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For independent studies comparing treatments $A$ and $B$ to common control $C$, a consistent log odds ratio comparison uses ${\hat{\delta }}_{AB}={\hat{\delta }}_{AC}-{\hat{\delta }}_{BC}$, with variance $v_{AC}+v_{BC}$. This follows from subtracting two estimates of additive effects relative to the same reference. With overlapping evidence, subtract twice their covariance. Causal comparability across the studies is supplied by transitivity in network meta-analysis, not by the variance calculation.

A network meta-analysis jointly compares several treatments using a graph of direct Randomized controlled trials and indirect treatment comparisons. If effects are measured relative to a reference treatment, consistency means ${\delta }_{ij}={\delta }_{i0}-{\delta }_{j0}$. A connected graph identifies all relative effects under that model. Transitivity in network meta-analysis is the substantive comparability assumption supporting these relations; inconsistency can be assessed when the network has loops.

Write $y_{WP}$ and $y_{AP}$ for the estimated log odds ratios relative to the common control. Indirect treatment comparison uses the consistency relation ${\delta }_{WA}={\delta }_{WP}-{\delta }_{AP}$. For independent trials, subtraction adds their variances, givingThe substantive assumption is transitivity in network meta-analysis: the study populations, outcome definitions and follow-up are sufficiently comparable, particularly in their distributions of effect modifiers, that the two control comparisons estimate effects applicable to the same target population. The common control's name alone does not justify that assumption. This also assumes comparable marginal odds ratios; these need not equal covariate-adjusted odds ratios even without confounding.

Using the supplied rounded variances and the given standard normal quantile, the approximate confidence interval isOn the odds ratio scale this is $e^{-1.26}\simeq 0.284$, with approximate limits $(0.104,0.771)$. The estimated odds of recurrent stroke under treatment $W$ are about 28% of those under $A$, a reduction of about 72% in odds. The interval excludes equal odds, so this comparison provides evidence of lower odds under $W$ if the indirect treatment comparison assumptions hold. These are odds, not a 72% reduction in probability; a confidence interval describes the long-run coverage of the procedure, not a posterior probability for this particular interval.

Under the null hypothesis of consistency, ${\hat{\delta }}_D-{\hat{\delta }}_I$ has approximately zero expectation and variance $v_D+v_I$, since the displayed trials are independent. Thus a two-sided Wald test usesThe observed difference is $-0.07$, and $Z=-0.07/\sqrt{0.29}\simeq -0.130$. Its two-sided p-value is approximately $0.897$, so there is no detectable discrepancy here. Failure to reject is not proof of consistency: this test has limited statistical power, and transitivity in network meta-analysis also needs substantive assessment. If evidence sources overlap, replace $v_D+v_I$ by $v_D+v_I-2\mathrm{Cov}({\hat{\delta }}_D,{\hat{\delta }}_I)$.

A network meta-analysis can offer three distinct benefits, assuming a connected treatment network and credible transitivity in network meta-analysis.

These are potential benefits, not guarantees: heterogeneous effect modifiers, inconsistent evidence, disconnected networks or imprecise rankings can defeat them.