# Is $\sum_{m}P_{M|A}(m,a)$ the probability mass function of random variable $A$?

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?

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.)

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.