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KL divergence (gentle)

Prereqs: Independence heuristics · Probability
Next: Paper: 2025/018 explainer · Gaps · DFR
Tracks: code-based-kems

Learning goals. Read \(D(P\|Q)\) as “how badly \(Q\) models \(P\),” compute a tiny discrete example by hand, connect the independence case to an entropy gap, and know that large KL is about distributions — not an automatic attack.

Symbols on this page

Symbol English
\(P\) True distribution (reality)
\(Q\) Reference / model distribution (often “independent pretend”)
\(D(P\|Q)\) KL divergence — penalty for using \(Q\) when truth is \(P\)
\(H(P)\) Entropy of \(P\) (uncertainty; appears in the independence special case)
\(\Theta(\cdot)\) Growth-rate slogan in asymptotics (order of magnitude)

Why this card exists

When a paper says “the KL divergence to the product-Bernoulli idealization is \(\Theta(n\cdot 2^{-2\omega})\),” you need a gentle meaning of KL before diving into the 2025/018 explainer. This card follows independence heuristics: there \(P\) was reality and \(Q\) the independent pretend model; here we measure the gap.

Intuition — a score for “bad model”

Kullback–Leibler (KL) divergence \(D(P\|Q)\) measures how different distribution \(P\) is from a reference model \(Q\), in information units (bits or nats depending on the log base).

Use it to say: “the independence model \(Q\) is a bad description of reality \(P\).”

It is not a symmetric distance: \(D(P\|Q)\) need not equal \(D(Q\|P)\). Think “penalty for coding samples from \(P\) as if they came from \(Q\),” not “Euclidean length between two points.”

Symbol Role in independence audits
\(P\) True joint law of noise (or relevant bits)
\(Q\) Product of \(P\)’s marginals (independent idealization)
\(D(P\|Q)\) How badly that idealization fits

Quick self-check

If \(P=Q\), what is \(D(P\|Q)\)?

Answer Zero (when the divergence is defined). Perfect match ⇒ no penalty.

Formal definition

For discrete laws (when defined — need \(Q(x)>0\) whenever \(P(x)>0\)),

\[ D(P\|Q)=\sum_x P(x)\log\frac{P(x)}{Q(x)}. \]

Special case used in independence audits: if \(Q\) is the product of \(P\)’s marginals, then under mild conditions

\[ D(P\|Q)=H(Q)-H(P), \]

an entropy gap — dependence reduces entropy below the independent ideal. Here \(H\) denotes Shannon entropy (same log base as in the KL sum).

Pinsker / reverse-Pinsker (postcard): inequalities relate KL to total variation (statistical) distance. Reverse forms need extra assumptions — used carefully in 2025/018-style arguments. You do not need the full inequality list to read this curriculum card.

Tiny worked examples

Example A — identical laws

If \(P=Q\), every term has \(\log 1=0\), so \(D=0\).

Example B — point mass vs uniform

\(P\) puts mass \(1\) on one point; \(Q\) is uniform on \(N\) points. Then

\[ D(P\|Q)=\log N \]

(up to log-base convention). \(Q\) is a terrible model for a deterministic \(P\) when \(N\) is large.

Example C — always-equal bits (from the independence card)

Recall \(P\): always \(b_1=b_2\), fair; \(Q\): independent fair bits.

On the support of \(P\), the outcomes are \((0,0)\) and \((1,1)\) with probability \(1/2\) each, and

\[ \frac{P(0,0)}{Q(0,0)}=\frac{1/2}{1/4}=2,\qquad \frac{P(1,1)}{Q(1,1)}=2. \]

So

\[ D(P\|Q)=\tfrac12\log 2+\tfrac12\log 2=\log 2 \]

(one bit of divergence if \(\log=\log_2\)). The independent model is wrong by a clear, computable gap — even though each bit looked fair.

Quick self-check

Is \(D(P\|Q)\) always equal to \(D(Q\|P)\)?

Answer No. KL is directed; swap $P$ and $Q$ and you generally get a different number (and $D(Q\|P)$ may even be infinite if supports disagree).

How papers use the language (without inventing attacks)

A statement like “\(D(P\|Q)=\Theta(n\cdot 2^{-2\omega})\)” is a distribution claim: as parameters grow, the independence idealization drifts from \(P\) at that asymptotic scale.

What it does not automatically say:

  • that a concrete KEM is broken,
  • that DFR is large,
  • that an attacker has a key-recovery algorithm.

Turning a modeling gap into an attack needs more work. Small KL suggests closeness; it is not by itself a security proof either. At first reading, asymptotics \(\Theta(\cdot)\) matter more than the log-base constant.

Reading OK takeaway Overreach
Large \(D(P\|Q)\) \(Q\) is a poor model for \(P\) “Therefore the KEM falls”
Small \(D(P\|Q)\) Idealization is close as a distribution “Therefore the scheme is secure”
Entropy gap form Dependence costs entropy vs \(Q\) “Entropy gap = stolen key bits”

Where it appears in PQC

Exact KL for invertible sparse polynomial products vs product-Bernoulli noise in quasi-cyclic settings — see the analysis route in the 2025/018 explainer. This lab’s pedagogy stops at “what KL means” and “dependence ≠ break”; the paper digests the structured calculation.

Link Why
Independence heuristics Defines \(P\) vs \(Q\)
DFR Why people wanted independence for tails
2025/018 explainer Research-shaped KL / TV statements
Gaps Open research edges (not pedagogy holes)

Common confusions

Confusion Clearer picture
\(D=0\) always means “secure” \(D=0\) means \(P=Q\) (when defined) — a modeling statement.
Large KL = key recovery Modeling gap ≠ attack by itself.
Small KL = security proof Closeness of distributions is not a full KEM proof.
Symmetric distance KL is directed; use TV if you need a metric slogan.

Check yourself

  1. In one sentence, what is \(D(P\|Q)\) measuring?
  2. If \(Q\) is the independent model for \(P\), what does a large \(D(P\|Q)\) mean?
  3. Does large KL automatically break a KEM?
  4. In the always-equal example with \(\log_2\), what is \(D(P\|Q)\)?
  5. What entropy identity links \(D(P\|Q)\) to \(H(Q)-H(P)\) in the independence case?
Answers 1. How poorly $Q$ describes $P$ (information divergence). 2. Independence is a bad description of the true joint law. 3. No — modeling gap ≠ key recovery by itself. 4. $1$ bit ($\log_2 2$). 5. When $Q$ is the product of $P$’s marginals, under mild conditions $D(P\|Q)=H(Q)-H(P)$.

Big practice set

  1. Write the discrete KL sum from memory, then check against the formula above.
  2. True or false: if coordinates are dependent, \(D(P\|Q)\) must be infinite.
  3. Point to the card that introduced the always-equal counterexample before KL.
Answers 1. $D(P\|Q)=\sum_x P(x)\log\bigl(P(x)/Q(x)\bigr)$. 2. False — the always-equal example has finite $D=\log 2$. Infinity appears when $P$ puts mass where $Q$ is zero. 3. [Independence heuristics](independence-heuristics.md).

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