The nodes are B (discrete) and C (continuous); C being a child node of B.

The distribution of B is:

B=stable 0.85

B=unstable 0.15

The distribution of C is:

C | B=stable \(N(-2,0.1)\)

C | B=unstable \(N(-1,0.3)\)

If we assume a clique of \(\left \{ B,C \right \}\) with the potentials \(P(B)\) and \(P(C|B)\) assigned to it. By the literature being followed, we need to initialise these potentials in a canonical form.

Therefore the initialisation of \(B\) is:

\(g_{B}(stable) = log(0.85) = -0.16252\)

\(h_{B}(stable) = 0\)

\(k_{B}(stable) = 0\)

\(g_{B}(stable) = log(0.15) = -1.89712\)

\(h_{B}(stable) = 0\)

\(k_{B}(stable) = 0\)

The initialisation of \(C\) is:

\(g_{C}(stable) = -\frac{(-2)^{2}}{2*0.1}-\left \{ log(2\pi *0.1) \right \}/2=-19.76765\)

\(h_{C}(stable) = -2/0.1=-20\)

\(k_{C}(stable) = 1/0.1=10\)

\(g_{C}(unstable) = -\frac{(-1)^{2}}{2*0.3}-\left \{ log(2\pi *0.3) \right \}/2=-1.98362\)

\(h_{C}(unstable) = -1/0.3=-3.33333\)

\(k_{C}(unstable) = 1/0.3=3.33333\)

We add the canonical forms together to produce the canonical form for \(B,C\):

\(g_{B,C}(stable) = -19.9302\)

\(h_{B,C}(stable) = -20\)

\(k_{B,C}(stable) = 10\)

\(g_{B,C}(unstable) = -3.88074\)

\(h_{B,C}(unstable) = -3.33333\)

\(k_{B,C}(unstable) = 3.33333\)

When I use Hugin I get the result for \(C\) - \(\mu_{C}=-1.85, \sigma ^{2}=0.257049\). Now,I can achieve those same values without performing the canonical transformation, but what I want to know is how do I get those result values from the canonical forms in the potentials?