Md. Asif Uddin

Book 0 · PR

Probability

distributions, expectation, variance, Bayes, conditional independence

0.PR.01

Distributions, discrete and continuous

Statement
A discrete distribution assigns p(x) ≥ 0 with Σ p(x) = 1. A continuous one has a density f with ∫f = 1, and f may exceed 1 while a probability may not.
Shape check
A distribution over a vocabulary of size |V| is a vector in the |V|-simplex: |V| non-negative numbers summing to one.
Worked line
Over {a, b, c} with p = (0.5, 0.3, 0.2): the entries are non-negative and sum to 1, so this is a distribution.

Every model output in Books I, III and IV is a point in a simplex. Most of what goes wrong with a model output is a claim about that point that the point does not support.

Used byII.1 · II.7 · III.6 · IV.1 · IV.2 · IV.8 · IV.9 · V.6 · VI.1 · VI.2 · VII.2 · VII.3 · I.3.P04

0.PR.02

Expectation

Statement
𝔼[X] = Σ x p(x), or ∫ x f(x) dx. It is linear: 𝔼[aX + bY] = a𝔼[X] + b𝔼[Y], whether or not X and Y are independent.
Shape check
The expectation of a vector is taken component-wise and keeps the shape.
Worked line
X uniform on {1, 2, 3, 4}: 𝔼[X] = (1 + 2 + 3 + 4)/4 = 2.5.

Linearity holding without independence is the most useful fact in this part of the Apparatus, and the one most often assumed to need independence.

Used byI.1 · III.1 · VI.5

0.PR.03

Variance and covariance

Statement
Var(X) = 𝔼[X²] − 𝔼[X]². Var(aX) = a²Var(X). For independent X and Y, Var(X + Y) = Var(X) + Var(Y); in general the covariance term is added twice.
Shape check
For a random vector, the covariance is a symmetric positive semi-definite n×n matrix.
Worked line
X uniform on {1, 2, 3, 4}: 𝔼[X²] = 7.5, 𝔼[X]² = 6.25, so Var(X) = 1.25.

The a² is where the √d_k scaling in attention comes from, and the 1/B in the variance of a minibatch gradient.

Used byI.7 · I.8 · II.3 · II.8 · IV.7 · V.3 · VI.6 · VII.2 · VII.4 · VIII.4

0.PR.04

Bayes' rule

Statement
P(A|B) = P(B|A)P(A) / P(B), with P(B) = Σ P(B|Aᵢ)P(Aᵢ).
Shape check
Posterior, likelihood, prior, evidence — four scalars for a binary question, four vectors of the same length for a categorical one.
Worked line
Se = 0.95, Sp = 0.95, prevalence 0.01. P(D|+) = (0.95)(0.01) / [(0.95)(0.01) + (0.05)(0.99)] = 0.0095/0.0590 ≈ 0.161.

That worked line is the single most consequential arithmetic in Book II. A test that is right 95% of the time, applied to a rare disease, is wrong about five times out of six when it says yes.

Used byIII.7 · IV.3 · IV.6 · VI.1 · VI.3 · VI.4 · VI.5 · VI.7 · VI.9 · VII.6 · VIII.3

0.PR.05

Conditional independence

Statement
X ⫫ Y | Z means P(X, Y | Z) = P(X | Z)P(Y | Z). It neither implies nor is implied by marginal independence.
Shape check
For binary variables this is four equalities, one for each value of the pair, at each value of Z.
Worked line
X and Y independent fair coins, Z = X ⊕ Y. Then X ⫫ Y, but conditioning on Z makes each determine the other exactly.

That worked line is collider bias with the labels removed. Book V spends a chapter on why conditioning can create a dependence that was not there.

Used byVI.1 · VI.2 · VI.3 · VI.4

0.PR.06

The exponential family

Statement
p(x | η) = h(x) exp(ηᵀ T(x) − A(η)). The log-partition A generates the moments: ∇A(η) = 𝔼[T(X)].
Shape check
η and T(x) have the same length; A(η) is a scalar.
Worked line
Bernoulli: η = log(p/(1 − p)), T(x) = x, A(η) = log(1 + e^η). Then A′(η) = σ(η) = p, which is 𝔼[X].

A logit is a natural parameter. That is why the last layer of a classifier is linear and the loss is what it is.

Used byI.4 · I.13 · IV.5 · VII.2 · VII.5