Conditional probability of a joint independent distribution

Hi :)

From the definition of independence I know that if A and B are independent .

But if I have a conditional probability of a joint probability, such as and I make the same assumption that A and B are independent, can I now say that ?

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

Thanks!

Re: Conditional probability of a joint independent distribution

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,

and since we can further expand to

and finally from the chain rule of probability we know that

which by our assumption of independence between A and B reduces to

, so by the chain rule the two are equivalent and therefore when A and B are independent

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

.

But if I have a conditional probability of a joint probability, such as

and I make the same assumption that A and B are independent, can I now say that

?

Consider the set of digits: .

Select one digit at random. Let be the event that the digit is prime, be the event that the digit is less than five, and be the event that the digit is even.

Are independent? Why or why not?

Now what about

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 is equal to 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?

Re: Conditional probability of a joint independent distribution

Believe I have it figured out. There are a few equivalent answers

which doesn't rely on the assumption of independence between A and B, and