Uncertainty formulas

Each conditional is defined only when the event it conditions on has positive probability ($P(B)>0$ for $P(A\mid B)$).

Sum over every value of each variable being removed.

Requires $P(B)>0$. Turns a likelihood $P(B\mid A)$ into a posterior $P(A\mid B)$; the denominator comes from marginalisation.

Requires $P(e)>0$. Compute the unnormalised values $P(x,e)$ for each $x$, then divide by their sum. $\alpha$ is just $1/P(e)$.

Holds for any distribution and any variable ordering (wherever the conditionals are defined); it is the starting point for the Bayesian-network factorisation.

Must hold for every pair of values of $A$ and $B$; the last form is for values with $P(B)>0$.

Must hold for every value of $A$, $B$ and $C$ with $P(C)>0$ (and $P(B,C)>0$ for the last form).

For any choice of CPTs this product is a valid joint distribution, and a joint factorises this way exactly when it satisfies the local Markov property for the DAG. Any joint entry is the product of one CPT entry per node.

For all-binary variables this is ${\sum }_i{2}^{|\mathrm{Pa}(X_i)|}$, compared with $2^n-1$ for the full joint.

Parents, children and the children's other parents (co-parents). Given its Markov blanket, $X$ is conditionally independent of every other variable in the network.

Sum the network's joint over all hidden-variable assignments with evidence fixed, then normalise ($P(e)>0$). Cost is exponential in the number of hidden variables.

Multiply only the factors that mention $X$, then sum $X$ out; factors not mentioning $X$ are left untouched and can be pulled outside the sum.

Each sample is drawn from the full joint, ignoring evidence; $N(\cdot )$ counts matching samples out of $N$.

Generate prior samples, discard any that disagree with the evidence, and use the counts among the kept ones (undefined if $N(e)=0$). The fraction kept is about $P(e)$, so rare evidence wastes almost all samples.

Evidence variables are fixed, not sampled; each sample is weighted by the probability of the evidence given its parents' values in that same sample (a parent may itself be an evidence variable, fixed at its observed value). Non-evidence variables are sampled as usual, so no sample is rejected. The method is consistent because the probability of drawing the non-evidence values $z$ times the weight equals $P(z,e)$.

All other variables are held at their current values; $\alpha$ normalises over the values $x_i$. Only the CPTs of $X_i$ and of its children are needed, because $X_i$ is independent of everything else given its Markov blanket.