Change-of-measure pigeonhole with separate deficit and penalty totals

Let $\iota$ be a finite index set of cardinality $k\ge 1$, and let $V,V^t,W:\iota\to\mathbb R$ with $V\ge 0$, $\sum_j V_j=T_0^{\mathrm{full}}$ and $\sum_j V^t_j=T_0$. Suppose that for every $j$

Then some index $j$ satisfies

This is the change-of-measure summation step of Lattimore--Szepesv'ari, Bandit Algorithms, section 38.7 (eq. 38.24), with the deficit and the Pinsker penalty carried by two different count vectors. In the intended application the observable fed to the bounded change-of-measure inequality is a truncated count, because Step 2 needs it bounded by $\Theta(n/D)$ rather than by $n$; the divergence, however, is controlled by the full count of the $j$-th pair, since the two Markov decision processes differ in that pair's transition row at every round. Hence the deficit $T_0-V^t_j$ uses the truncated counts, whose total $T_0$ is bounded below, while the Cauchy--Schwarz penalty uses the full counts, whose total $T_0^{\mathrm{full}}$ is only bounded above. The single-total form is the special case $V^t=V$, $T_0^{\mathrm{full}}=T_0$.