Probability models
도입됨
어휘
| English | 의미 |
|---|---|
| Bayesian network | A directed acyclic graph representing a joint probability distribution through conditional probabilities. |
English
The network and the numbers
- A Bayesian network is a DAG of variables with conditional probability tables; inference multiplies along the graph and normalises. Course net: Storm → Delay → {Late, Complaint}.
{exercise:uni-cs1303-net}
연습
A Bayesian network factorises —
난이도 1/5 · 100코인 + 오늘 첫 완료 시 작은 보너스 추가
Each node keeps a CPT given its parents; the joint is their product.
English
Bayes, in the small
- $P(S) = 0.3$, $P(D \mid S) = 0.8$, $P(D \mid \neg S) = 0.1$, $P(C \mid D) = 0.6$, $P(C \mid \neg D) = 0.05$: marginalising D, $P(C \mid S) = 0.49$; $P(C \mid \neg S) = 0.105$; $P(C) = 0.2205$; $P(S \mid C) = \mathbf{0.667}$.
- The 4-node joint has 16 rows; the suite enumerates every one and confirms 0.667 — exact inference on small nets is bookkeeping, and the bookkeeping is the derivation.
{exercise:uni-cs1303-bayes}
연습
With the course numbers, P(Storm | Complaint) = —
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0.49 x 0.3 / 0.2205 = 0.667 - marginalise D both ways first.
English
The base rate is not a footnote
- With $P(S) = 0.03$ (rare storms): $P(C) = 0.11655$ and $P(S \mid C) = \mathbf{0.126}$ — same evidence, belief collapsing to one-eighth. Rare causes need strong evidence.
- Watch the two fallacies: dropping the normaliser $P(C)$, and swapping $P(C \mid S)$ for $P(S \mid C)$ (the prosecutor's fallacy). D-separation, not missing edges, decides dependence.
{exercise:uni-cs1303-base}
연습
Halving the storm prior to 0.03 moves the posterior to —
난이도 1/5 · 100코인 + 오늘 첫 완료 시 작은 보너스 추가
Rare causes need strong evidence: 0.667 falls to 0.126.