Recipe

Test a candidate surprise function for additivity

Use when: you have a candidate function $s(p)$ meant to measure information or surprise, and need to check whether it satisfies $s(pq)=s(p)+s(q)$ for independent events

Goal: a definite pass or fail on the additivity requirement, using one concrete numeric pair as a fast screen

Taught in Why logarithms

Quick recipe

Pick one concrete pair of probabilities to test
Compute $s(p)+s(q)$ using the candidate's own formula
Compute $s(pq)$ , the candidate applied to the joint probability
Compare the two results

1
Choose specific values for $p$ and $q$ , representing two independent events, keeping the arithmetic simple.

Example

$p=0.25$ , $q=0.5$ , testing the reciprocal candidate $s(p)=\dfrac1p-1$ .

Why: additivity is a claim about every pair of probabilities, but a single well-chosen pair is enough to disprove it if it fails.
2
Evaluate the candidate function separately at $p$ and at $q$ , then add the two results.

Example

$s(0.25)+s(0.5) = \left(\dfrac{1}{0.25}-1\right)+\left(\dfrac{1}{0.5}-1\right) = 3+1 = 4$ .

Why: this is the value additivity claims the joint surprise should equal.
3
Multiply $p$ and $q$ together first, then evaluate the candidate function at that single joint probability.

Example

$pq=0.125$ , so $s(0.125) = \dfrac{1}{0.125}-1 = 7$ .

Why: this is the value additivity is actually testing against, the surprise the candidate assigns to the combined event.
4
Check whether the sum from Step 2 and the value from Step 3 come out equal.

Example

$4 \neq 7$ : the reciprocal candidate fails additivity.
- the two results match — the candidate survives this particular test, though a genuine proof would still need to hold for every pair, not just one.
- the two results don't match — the candidate has definitively failed additivity, and no other requirement can rescue it.
Why: the two computed numbers not matching is a complete, immediate disproof; a single counterexample is all it takes to fail a claim that's supposed to hold for every pair.

Common mistakes

Concluding a candidate passes additivity from one matching pair alone, without recognizing that's only a screen, not a full proof.
Computing $s(p) \cdot s(q)$ or $s(p+q)$ instead of $s(p)+s(q)$ , testing the wrong algebraic combination entirely.
Multiplying $p$ and $q$ incorrectly before evaluating $s(pq)$ , especially when the probabilities are already decimals.
Assuming a candidate that passes every single-outcome requirement (decreasing, zero at certainty, unbounded) must automatically be additive too; additivity is a genuinely separate, pairwise check.