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Suppose I have a conditional discrete distribution $P_{M|A}(m,a)$. If I take sum over $m$, i.e. $\sum_{m}P_{M|A}(m,a)$, do I get $f_A(a)$, where $f_A(a)$ is the pmf of the random variable A?


Thanks for clarifying. I am trying to read about graphical models from http://ai.stanford.edu/~paskin/gm-short-course/lec2.pdf. Here are the equations I am confused about:
\begin{align} &∑_e∑_a∑_{mp}E(e)p_B(b)p_{A|EB}(a,e,b)p_{J|A}(true,a)p_{M|A}(m,a) \\ &\qquad = ∑_e∑_{ap}E(e)p_B(b)p_{A|E}B(a,e,b)p_{J|A}(true,a)∑_mp_{M|A}(m,a) \\ &\qquad = ∑_e∑_{ap}E(e)p_B(b)p_{A|EB}(a,e,b)p_{J|A}(true,a)ψ_A(a) \end{align}
I am assuming $ψ_A(a)$ is pmf of $A$, and there comes the question, that how to marginalize a conditional density function. Ideally, $\sum_mp_{M|A}(m,a)$ should be equal to 1. What am I missing?

(I have asked this as a new question here: http://stats.stackexchange.com/questions/45622/marginalizing-a-conditional-distribution-for-bayesian-networks.)

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For each $a$, $P_{M|A}(m|A=a)$ is a pmf, so if you sum over all $m$ in its domain, you should get

$$ \sum_m P_{M|A}(m|A=a) = 1 $$

Otherwise, the function is not a valid pmf. The pmf of $A$ is the sum of joint pmf $P_{A,M}(a,m)$ over all $m$,

$$ f_{A}(a)= \sum_{m} P_{A,M}(a,m) $$

which cannot be recovered from $P_{M|A}(m|A=a)$ without already having the marginal pmf.

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