# When does $\sum_l p(e|l)p(l|s)=p(e|s)$ holds?

I'm working with Bayesian Networks and conditional probability. Let $e$, $l$ and $s$ be random variables, the expression $$\sum_l p(e|l)p(l|s)=p(e|s)$$

holds always? If so, how can I derive it using the definition of conditional probability and marginalization?

If I understand correctly, $s$ is the parent of $l$ and $l$ is the parent of $e$ in the Bayesian network. Therefore, $$p(e \mid l, s) = p(e \mid l)$$ since the probability of $e$ depends only on $l$. Then \begin{align} \sum_l p(e\mid l)p(l\mid s) = \sum_l p(e\mid l, s)p(l\mid s) = \sum_l p(e, l \mid s) = p(e \mid s) \end{align} where the second equality is due to $$p(A, B \mid C) = p(A \mid B, C) \cdot p(B \mid C)$$ and the last equality is due to the law of total probability.
• That is one situation. More generally, the equality, $p(e\mid \ell, s)=p(e\mid \ell)$, means that $e$ and $s$ are conditionally independent when given $\ell$, and vice versa. \begin{align}p(e\mid s)~&= \sum_\ell p(e,\ell\mid s) && \text{Law of Total Probability}\\[1ex] &=~\sum_\ell p(e\mid\ell,s)~p(\ell\mid s) &&\text{Definition of Conditional Probability}\\[1ex] &=~\sum_\ell p(e\mid \ell)~p(\ell\mid s) && e\perp s\mid\ell\end{align} Mar 21 '17 at 8:08
• @GrahamKemp Thanks fo the comment, but I don't see any difference with the answer. The first equation of the answer $p(e|l,s)=p(e|l)$ means that $e$ and $s$ are conditionally independent given $l$. Mar 21 '17 at 15:00