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2 deleted 89 characters in body I'm trying to check if my understanding of variable elimination is correct.

Assume the above Bayesian network is factorized as:

$$p(a,b,d,e,l,s,t,x) = p(a)p(t|a)p(e|t,l)p(x|e)p(l|s)p(b|s)p(d|b,e)p(s)$$

Suppose I want to find $$p(e|s)$$, this means performing $$p(e|s) = \sum_{t,l}p(e|l,t)p(t)p(l|s)$$

Now, $$p(e|l,t)p(t)= \frac{p(e,l,t)}{p(l,t)}p(t) = \frac{p(e,l,t)}{p(l|t)} = \frac{p(e,l,t)}{p(l)} = p(e,t|l)$$. The third equality is due to the unconditioned collider $$e$$. Marginalizing over $$t$$ gives us $$p(e|l)$$.

Subsequently, $$p(e|l)p(l|s) = p(e|l,s)p(l|s) = \frac{p(e|l,s)p(s)}{p(s)}p(l|s) = \frac{p(e,l,s)}{p(s)} = p(e,l|s)$$, where the first equality is because conditioning on $$l$$ makes $$s$$ irrelevant, i.e. $$p(e|l,s) = p(e|l)$$. MarginalizingAnd marginalizing over $$l$$ finally gives $$p(e|s)$$.

Finally, combining the 2 steps above gives us the desired $$\sum_{t,l}p(e|l,t)p(t)p(l|s) = p(e|s)$$

Is this correct?

Also and on to the larger picture, is there a faster way of deriving or 'seeing' this equality? It seems to me that the faster approach is moving the conditioned variables over the conditioning bar, and then marginalizing, that is $$\sum_{t,l}p(e|l,t)p(t)p(l|s) = \sum_{t,l}p(e,l,t|s) = p(e|s)$$. Is this valid?

Thank you. I'm trying to check if my understanding of variable elimination is correct.

Assume the above Bayesian network is factorized as:

$$p(a,b,d,e,l,s,t,x) = p(a)p(t|a)p(e|t,l)p(x|e)p(l|s)p(b|s)p(d|b,e)p(s)$$

Suppose I want to find $$p(e|s)$$, this means performing $$p(e|s) = \sum_{t,l}p(e|l,t)p(t)p(l|s)$$

Now, $$p(e|l,t)p(t)= \frac{p(e,l,t)}{p(l,t)}p(t) = \frac{p(e,l,t)}{p(l|t)} = \frac{p(e,l,t)}{p(l)} = p(e,t|l)$$. The third equality is due to the unconditioned collider $$e$$. Marginalizing over $$t$$ gives us $$p(e|l)$$.

Subsequently, $$p(e|l)p(l|s) = p(e|l,s)p(l|s) = \frac{p(e|l,s)p(s)}{p(s)}p(l|s) = \frac{p(e,l,s)}{p(s)} = p(e,l|s)$$, where the first equality is because conditioning on $$l$$ makes $$s$$ irrelevant, i.e. $$p(e|l,s) = p(e|l)$$. Marginalizing over $$l$$ gives $$p(e|s)$$.

Finally, combining the 2 steps above gives us the desired $$\sum_{t,l}p(e|l,t)p(t)p(l|s) = p(e|s)$$

Is this correct?

Also and on to the larger picture, is there a faster way of deriving or 'seeing' this equality? It seems to me that the faster approach is moving the conditioned variables over the conditioning bar, and then marginalizing, that is $$\sum_{t,l}p(e|l,t)p(t)p(l|s) = \sum_{t,l}p(e,l,t|s) = p(e|s)$$. Is this valid?

Thank you. I'm trying to check if my understanding of variable elimination is correct.

Assume the above Bayesian network is factorized as:

$$p(a,b,d,e,l,s,t,x) = p(a)p(t|a)p(e|t,l)p(x|e)p(l|s)p(b|s)p(d|b,e)p(s)$$

Suppose I want to find $$p(e|s)$$, this means performing $$p(e|s) = \sum_{t,l}p(e|l,t)p(t)p(l|s)$$

Now, $$p(e|l,t)p(t)= \frac{p(e,l,t)}{p(l,t)}p(t) = \frac{p(e,l,t)}{p(l|t)} = \frac{p(e,l,t)}{p(l)} = p(e,t|l)$$. The third equality is due to the unconditioned collider $$e$$. Marginalizing over $$t$$ gives us $$p(e|l)$$.

Subsequently, $$p(e|l)p(l|s) = p(e|l,s)p(l|s) = \frac{p(e|l,s)p(s)}{p(s)}p(l|s) = \frac{p(e,l,s)}{p(s)} = p(e,l|s)$$, where the first equality is because conditioning on $$l$$ makes $$s$$ irrelevant, i.e. $$p(e|l,s) = p(e|l)$$. And marginalizing over $$l$$ finally gives $$p(e|s)$$.

Is this correct?

Also and on to the larger picture, is there a faster way of deriving or 'seeing' this equality? It seems to me that the faster approach is moving the conditioned variables over the conditioning bar, and then marginalizing, that is $$\sum_{t,l}p(e|l,t)p(t)p(l|s) = \sum_{t,l}p(e,l,t|s) = p(e|s)$$. Is this valid?

Thank you.

1

# Variable elimination in Bayesian network I'm trying to check if my understanding of variable elimination is correct.

Assume the above Bayesian network is factorized as:

$$p(a,b,d,e,l,s,t,x) = p(a)p(t|a)p(e|t,l)p(x|e)p(l|s)p(b|s)p(d|b,e)p(s)$$

Suppose I want to find $$p(e|s)$$, this means performing $$p(e|s) = \sum_{t,l}p(e|l,t)p(t)p(l|s)$$

Now, $$p(e|l,t)p(t)= \frac{p(e,l,t)}{p(l,t)}p(t) = \frac{p(e,l,t)}{p(l|t)} = \frac{p(e,l,t)}{p(l)} = p(e,t|l)$$. The third equality is due to the unconditioned collider $$e$$. Marginalizing over $$t$$ gives us $$p(e|l)$$.

Subsequently, $$p(e|l)p(l|s) = p(e|l,s)p(l|s) = \frac{p(e|l,s)p(s)}{p(s)}p(l|s) = \frac{p(e,l,s)}{p(s)} = p(e,l|s)$$, where the first equality is because conditioning on $$l$$ makes $$s$$ irrelevant, i.e. $$p(e|l,s) = p(e|l)$$. Marginalizing over $$l$$ gives $$p(e|s)$$.

Finally, combining the 2 steps above gives us the desired $$\sum_{t,l}p(e|l,t)p(t)p(l|s) = p(e|s)$$

Is this correct?

Also and on to the larger picture, is there a faster way of deriving or 'seeing' this equality? It seems to me that the faster approach is moving the conditioned variables over the conditioning bar, and then marginalizing, that is $$\sum_{t,l}p(e|l,t)p(t)p(l|s) = \sum_{t,l}p(e,l,t|s) = p(e|s)$$. Is this valid?

Thank you.