Using the above DAG I need to simplify the following conditional probabilities:

$$i) \quad p(x_4|x_1,x_2)$$

For this one I guess I can just remove the conditioning on $x_1$ (using the DAG) and the most simplified conditional probability would be $p(x_4|x_2)$

$$ii) \quad p(x_4,x_5|x_1,x_2)$$

Again removing the condition on $x_1$ we get $p(x_4,x_5|x_2)$ which can be further simplified as:

$$ p(x_4,x_5|x_2) = \frac{p(x_2,x_4,x_5)}{p(x_2} = \frac{p(x_2)p(x_4|x_2)p(x_5|x_4)}{p(x_2)} $$

Here $p(x_2)$ cancel and we are left with:

$$p(x_4,x_5|x_2) = p(x_4|x_2)p(x_5|x_4)$$

Essentially my question is can we ignore the conditioning on an ancestor that is above another ancestor that is being conditioned on?


1 Answer 1


From the graph: $$p(x_1,x_2,x_3,x_4,x_5) = p(x_1)p(x_2|x_1)p(x_3|x_2)p(x_4|x_3)p(x_5|x_4)$$

Using conditional probability:

Part $i)$

$$p(x_4|x_1,x_2) = \frac{p(x_1,x_2,x_4)}{p(x_1,x_2)} = \frac{\sum_{x_3x_5}p(x_3|x_2)p(x_4|x_3)p(x_5|x_4)}{\underbrace{\boxed{\sum_{x_3x_4x_5}p(x_3|x_2)p(x_4|x_3)p(x_5|x_4)}}_{1}}\cdot\underbrace{\frac{p(x_1)p(x_2|x_1)}{p(x_1)p(x_2|x_1)}}_{\text{cancels}}$$

$$p(x_4|x_1,x_2) = \sum_{x_3}p(x_3|x_2)p(x_4|x_3)$$

Part $ii)$

$$p(x_4,x_5|x_1,x_2) = \frac{p(x_1,x_2,x_4,x_5)}{p(x_1,x_2)} = \frac{\sum_{x_3}p(x_3|x_2)p(x_4|x_3)}{\underbrace{\boxed{\sum_{x_3x_4x_5}p(x_3|x_2)p(x_4|x_3)p(x_5|x_4)}}_{1}}\cdot \frac{p(x_1)p(x_2|x_1)p(x_5|x_4)}{\underbrace{p(x_1)p(x_2|x_1)}_{\text{cancels leaving }p(x_5|x_4)}}$$

$$p(x_4,x_5|x_1,x_2) = p(x_5|x_4) \cdot \sum_{x_3}p(x_3|x_2)p(x_4|x_3)$$

  • $\begingroup$ Understood! Thanks a lot. $\endgroup$ Jun 5 at 12:44

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