The random walk in the proof of Lemma S3 #
This file proves the bound on the random walk (S_t) of Lemma 2 in the proof of Lemma S3
(Section S1.6 of the supplementary material). The random walk is realized under stdGaussSeq on
the space of its increments, and pr denotes probability. Let s ∈ (0, 1), a = 1 − s²,
q ∈ (1, 1/a), and M = max[1, max_{1 ≤ t ≤ T} exp{aS_t²/(2t)}]. Then h_t ≤ M for
0 ≤ t ≤ T, and E(M) ≤ N'^{aq}/(1 − aq) ≤ exp[aq log(2 + 2 log T/log q) + log{1/(1 − aq)}],
where N = 1 + ⌊log T/log q⌋ and N' = 2N ≤ 2 + 2 log T/log q. Together with Lemma 2(ii), which
gives η_T = sup_τ E(h_{τ∧T}), these bounds give Lemma S3 (log_le_of_le_iSup_lintegral_reward).
walkMax s T: the maximumM, as a function of the increments of the random walk;walkBlock q T k: the blockB_k = {t ≤ T : q^k ≤ t < q^{k+1}};walkBlockCount q T: the numberN = 1 + ⌊log T/log q⌋; the blocksB_kwithk ≥ Nare empty (walkBlock_eq_empty);submartingale_exp_mul_walk:(exp(cS_t))_{t ≥ 0}is a submartingale for everyc ∈ ℝ;measure_exists_le_walk_le: eq. (S5), by Doob's maximal inequality (maximal_ineq);measure_exists_le_abs_walk_le: eq. (S5) with|S_t|in place ofS_tand an additional factor 2;measure_walkBlock_le: eq. (S6);measure_exists_le_sq_walk_div_le: the union bound over the nonempty blocks;lt_walkMax_iffandmeasure_lt_walkMax_le: forx ≥ 1,M > xexactly whenmax_{1 ≤ t ≤ T} S_t²/t > (2/a) log x, andpr(M > x) ≤ min{1, 2N x^{−1/(aq)}};lintegral_walkMax_eq,one_add_lintegral_min_rpow, andlintegral_walkMax_le:E(M) = 1 + ∫_1^∞ pr(M > x) dx ≤ 1 + ∫_1^∞ min(1, N'x^{−p}) dx = pN'^{1/p}/(p − 1)withp = 1/(aq), so thatE(M) ≤ N'^{aq}/(1 − aq);log_lintegral_walkMax_le:log E(M) ≤ aq log(2 + 2 log T/log q) + log{1/(1 − aq)};iSup_lintegral_reward_le_lintegral_walkMaxandlog_le_of_le_iSup_lintegral_reward:sup_τ E(h_{τ∧T}) ≤ E(M), so that everyη ≤ sup_τ E(h_{τ∧T})satisfies the bound of Lemma S3.
We prove eq. (S5) for −S_t by Doob's maximal inequality applied to the submartingale
exp(−xS_t/u). In the paper, it follows from eq. (S5) for S_t by symmetry. The statements hold
for every T ≥ 0, with log 0 = 0 as in Mathlib, while Lemma S3 assumes T ≥ 1.
The increments of the random walk #
The submartingale exp(cS_t) #
For every c ∈ ℝ, (exp(cS_t))_{t ≥ 0} is a submartingale with respect to the filtration
generated by (S_t). For an event A of this filtration at time t, the increment S_{t+1} − S_t
is N(0, 1) and independent of 1(A) exp(cS_t), so that
E{1(A) exp(cS_{t+1})} = E{1(A) exp(cS_t)} exp(c²/2) ≥ E{1(A) exp(cS_t)}.
Doob's maximal inequality, eq. (S5) #
Doob's maximal inequality applied to the submartingale exp(cS_t): for all c, y ∈ ℝ,
pr(max_{t ≤ u} cS_t ≥ y) ≤ e^{−y} E{exp(cS_u)} = exp(uc²/2 − y).
Eq. (S5): for x ≥ 0 and u ≥ 1, pr(max_{t ≤ u} S_t ≥ x) ≤ exp{−x²/(2u)}, by Doob's
maximal inequality applied to the submartingale exp(xS_t/u).
Eq. (S5) for −S_t: for x ≥ 0 and u ≥ 1, pr{max_{t ≤ u} (−S_t) ≥ x} ≤ exp{−x²/(2u)},
by Doob's maximal inequality applied to the submartingale exp(−xS_t/u).
Eq. (S5) with |S_t| in place of S_t and an additional factor 2: for x ≥ 0 and u ≥ 1,
pr(max_{t ≤ u} |S_t| ≥ x) ≤ 2 exp{−x²/(2u)}.
The blocks, eq. (S6), and the union bound #
The number N = 1 + ⌊log T/log q⌋ of the proof of Lemma S3. For q > 1, the blocks B_k with
k ≥ N are empty (walkBlock_eq_empty). The floor is taken in ℕ; it agrees with the integer
floor of the paper since log T/log q ≥ 0 for T ≥ 1 and q > 1.
Instances For
For q > 1, the block B_k is empty for k ≥ N: an element t of B_k satisfies
q^k ≤ t ≤ T, so that k ≤ log T/log q.
Eq. (S6): for z ≥ 0, pr(max_{t ∈ B_k} S_t²/t ≥ z²) ≤ pr(max_{t ≤ u} |S_t| ≥ zm^{1/2}) ≤ 2 exp{−z²m/(2u)} ≤ 2 exp{−z²/(2q)}, where m and u are the smallest and the largest elements of
a nonempty block B_k, and u/m < q. If B_k is empty, the event is empty.
The union bound in eq. (S6) over the nonempty blocks: for z ≥ 0,
pr(max_{1 ≤ t ≤ T} S_t²/t ≥ z²) ≤ 2N exp{−z²/(2q)}. Every t with 1 ≤ t ≤ T lies in the block
B_k with k = ⌊log t/log q⌋, and k < N since this block is nonempty.
The maximum M and its tail #
The maximum M = max[1, max_{1 ≤ t ≤ T} exp{aS_t²/(2t)}] of the proof of Lemma S3, with
a = 1 − s², as a function of the increments ω of the random walk: the maximum over 0 ≤ t ≤ T
of the terms 1 for t = 0 and exp{aS_t²/(2t)} for t ≥ 1.
Equations
Instances For
For x ≥ 1, pr(M > x) ≤ min{1, 2N x^{−1/(aq)}}, with a = 1 − s²: by lt_walkMax_iff,
the union bound with z² = (2/a) log x applies.
The expectation of M #
For N' ≥ 1 and p > 1, 1 + ∫_1^∞ min(1, N'x^{−p}) dx = pN'^{1/p}/(p − 1): the integrand
equals one on (1, x₀] and N'x^{−p} on (x₀, ∞), where x₀ = N'^{1/p}.
E(M) ≤ N'^{aq}/(1 − aq), with a = 1 − s² and N' = 2N. With p = 1/(aq) > 1,
lintegral_walkMax_eq, measure_lt_walkMax_le, and one_add_lintegral_min_rpow give
E(M) = 1 + ∫_1^∞ pr(M > x) dx ≤ 1 + ∫_1^∞ min(1, N'x^{−p}) dx = pN'^{1/p}/(p − 1), and
pN'^{1/p}/(p − 1) = N'^{aq}/(1 − aq).
The supremum of Lemma 2(ii) #
Since h_t ≤ M for 0 ≤ t ≤ T, sup_τ E(h_{τ∧T}) ≤ E(M), where τ ranges over the stopping
times of the filtration generated by (S_t), as in Lemma 2(ii).
The bound of Lemma S3 for every η ≤ sup_τ E(h_{τ∧T}). Since Lemma 2(ii) gives
η_T = sup_τ E(h_{τ∧T}), the case η = η_T is Lemma S3.