Lecture 2 updated one unknown. A Bayesian network connects many interacting variables: represent a system, infer its hidden parts, then decide.

Bring: Bayes' rule and marginalization from Lecture 2.

A directed acyclic graph (DAG) has no directed cycle. On the left, $A\to B\to C\to A$ returns to $A$; on the right it cannot.

A Bayesian network specifies $p(x_1,\ldots,x_n)=\prod_i p(x_i\mid\mathrm{pa}_i)$ on a DAG.

Free parameters: $1+1+4+2+2=10$, compared with $2^5-1=31$ in the full joint.

Let $F$ mean a machine fault and $A$ mean an alarm: $F\to A$. Use $P(F)=0.1$, $P(A\mid F)=0.8$, $P(A\mid\neg F)=0.2$.

The alarm probability sums both possible causes: $P(A)=0.08+0.18=0.26$.

The chain rule already writes any joint as a product of conditionals:

which is a Bayesian network — on the fully connected DAG. It costs exactly $2^n-1$ numbers and buys nothing.

Chain any two variables through a third and there are only three shapes. Whether $X$ and $Y$ are independent depends on the shape and on whether the middle node is observed.

Before: $B$ and $S$ are independent. Observe $E$: the electrical system failed. Then learn $B$ failed: in a competing-causes model, less evidence is needed for $S$ to explain $E$.

Keep $B$ as the query. Fix $D,C$ at their observed values. Sum over $E,S$, then normalise.

Marginalise the joint over the hidden variables and normalise:

Rejection threw away six rows in nine. The waste is in sampling $D$ and $C$ at all, only to check them against values we already know. So do not sample them — write $D=1$, $C=1$ straight into every row.

Give $E\to D$ and $E\to C$ illustrative tables: $P(D{=}1\mid E{=}1)=0.9$, $P(C{=}1\mid E{=}1)=0.8$, and $P(D{=}1\mid E{=}0)=0.1$, $P(C{=}1\mid E{=}0)=0.2$.

Sample $B,S,E$ honestly from $P(B),\,P(S),\,P(E\mid B,S)$ and their probabilities are already built into how many rows carry each value. With $P(B{=}1)=0.1$, roughly one row in ten already has $B=1$ — not because a weight said so, but because that is how often the sampler drew it.

Two facts, and the method follows from their product. A row $(b,s,e)$ is drawn with the prior probability of its unclamped part, and it is then weighted by the evidence's probability given that part:

But likelihood weighting still proposes $C=1$ with probability 0.001. In 1,000 draws, the probability of missing it entirely is $0.999^{1000}\approx0.368$.

Given class $C$, the features are conditionally independent:

$$P(C=c\mid o_{1:n})\propto P(C=c)\prod_iP(o_i\mid C=c).$$

For an illustrative two-class, two-feature problem:

Thus $P(A\mid o_1,o_2)=0.16/0.22\approx0.727$. Choose $A$ under equal misclassification costs.

For complete discrete data, count each node separately for each configuration of its parents:

In the illustrative fault model, suppose 8 of 10 faulty machines and 18 of 90 healthy machines trigger the alarm.

Nothing so far said the variables were simultaneous. Index them by time and the identical machinery becomes a model of a system evolving.

Use the source convention $M_{ij}=P(S_{t+1}=i\mid S_t=j)$: column $j$ is the distribution of the next state given the current state.

For the illustrative observed sequence $A,A,B,A,B,B$, the four transition counts are $N_{AA}=1$, $N_{BA}=2$, $N_{AB}=1$, $N_{BB}=1$.

Starting at $A$ gives $p_t=[1,0]^\top$, so the next-state probabilities are $[1/3,2/3]^\top$. Every column sums to one.

Time-homogeneous transitions: the same matrix $M$ is used at each step. Stationary distribution: a probability vector $p_\infty$ satisfying $Mp_\infty=p_\infty$.

For a separate illustrative two-state model,

From $p_0=[1,0]^\top$, the first two distributions are $p_1=[0.9,0.1]^\top$ and $p_2=[0.83,0.17]^\top$: the marginal distribution changes even though $M$ stays fixed.

Hide the state and observe an emission — $X_t \to Y_t$ over a chain $X_{t-1}\to X_t$ — and you have a hidden Markov model. The standard query is filtering, the belief about now given everything seen so far:

Yesterday's posterior fault probability is $0.2$. A fault persists with probability $0.8$; a healthy machine develops a fault with probability $0.1$.

Predict today's state:

Now an alarm occurs, with $P(A_t\mid F_t)=0.9$ and $P(A_t\mid\neg F_t)=0.1$.

Correct using the observation:

In a two-mode variant, the heater is good ($G$) or weak ($W$). Initially $P(G)=0.6$. Next-step probabilities are $P(G'\mid G)=0.9$ and $P(G'\mid W)=0.3$. A warm reading has likelihood 0.8 under $G'$ and 0.2 under $W'$.

Predict: First predict the next mode without the reading. Then decide whether a warm reading should raise or lower its good-mode probability.

Keep the same predicted probabilities (0.66, 0.34). Instead observe “cool,” the complement of “warm.” Its likelihoods are therefore 0.2 under $G'$ and 0.8 under $W'$. Calculate the new posterior.

The update carries two objects: where the state is likely to be, and how uncertain that estimate remains. The next slide writes these quantities as $m_t,P_t$.

With a Gaussian initial belief and independent Gaussian noises, the posterior remains Gaussian. Track its mean and covariance instead of a table.

A scalar example with $m^-=10$, $P^-=4$, $C=1$, $R=1$, and $y=12$ gives $K=0.8$, $m=11.6$ and $P=0.8$.

Probabilities describe how likely outcomes are. Utilities describe how desirable those outcomes are to the decision maker.

Utility need not equal money. A concave utility of money represents risk aversion; the same expected-utility calculation still applies.

Compare A: receive 1 dollar for sure with B: receive 100 dollars with probability 0.01, otherwise 0. Both have expected money 1 dollar.

Same money on average, three different decisions — the shape of $U$ is doing all the work.

The original example distinguishes horizontal closeness $H$, vertical closeness $V$, and whether an alarm is raised $A$.

The first term represents the consequence of unsafe separation; the second can represent alarm cost. This additive preference assumption is separate from any independence assumption about the random variables.