Does a Bayesian network include the CPTs? I'm preparing slides for a lecture, and I require some guidance.
I'm only talking about discrete variables.
How would you formally define the concepts surrounding Bayesian networks?


*

*A Bayesian network is just the graph (without CPTs).

*A Bayesian network is a graph, together with a set of compatible CPTs.

*A Bayesian network is a set of CPTs (inducing the graph).


I find support for the first two options within the literature. 
E.g. the books by Pearl (1986, Probabilistic reasoning...) or Koller and Friedman (2009, Probabilistic Graphical Models...) define a BN as just the graph/DAG that has to be compatible with some distribution.
Then many resources say that a BN defines a distribution (e.g. Russel and Norvig, 2009, AI - A modern Approach).
What is the more common, or more accessible way of defining BNs?
 A: If you take a Bayesian network structure and parameterize it, then it defines a probability distribution. However before adding the parameterization, it only defines constraints on the distribution. And in general, a BN may include continuous variables - so there might be no CPTs at all! 
The third option - "A Bayesian network is a set of CPTs (inducing the graph)" - won't work unless you limit the possible sets of CPTs, in a way that would seem arbitrary. Not all sets of CPTs induce BN structures; there are sets of conditional independences that cannot be fully represented by a DAG. (For example, say we have variables $A, B, C$ and $D$, and only two independences: $A \perp\!\!\!\perp C | \{B,D\}$ and $B \perp\!\!\!\perp D | \{A,C\}$.) 
Furthermore, even if the CPTs are compatible with a DAG, they won't necessarily identify a unique model; they will only tell you the Markov equivalence class of the generating DAG.
[Edit: Here is an example, which hopefully will clarify this point. Say we have the DAG $A \to B \to C$, and the following three CPTs:
\begin{align}
\text{CPT1:} \qquad P(A=1) = .2
\end{align}
\begin{align}
\text{CPT2:} \qquad P(B=1|A=1) &= .7 \\
P(B=1|A=0) &= .4
\end{align}
\begin{align}
\text{CPT3:} \qquad P(C=1|B=1) &= .75 \\
P(C=1|B=0) &= .5
\end{align}
CPT1 and CPT2 can be combined to produce the joint pmf $P(A,B)$, like so:
\begin{align}
P(A,B): \qquad P(A=1\; \& \;B=1) &= .14 \\
P(A=0\; \& \;B=1) &= .32 \\
P(A=1\; \& \;B=0) &= .06 \\
P(A=0\; \& \;B=0) &= .48 \\
\end{align}
The joint pmf can then be factored using the chain rule into CPT1' and CPT2':
\begin{align}
\text{CPT1':} \qquad P(B=1) = .46
\end{align}
\begin{align}
\text{CPT2':} \qquad P(A=1|B=1) &= \frac{.14}{.46} \\
P(A=1|B=0) &= \frac{.06}{.54}
\end{align}
Clearly $P(A,B,C)$ can be represented by either CPT1, CPT2 and CPT3, or alternatively by CPT1', CPT2' and CPT3. This illustrates that the distribution is compatible with two distinct DAGs: $A \to B \to C$, and $A \leftarrow B \to C$. This is because these two DAGs entail the same conditional independences (i.e. they are members of the same Markov equivalence class).]
So I would stick with the first option. In more detail:
A Bayesian network induces conditional independence constraints on all probability distributions over the nodes of the graph. In particular, a graph $G$ over a set of variables $\mathbf{X}$ entails that the probability distribution $P(\mathbf{X})$ should factor into $\Pi_i P(X_i|\mathbf{Pa}_i)$, where $\mathbf{Pa}_i$ is the set of parents of node $X_i$ in $G$. If all the variables in $\mathbf{X}$ are discrete, this means $P(\mathbf{X})$ can be characterized by a set of small CPTs. A fully parameterized discrete BN is a graph together with a set of CPTs, one for each factor in the distribution.
(If you are looking for good teaching resources I recommend David Heckerman's classic A Tutorial on Learning with Bayesian Networks (pdf). There is an associated set of slides that might save some time.)
