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Closed form for the expected hit count SnS_nSn​

Proved
SpecActions.hits_closed_form

by naimengye · Sep 10, 2026 · Mathlib 0df444a (Lean v4.33.1)

machine-learningprobabilitytheoretical-computer-science

Let SnS_nSn​ be the expected number of speculation hits by round nnn. Because a correct guess leaves the next call already cached, that round opens no speculation window, which gives the two-term recursion

S0=0,S1=p,Sn=p (1+Sn−2)+(1−p) Sn−1.S_0=0,\qquad S_1=p,\qquad S_n=p\,(1+S_{n-2})+(1-p)\,S_{n-1}.S0​=0,S1​=p,Sn​=p(1+Sn−2​)+(1−p)Sn−1​.

For every p≥0p\ge 0p≥0 and every nnn,

Sn=p1+p n+p2(1+p)2(1−(−p)n).S_n=\frac{p}{1+p}\,n+\frac{p^2}{(1+p)^2}\bigl(1-(-p)^n\bigr).Sn​=1+pp​n+(1+p)2p2​(1−(−p)n).

The characteristic polynomial r2−(1−p)r−pr^2-(1-p)r-pr2−(1−p)r−p has roots 111 and −p-p−p; since r=1r=1r=1 is a root, a constant particular solution collides with the homogeneous family and the particular solution is linear in nnn.

Preamble
import Definitions.Def_SpecActions_model
Formal statement
import Definitions.Def_SpecActions_model

namespace SpecActions
theorem hits_closed_form (p : ℝ) (hp : 0 ≤ p) (n : ℕ) :
    hits p n = p / (1 + p) * n + p ^ 2 / (1 + p) ^ 2 * (1 - (-p) ^ n) := by sorry
end SpecActions
Source
Ye, Ahuja, Liargkovas, Lu, Kaffes, Peng, "Speculative Actions: A Lossless Framework for Faster Agentic Systems", ICLR 2026, arXiv:2510.04371, https://arxiv.org/abs/2510.04371, Appendix A (pp. 13–14), the recursion Sn=p(k)(1+Sn−2)+(1−p(k))Sn−1S_n=p(k)(1+S_{n-2})+(1-p(k))S_{n-1}Sn​=p(k)(1+Sn−2​)+(1−p(k))Sn−1​ and its closed-form solution
Read-back

What the Lean code literally says, in plain math · claude-opus-5

Read-back — SpecActions.hits_closed_form

The declaration is an equational claim about one real-valued sequence. Fix a real number ppp subject to the single hypothesis 0≤p0 \le p0≤p (no upper bound is imposed: ppp may be 000, may equal 111, and may be arbitrarily large; nothing in the statement asserts that ppp is a probability), and fix an arbitrary natural number nnn (the quantifier includes n=0n = 0n=0).

The symbol Sp(n)S_p(n)Sp​(n) — written hits p n in the code under audit — is not a standard notion; it is the sequence introduced in the imported bundle by the two-step recursion

Sp(0)=0,Sp(1)=p,Sp(n+2)  =  p(1+Sp(n))  +  (1−p) Sp(n+1),S_p(0) = 0, \qquad S_p(1) = p, \qquad S_p(n+2) \;=\; p\bigl(1 + S_p(n)\bigr) \;+\; (1 - p)\,S_p(n+1),Sp​(0)=0,Sp​(1)=p,Sp​(n+2)=p(1+Sp​(n))+(1−p)Sp​(n+1),

i.e. each term two steps ahead is a fixed real combination of the two preceding terms, with the coefficients ppp and 1−p1-p1−p taken as literal real numbers built from the same ppp (they are not assumed to lie in [0,1][0,1][0,1] or to sum to a probability weighting beyond the algebraic fact that they sum to 111). The recursion is total: it defines a real number for every natural nnn and every real ppp, whatever the sign or size of ppp.

For every such ppp and nnn, the theorem asserts the exact real-number identity

Sp(n)  =  p1+p⋅n  +  p2(1+p)2(1−(−p)n).S_p(n) \;=\; \frac{p}{1+p}\cdot n \;+\; \frac{p^{2}}{(1+p)^{2}}\Bigl(1 - (-p)^{n}\Bigr).Sp​(n)=1+pp​⋅n+(1+p)2p2​(1−(−p)n).

Points of literal reading:

  • The two coefficients are written exactly as displayed: a first-power quotient p1+p\dfrac{p}{1+p}1+pp​ multiplying nnn, and a quotient of squares p2(1+p)2\dfrac{p^{2}}{(1+p)^{2}}(1+p)2p2​ multiplying the bracket. The factor nnn is the natural number nnn regarded as a real number, and (−p)n(-p)^{n}(−p)n is the nnn-th natural-number power of the negated quantity −p-p−p (so it alternates in sign with the parity of nnn when p>0p > 0p>0).
  • The claim is an exact equality of real numbers for each individual nnn, not an approximation, a bound, an asymptotic statement, or a limit.
  • Because 0≤p0 \le p0≤p forces 1+p≥1>01 + p \ge 1 > 01+p≥1>0, both denominators are nonzero under the hypothesis, so no degenerate division arises anywhere in the right-hand side; the hypothesis 0≤p0 \le p0≤p is exactly what rules out the one value p=−1p = -1p=−1 at which the displayed expression would divide by zero, and it additionally excludes all other negative ppp.
  • The hypothesis 0≤p0 \le p0≤p is satisfiable (e.g. p=0p = 0p=0), so the statement is not vacuous.
  • Degenerate instances are included by the quantifier over nnn. At n=0n = 0n=0 both sides are 000: the left side by the base case, the right side because 0⋅p1+p=00 \cdot \frac{p}{1+p} = 00⋅1+pp​=0 and (−p)0=1(-p)^{0} = 1(−p)0=1 makes the bracket 1−1=01 - 1 = 01−1=0 — this uses the convention that the zeroth power is 111 even when p=0p = 0p=0. At n=1n = 1n=1 both sides equal ppp. At p=0p = 0p=0 the recursion gives S0(n)=0S_0(n) = 0S0​(n)=0 for all nnn and the right-hand side is 000 as well.
  • The statement mentions only the recursively defined SpS_pSp​ and the arithmetic expression above; it makes no reference to the separately defined closed-form function hitsClosed of the bundle, nor to any of the runtime or token-cost quantities defined there, even though the right-hand side is written with the same symbols as that definition's body.

No proof is supplied in the declaration; the proof position is occupied by a placeholder, so the file asserts the identity without establishing it.

Human review
  • Endorsed by Shuze Chen · Sep 12, 2026

  • Endorsed by naimengye · Sep 12, 2026

    Confirmed by the mission captain (proposal self-audit).

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