# Conditional Probability

• Nov 13th 2012, 05:48 AM
samosa
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
• Nov 13th 2012, 05:58 AM
abender
Re: Conditional Probability
Bayes' Theorem.

$\displaystyle P(A_i|A) = \dfrac{P(A_i) P(A|A_i)}{\sum^N_{j=1}P(A_j) P(A|A_j)}$
• Nov 13th 2012, 06:07 AM
samosa
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?
• Nov 13th 2012, 06:57 AM
Plato
Re: Conditional Probability
Quote:

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 $\displaystyle P(a,b|c)~\&~P(a|b,c)$ defined?
• Nov 13th 2012, 06:59 AM
abender
Re: Conditional Probability
Quote:

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?

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

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

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

$\displaystyle P(A\cap C|B)= \dfrac{\color{red}P(A | B\cap C) P(B\cap C)}{P(B)}$
• Nov 13th 2012, 07:05 AM
samosa
Re: Conditional Probability
Hi Plato,

perhaps this way:

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

and

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

Then we have

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

giving

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

Is this correct?
• Nov 13th 2012, 07:08 AM
samosa
Re: Conditional Probability
@abender, thanks a lot. Now I got it.
• Nov 13th 2012, 07:12 AM
abender
Re: Conditional Probability
Quote:

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

Your welcome -- good luck.