Short answer: Your statements A and B are false. You can construct sequences where the crossing happens in fewer than n steps but the log-return is negative, and also sequences where the crossing takes at least n steps but the log-return is positive. So there’s no such deterministic relation in general.

First, a quick clarification: you wrote winnum = log(tstarstar) − log(tstar). If tstarstar > tstar (the game ends after it starts), that’s always positive, contradicting your note that winnum could be negative. I assume you meant winnum = log(x_{tstarstar}) − log(x_{tstar}) (i.e., a log-return). That’s what I use below.

Notation: let y_t = log x_t and let the n-sample moving average be A_t = (1/n)∑{k=0}^{n−1} y{t−k}. Note the identity y_t ≥ A_t iff y_t ≥ average of the previous n−1 logs, because A_t = (y_t + ∑{k=1}^{n−1} y{t−k})/n, so y_t ≥ A_t ⇔ (n−1)y_t ≥ ∑{k=1}^{n−1} y{t−k}.

This shows the “cross from below at time t” depends only on y_t relative to the average of the previous n−1 values, not on y_{t*} specifically. That freedom lets you force counterexamples by choosing intermediate values suitably.

Counterexample 1 (violates: “if t**−t* < n then winnum must be positive”):

• Take n = 3, d = 1.
• Logs: y1 = 0, y2 = 10, y3 = 0, y4 = −100, y5 = −40.
• At t* = 3: A3 = (0 + 10 + 0)/3 = 10/3 ≈ 3.33, and y3 = 0 ≤ A3 − d (0 ≤ 2.33), so the game starts.
• At t = 4: A4 = (−100 + 0 + 10)/3 = −30, and y4 = −100 < A4 (still below).
• At t** = 5: A5 = (−40 − 100 + 0)/3 ≈ −46.7, and y5 = −40 ≥ A5, so we crossed from below at t** = 5.
• Duration t** − t* = 2 < n.
• Winnum = y5 − y3 = −40 − 0 = −40 < 0 (negative). So duration < n but negative return, contradicting A (and also contradicts the “only if” direction B: “if winnum not positive then duration ≥ n”).

Counterexample 2 (violates: “if t**−t* ≥ n then winnum must be non-positive” and also B’s “if winnum positive then duration < n”):

• Take n = 3, d = 1.
• Logs: y1 = 0, y2 = 10, y3 = 0, y4 = −100, y5 = −100, y6 = 1.
• At t* = 3: same start condition as above holds.
• At t = 4: A4 = (−100 + 0 + 10)/3 = −30, y4 = −100 < A4 (below).
• At t = 5: A5 = (−100 − 100 + 0)/3 = −66.7, y5 = −100 < A5 (still below).
• At t** = 6: A6 = (1 − 100 − 100)/3 ≈ −66.3, and y6 = 1 ≥ A6, so we cross at t** = 6.
• Duration t** − t* = 3 ≥ n.
• Winnum = y6 − y3 = 1 − 0 = +1 > 0 (positive). So duration ≥ n but positive return, contradicting A and B.

Why this works (intuition):

• The ending condition y_{t**} ≥ A_{t**} is equivalent to y_{t**} being at least the average of the previous n−1 logs. By making the intermediate y’s (between t* and t**) very small, you can make that average arbitrarily low, allowing a crossing with y_{t**} well below or well above y_{t*}, at practically any chosen duration. Conversely, by keeping intermediate values low, you can delay the crossing past n steps while making y_{t**} > y_{t*}. There’s no inherent constraint tying the sign of y_{t**} − y_{t*} to whether the gap is less than n or not.

Is this a known result?

• Because the statements are false, there’s no positive result of this form in the literature. More broadly, “moving average crossover” rules are common in technical analysis, but there are no general deterministic implications like the ones you claim that hold for arbitrary sequences of positive numbers.

If you actually meant winnum = log(t**/t*), then it is always positive since t** > t*, making the discussion moot.