# Understanding the chain rule in probability theory

When my teacher told us about the chain rule I found it quite easy, but when I am trying to prove something based on this rule I kind of get confused about what are the allowed forms of this rule. For example, I can't understand why I can say:

$$p(x,y\mid z)=p(y\mid z)p(x\mid y,z)$$

I found this post useful for my question:

Is order of variables important in probability chain rule

• What is $p$? And what are $x,y$, and $z$? Nov 4 '12 at 11:08
• Nov 4 '12 at 11:09
• @StéphaneLaurent Thanks for reply. Reading your post I got one question. In there you defined the general rule for more than 2 RV. When I follow your definition for the second case in the question I come up with : p(x|z,y)p(z|y) which is different from p(z|x,y)p(x|y). My problem in the fist step is how these two are equivalent ?
– Moj
Nov 4 '12 at 11:36
• @Moj $p(x,z|y)=p(x|z,y)p(z|y)=p(z|x,y)p(x|y)$ These are the two possible decompositions of the conditional joint distribution $p(x,z|y)$ of $(x,z)$ given $y$. In the first decomposition you choose to condition $x$ given $z$ whereas in the second decomposition you condition $z$ given $x$. Nov 4 '12 at 11:49
• great!This is was my problem! I wanted to make sure that changing the decomposition is possible or not!Thanks
– Moj
Nov 4 '12 at 12:19

$$p(x,y|z) = \frac{p(x,y,z)}{p(z)} = \frac{p(x|y,z)p(y,z)}{p(z)} = p(x|y,z)p(y|z)$$

On the first step we use the definition of conditional probability. On the second step we use the same definition on the numerator to convert the joint probability $$p(x,y,z)$$ into a conditional $$p(x|y,z)$$ and a joint $$p(y,z)$$. Finally, we divide $$p(y,z)$$ by $$p(z)$$ applying once again the definition of conditional probability, and we obtain the result.

Another way of looking at it is that you can just ignore variables that are always on the right side of the conditional sign. In that case the expression is just the usual conditional probability:

$$p(x,y) = p(x|y)p(y)$$

You simply condition all of these probabilities on $$z$$ and you get your original formula.

• Thanks but I still confused about how many different equivalent are exist for this equation. For ex: p(x|y,z)= p(z|x,y) is it true?
– Moj
Nov 4 '12 at 11:26
• Obviously not. p(x|y,z) = p(x,y,z)/p(y,z) = p(z|x,y)p(x,y)/p(y,z) Nov 4 '12 at 12:01
• Right, my problem was just different ordering of this conditional dependencies!Thanks
– Moj
Nov 4 '12 at 12:21
• Ofcourse I accepted it!specially the second trick was really helpful! Thanks again
– Moj
Nov 4 '12 at 12:25
• Nov 4 '12 at 14:04