# How to simplify the following conditional probability distributions using the given DAG? 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?

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 $$ii)$$

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

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

• Understood! Thanks a lot. Jun 5 at 12:44