Theorem: Result C: complete-record gap and sample certificate

Randomize the four settings independently with probability $1/4$. Suppose their ideal mean ranges have gap $G_T>0$, including any certified mean-model errors. Suppose the uniform record second-moment bounds above hold.

If the split record is always available, define the predeclared bounded statistic

It belongs to $[0,1]$. If instead the split record is independently erased with common known probability $1-\gamma$, $0<\gamma \le 1$, replace the second term by $a_i{I}_{\mathrm{observed}}{c}_L(z)/(6L)$, scoring an erased record by $1/2$.

Let actual complete attempted-run laws be within total variation $e_h$ of the corresponding specified laws, uniformly over the joint families. Then the actual oriented expectation gap is at least

where $\gamma =1$ in the always-recorded case.

If $g_{\ast }>0$, the midpoint of the enlarged null and alternative expectation ranges gives a test with each error at most

after $N$ independent attempted runs from a fixed actual family member. Thus

suffices for each error to be at most $\alpha$. A simultaneous calibration coverage probability at least $1-\beta$ gives unconditional error at most $\alpha +\beta$.

Proof. Since ${\max }_i|a_i|=3$ and $|c_L|\le L$, the statistic is bounded. Randomization gives

The independent-erasure condition gives the factor $\gamma$. Equation (eq:clipping) reduces the ideal gap by at most $B/L$. For a statistic in $[0,1]$, total variation changes its expectation by at most $e_h$, giving (eq:g).

The midpoint is at least $g_{\ast }/2$ from each allowed expectation. Hoeffding's one-sided inequality gives $\exp [-2N(g_{\ast }/2{)}^2]$ Hoeffding1963. Uniformity follows from the same expectation bounds for every family member. Finally, combine test and calibration failure events by a union bound.