces to

$P(A\cap B\cap R) = P(A|R)*P(B|R)*P(R)$ , so by the chain rule the two are equivalent and therefore $P(A\cap B|R)=P(A|R)*P(B|R)$ when A and B are independent
• July 31st 2012, 04:45 PM
Plato
Re: Conditional probability of a joint independent distribution
Quote:

Originally Posted by Salain
From the definition of independence I know that if A and B are independent $P(A\cap B)=P(A)*P(B)$.
But if I have a conditional probability of a joint probability, such as $P(A\cap B|R)$ and I make the same assumption that A and B are independent, can I now say that $P(A\cap B|R)=P(A|R)*P(B|R)$?

Consider the set of digits: $\{0,1,2,3,4,5,6,7,8,9\}$.
Select one digit at random. Let $A$ be the event that the digit is prime, $B$ be the event that the digit is less than five, and $R$ be the event that the digit is even.
Are $A~\&~B$ independent? Why or why not?

Now what about $(A\cap R)~\&~(B\cap R)~?$
• July 31st 2012, 07:54 PM
Salain
Re: Conditional probability of a joint independent distribution
So I tried this and ended up getting a slight imbalance in the example which I don't think you intended. There are 3 even numbers and 2 odd numbers below 5, and 3 odd numbers and 2 evens above 5. So, I believe, none of the events A, B or R are independent of any of the others. If I changed the example to be 2..9 and ask <=5, then I get that A (even) and B (<=5) are independent of each other, but not of R (prime). In that case I can replicate the conditions of the original problem and ask if $P(A,B|R)$ is equal to $P(A|R)*P(B|R)$ and by my calculations they're not. The numbers were as close as they could be, but when you've only got 5 numbers which are not prime in a set it's simply not possible to pull any two probabilities from it, multiply, and have the number equal a single probability from it.

That's good to know, but it doesn't show me where the proof went wrong or what the real answer is. Any insights?
• August 1st 2012, 07:43 AM
Salain
Re: Conditional probability of a joint independent distribution
Believe I have it figured out. There are a few equivalent answers

$P(A\cap B|R)=P(R|A\cap B)*P(B|R)$ which doesn't rely on the assumption of independence between A and B, and

$P(A\cap B|R)=P(R|A\cap B)*\frac{p(A)*p(B)}{p(R)}$