Table of Contents
🌏 中文版
This is article 12 in Reading Stanford CS109, covering Summer 2026 Lecture 11: Inference on July 8 with Chris Gregg. Its Summer agenda follows the worksheet, answer key, LLM Learning Guide, and the shared Spring-dated reader's inference chapter. The Canvas recording is inaccessible, so spoken material is not reconstructed.
The original worksheet is two pages with complete numbering: P1–P2 appear on page one, and P3–P6 plus the challenge appear on page two. The central move is not a new Bayes formula but an executable data structure and update loop:
posterior[h] = prior[h] × likelihood(observation | h)
normalize(posterior)P1: One Bayes update
A bag is equally likely to contain a Fair or Trick coin, whose head probabilities are 0.5 and 0.9. After one head,
P(Trick | heads)
= P(heads|Trick)P(Trick)
/ [P(heads|Trick)P(Trick)+P(heads|Fair)P(Fair)]
= 0.9(0.5) / [0.9(0.5)+0.5(0.5)]
= 0.45/0.70 ≈ 0.643A head is more likely under the Trick model, so the posterior rises above the 0.5 prior. Inference reallocates belief according to how compatible the observation is with each hidden state.
P2: A belief dictionary and update loop
The Stanford Acuity Test discretizes vision ability A into three candidates:
a | 0.2 | 0.5 | 0.9 |
|---|---|---|---|
P(A=a) | 0.2 | 0.3 | 0.5 |
The probability of reading a letter correctly equals a, so after a wrong answer Y=0, the likelihood is 1-a. Multiplying each entry gives
a=0.2: 0.2×0.8 = 0.16
a=0.5: 0.3×0.5 = 0.15
a=0.9: 0.5×0.1 = 0.05Their sum, 0.36, is the evidence probability P(Y=0) and the Bayes denominator. Dividing each entry by it yields
P(A=0.2|Y=0) ≈ 0.444
P(A=0.5|Y=0) ≈ 0.417
P(A=0.9|Y=0) ≈ 0.139A mistake is easier to generate at low ability, so belief shifts downward. This dictionary is non-parametric: it stores mass for candidate values without forcing the posterior into a named distribution family.
P3: Two routes for multiple observations
The next, larger letter is read correctly, giving likelihood a. Use the P2 posterior as the next prior:
0.444×0.2 = 0.0889
0.417×0.5 = 0.2083
0.139×0.9 = 0.1250
sum ≈ 0.4222
posterior ≈ {0.211, 0.493, 0.296}Alternatively, multiply the original prior by both likelihoods at once:
0.2(0.8)(0.2)=0.032
0.3(0.5)(0.5)=0.075
0.5(0.1)(0.9)=0.045The sum is 0.152, and normalization again produces {0.211,0.493,0.296}. Under conditional independence of observations given ability, sequential updating and one product of likelihoods are equivalent. Intermediate normalization is merely a shared constant.
P4: Why normalize?
The normalization constant is the total probability of the observation:
P(obs) = Σa P(A=a)P(obs|a)It sums out hidden value a, so it is the same positive number for every candidate. Dividing by it makes the posterior sum to one and therefore a probability distribution.
If only the mode or argmax_a P(A=a|obs) is needed, normalization may be skipped because a shared positive divisor cannot change rankings. Reporting probabilities, sampling, or computing expectations still requires normalization.
P5: Discretize a continuous belief
In Bayesian carbon dating, sample age A is continuous, but code uses integer ages from 100 through 10,000 as dictionary keys. Each entry approximates P(A=i|observation). A fine enough grid approximates the continuous belief while retaining the same multiply-and-normalize loop.
calc_likelihood(m, age) represents P(M=m|A=age): the probability of observing m remaining C14 molecules if the sample truly has that age, computed from radioactive decay. A likelihood scores candidate ages through the generative model; it is not itself P(age|m).
P6: A full Mutation Clock posterior
Mitochondrial DNA has 10,000 base pairs, each mutating at annual rate r=6.67×10^-6. At age t, the probability one position has mutated at least once follows from a Poisson zero-event complement:
p(t) = 1-P(Poi(rt)=0) = 1-e^(-rt)With independent positions and exactly ten observed mutations,
P(X=10|T=t)
= C(10000,10) p(t)^10 [1-p(t)]^9990The prior is uniform on integer t∈{0,...,200}. The combination constant and uniform prior cancel during normalization:
P(T=150|X=10)
= L(150) / Σ(t=0..200)L(t)
L(t)=p(t)^10[1-p(t)]^9990The official prompt gives a quick check of about 0.011, with the mode at t=150. P6 is a pset4 item deliberately omitted from the public answer key; these expressions are built only from the public prompt.
Challenge: The Baby That Won't Arrive
Delivery day D has a prior PMF around day zero. Today is day -17, and no baby has arrived. A true delivery day d≤-17 makes this observation impossible, whereas d>-17 is fully compatible:
P(no baby yet at -17 | D=d) = 1[d>-17]Multiply the prior by this indicator and normalize. Every d≤-17 receives zero posterior mass. Surviving days retain their relative prior weights, but each rises after division by the remaining total mass. Negative information is still information: it eliminates hidden states and redistributes their probability.
How to use the LLM Learning Guide
The guide's six concepts are belief updating, the belief-table loop, normalization, multiple observations, continuous-variable discretization, and reading likelihoods from models. For each exercise, label the hidden and observed variables first, then choose an indicator, PMF, or PDF likelihood. After multiplication, branch explicitly: stop at unnormalized values when only the mode is required, or verify a sum of one when a posterior distribution is required.
Material boundaries
- This guide covers official P1–P6, the optional due-date challenge, and all six LLM-guide concepts; page-boundary numbering is complete.
- P6 is a pset4 item deliberately omitted from the public answer key; this article derives only its likelihood and posterior from the public prompt.
- The Canvas recording is inaccessible, so no additional spoken examples or claims are inferred.
- The worksheet and guide are two pages each. The short-material exception applies; the article remains
draft: truepending independent review.
References
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