Stanford CS109 Lecture 6 | Moments: Expectation, LOTUS, and linearity

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This is article 7 in Reading Stanford CS109, covering Summer 2026 Lecture 6: Moments (Expectation) on June 30 with Chris Gregg. Its Summer agenda comes from the schedule, worksheet, answer key, and LLM guide. The /spr26 reader is a shared, Spring-dated concept reference rather than evidence of the Summer lecture. Canvas video is inaccessible and is not reconstructed.

From a distribution to an operational summary

Expectation is the probability-weighted average,

E[X] = Σx x P(X=x)

and need not be attainable. A fair die has expectation 3.5 although no face shows 3.5.

Problem 1: reconnecting the binomial PMF

For five four-option questions guessed independently, X~Bin(5,.25) and

P(X=k) = C(5,k)(.25)^k(.75)^(5-k)

Exactly two substitutes k=2; at least one is 1-.75^5. This separates a fixed-trial success count from a waiting-time variable.

Problem 2: naming four classic variables

A single ad click is Bernoulli(.01); dates until the first partner are geometric(.2); attempts until Shaq’s third make are negative binomial with r=3 and p=.53; heads in twenty fair flips are Bin(20,.5). Identify what X counts and whether trials or required successes determine stopping.

Problem 3: sampling a class versus sampling a student

Choosing one of classes sized 5, 10, and 150 uniformly gives mean 55. Choosing one of 165 students uniformly weights classes by enrollment, giving (5²+10²+150²)/165≈137.1. The larger student-experienced average is size-biased sampling, not contradictory arithmetic.

Problem 4: linearity and LOTUS

For die result X and winnings W=2X-1, linearity gives E[W]=2E[X]-1=6. LOTUS handles nonlinear transformations:

E[g(X)] = Σx g(x)P(X=x)

Thus E[X²]=91/6, while E[X]²=12.25; averaging then squaring differs from squaring then averaging. That contrast prepares the variance topic without claiming it as this lecture's agenda.

Problem 5: expectations of classic variables

The formulas are E[Bern(p)]=p, E[Bin(n,p)]=np, and E[Geo(p)]=1/p. Shaq expects 265 makes in 500 attempts at .53; improving by ten percentage points adds 50 expected makes. Geometric(.2) has mean waiting time five, a long-run average rather than a fifth-trial guarantee.

Problem 6: daycare revenue and piecewise cost

With six babies independently attending with probability 5/6, X~Bin(6,5/6) and high attendance is P(X=5)+P(X=6). Revenue is linear:

E[R] = 50E[X] = 50 × 6 × 5/6 = 250

Staffing cost is piecewise: 200P(X≤4)+400P(X≥5). This contrasts a linear transform with a function requiring event-weighted LOTUS reasoning.

Problem 7: linearity without independence

Let Xi indicate whether position i contains its sorted card. Each is Bernoulli(1/52), though the indicators are dependent. Linearity still gives

E[X] = E[Σ Xi] = Σ E[Xi] = 52 × 1/52 = 1

so a random permutation has one fixed point on average. The challenge derives E[Bin(n,p)]=np by writing the count as a sum of n Bernoulli indicators—no binomial-PMF sum required.

Material gaps

• This guide covers all seven worksheet problems and the optional challenge without inventing inaccessible classroom examples.
• Recordings are Canvas-gated and were not used.

References

• CS109 Summer 2026 schedule
• Lecture 6 worksheet
• Lecture 6 answer key
• Lecture 6 LLM Learning Guide
• Probability for Computer Science