1. ## Conditional Probability

Hi,

I have a question concerning conditional probabilities. Suppose that P(a, b, c) is a discrete joint distribution. My problem is to express the conditional P(a, c | b) in terms of the conditional P(a | b, c).

My approach:

P(a, c| b) = P(a|b, c) P(c|b)

Is this correct?

Thanks and best wishes,

samosa

2. ## Re: Conditional Probability

Bayes' Theorem.

$P(A_i|A) = \dfrac{P(A_i) P(A|A_i)}{\sum^N_{j=1}P(A_j) P(A|A_j)}$

3. ## Re: Conditional Probability

Hi,

thanks for your help.But I do not find the desired relationship.

Variant I: Bayes Theorem gives

P(a | b, c) = P(b, c | a)P(a)/P(b,c)

How do I get P(a,c | b) into the right hand side?

Variant II: Bayes Theorem gives
P(a, c | b) = P(b|a, c) P(a, c)/P(b)

How do I get P(a | b, c) into the right hand side?

4. ## Re: Conditional Probability

Originally Posted by samosa
Variant I: Bayes Theorem gives
P(a | b, c) = P(b, c | a)P(a)/P(b,c)
How do I get P(a,c | b) into the right hand side?

In your class and/or textbook how are the symbols $P(a,b|c)~\&~P(a|b,c)$ defined?

5. ## Re: Conditional Probability

Originally Posted by samosa
Hi,

thanks for your help.But I do not find the desired relationship.

Variant I: Bayes Theorem gives

P(a | b, c) = P(b, c | a)P(a)/P(b,c)

How do I get P(a,c | b) into the right hand side?

Variant II: Bayes Theorem gives
P(a, c | b) = P(b|a, c) P(a, c)/P(b)

How do I get P(a | b, c) into the right hand side?

$P(X|Y) = \dfrac{P(X\cap Y)}{P(Y)}$

$P(A\cap C|B)= \dfrac{\color{red}P(A\cap B\cap C)}{P(B)}$

$P(A | B\cap C)= \dfrac{P(A\cap B\cap C)}{P(B\cap C)} \implies P(A\cap B\cap C) = \color{red}P(A | B\cap C) P(B\cap C)$

$P(A\cap C|B)= \dfrac{\color{red}P(A | B\cap C) P(B\cap C)}{P(B)}$

6. ## Re: Conditional Probability

Hi Plato,

perhaps this way:

$P(a,b|c) = P(a, b, c)/P(c)$

and

$P(a|b,c) = P(a, b, c)/P(b,c)$

Then we have

$P(a|b,c)P(b,c) = P(a,b|c)P(c)$

giving

$P(a|b,c) = P(a,b|c)P(c)/P(b,c)$

Is this correct?

7. ## Re: Conditional Probability

@abender, thanks a lot. Now I got it.

8. ## Re: Conditional Probability

Originally Posted by samosa
@abender, thanks a lot. Now I got it.