Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 207 / 1

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.

Let ${\hat{\delta }}_D$ and ${\hat{\delta }}_I$ be independent, approximately unbiased estimates of one consistent treatment effect, with variances $v_D,v_I>0$. The inverse-variance weighted mean givesIndeed, weights $w,1-w$ give variance $w^2{v}_D+(1-w{)}^2{v}_I$; differentiating gives $w=v_I/(v_D+v_I)$, the weight on the direct estimate. Here the three displayed trials give ${\hat{\delta }}_D=-1.33$, $v_D=0.04$, and hence $\hat{\delta }\simeq -1.3203$ with variance $0.03448$. This numerical pooling uses the displayed evidence, not the unavailable data from all twelve trials. A shared study or shared control would introduce covariance, requiring the corresponding correlated-estimator formula; inconsistency would invalidate the common-effect interpretation.

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.

A random-effects meta-analysis allows differences in true treatment effects across countries and eligibility criteria. For trial $i=1,\dots ,6$, let $y_i$ be its estimated log odds ratio and $v_i$ its estimated within-trial variance. A usual approximate statistical model isindependently across trials, with $b_i$ independent of sampling errors. Here $\mu$ is the mean true log odds ratio in the population of comparable trials, and ${\tau }^2\ge 0$ is between-trial heterogeneity, not additional sampling error. Therefore $y_i\sim N(\mu ,v_i+{\tau }^2)$, and for a fitted heterogeneity value,One can estimate ${\tau }^2$ by restricted maximum likelihood or another justified method; $({\sum }_i{w}_i{)}^{-1}$ is a plug-in conditional variance and does not fully account for estimating heterogeneity. With only six studies that uncertainty matters. Different eligibility rules motivate this random effect but do not by themselves establish exchangeability or remove bias; known effect modifiers may warrant stratification or regression.

Use leave-one-study-out influence analysis: fit all six trials, then omit Cohen 1989 and refit both $\mu$ and ${\tau }^2$. Report $\hat{\mu }-{\hat{\mu }}_{(-C)}$, the corresponding change in the odds ratio, and changes in the confidence interval and heterogeneity. As a diagnostic with heterogeneity held fixed, writing $W={\sum }_i{w}_i$ givesThe fitted Cohen weight would be $(0.15+{\hat{\tau }}^2{)}^{-1}$ and $y_C=-1.95$. A precise but discordant trial can have large influence; refitting assesses additional influence through heterogeneity. The other five trials' data are absent, so a numerical six-trial influence assessment is not identifiable from the displayed table.