Marginalising probability distribution I am reading a book on probability theory and there is an example to denote conditional independence. A joint distribution is given as follows
$$
p(a, b, c) = p(a) p(b) p(c|a,b)
$$
Now, assuming none of the variables are observed, the point is to show $a$ and $b$ are conditionally independent. They marginalise over $c$ to get
$$
p(a, b) = p(a) p(b)
$$
Is this because when you sum over $c$, $p(c|a, b)$ is a valid distribution and sums to 1? 
 A: In general,
$$f_{X,Y,Z}(x,y,z) = f_{Z\mid X,Y}(z \mid x,y)\cdot f_{X,Y}(x,y) 
= f_{Z\mid X,Y}(z \mid x,y)\cdot f_{Y\mid X}(y \mid x)\cdot f_X(x).$$
In your instance, you are told that for your particular variables
$$f_{X,Y,Z}(x,y,z)  
= f_{Z\mid X,Y}(z \mid x,y)\cdot f_{X}(x)\cdot f_Y(y)$$
and so  
$$f_{X,Y}(x,y) = f_{X}(x)\cdot f_Y(y),$$
that is, $X$ and $Y$ are unconditionally independent, not conditionally
independent as your book claims.
To answer your specific questions


*

*Yes, $p(c\mid a,b)$ is a valid distribution and so $\sum_c p(c \mid a,b) = 1$. 

*Marginalizing $p(a,b,c)$ with respect to $c$ always
gives $\sum_c p(a,b,c) = p(a,b).$
In your special case, $p(a,b,c)$ happens
to equal $p(a)p(b)p(c\mid a,b)$ and so marginalizing $p(a,b,c)$ expressed
in the form $p(a)p(b)p(c\mid a,b)$  is the same as marginalizing $p(c\mid a,b)$ after pulling $p(a)p(b)$
out of the sum as a constant not depending on $c$. Thus we are left with
$$p(a,b) = p(a)p(b)$$ which shows that $a$ and $b$ are unconditionally
independent variables.  
Conditional independence would require something like $p(a,b\mid c) = p(a\mid c)p(b\mid c)$ to hold, and this may or may not be true in this particular
instance: there is not enough information to make a determination.

*Note that it is possible for variables to be unconditionally
independent but conditionally dependent.  A standard example is that
of Poisson splitting:  Color each arrival in a Poisson process 
with arrival rate $\lambda$ as red or blue
with probabilities $p$ and $1-p$ and independently of all other colorings.
Then the red arrivals and blue arrivals are independent Poisson processes
with arrival rates $\lambda p$ and $\lambda(1-p)$ respectively.  In
particular, the number of red arrivals in $(0,T]$ and the number of blue arrivals
in $(0,T]$ are independent Poisson random variables with parameters
$\lambda pT$ and $\lambda(1-p)T$ respectively, but conditioned
on a total number of $N$ arrivals in $(0,T]$, the numbers of red and
blue arrivals are dependent binomial random variables with parameters
$(N,p)$ and $(N,(1-p))$ respectively: dependent because their sum must equal $N$.
