Here is a very simple “by‐hand” counter–example which shows that neither of the two implications

(i) if t**–t* < n then winnum>0, and t**–t* ≥ n then winnum≤0
(ii) if winnum>0 then t**–t* < n, and if winnum≤0 then t**–t* ≥ n

holds in general. In fact one can do everything in log‐space so that xᵢ>0 is automatic.

— Take n=2. Let L₁=L₂=0 so that the 2-point moving average at t=2 is M₂=0.
— At t*=3 we must choose L₃ so that it sits exactly d below the average of the preceding 2 logs:

 M₃ = (L₂+L₃)/2, and we force L₃ = M₃ – d.  
 

For instance pick d=0.5; then

 L₃ = (0 + L₃)/2 – 0.5  
 ⇒ 2 L₃ = L₃ – 1  
 ⇒ L₃ = –1,  
 ⇒ M₃ = (0 + (–1))/2 = –0.5.  
 

So at t*=3 we have “initial” log‐value L₃ = –1 = M₃ – 0.5.

— Now for t=4 and t=5 we keep L₄ and L₅ just below their 2-point averages so that no crossing occurs yet. For example

 L₄ = –1.1  ⇒  M₄ = (L₃+L₄)/2 = (–1 + (–1.1))/2 = –1.05,   L₄ < M₄ ⇒ no crossing  
 L₅ = –1.2  ⇒  M₅ = (L₄+L₅)/2 = (–1.1 + (–1.2))/2 = –1.15,  L₅ < M₅ ⇒ no crossing.  

— Finally at t=6 let L₆ = 0. Then

 M₆ = (L₅ + L₆)/2 = (–1.2 + 0)/2 = –0.6,  and  L₆ = 0 > –0.6 ⇒ we do get a crossing from below.  
 

The game ends at t**=6, so

 t** – t* = 6 – 3 = 3  which is ≥ n = 2,  
 winnum = L₆ – L₃ = 0 – (–1) = +1.  

That single numerical example already shows:
• t–t* ≥ n** but winnum>0, so the one‐way implication “if (t**–t*)≥n then winnum≤0” fails.
• Conversely one also easily flips signs to break “if winnum>0 then t**–t*<n.”

So there is no need to look any further for a “tight” counterexample—the statement simply does not hold in general.

As for part B of your question (“if it were true, is it known?”), the answer is that because the equivalence is false no standard textbook is going to record it under that heading. It is just a little algebraic accident of the sliding‐window sum, and as soon as you choose the interleaving values cleverly (as above) you can make the “breakout” happen after the initial point has dropped out of the window and still end up with a positive log‐gain.

In short:

1. Counter‐examples exist for all n≥2 by exactly the same construction (hang below the moving average for n–1 steps, then jump above).
2. No classical reference for the false statement—one only finds varying sufficient-condition theorems when extra hypotheses (monotonicity, convexity, distributional assumptions, etc.) are imposed.