Deriving Inverse-Variance Fixed-Effect Meta-Analysis

Why Combine Cohort Estimates by Precision?

A genetic association may be estimated in several cohorts after their alleles, phenotype definitions, and effect scales have been aligned. The cohorts may differ in sample size and therefore in the precision of their estimates. Fixed-effect meta-analysis combines those estimates under the assumption that they estimate one shared effect. Its central idea is to give more influence to estimates with smaller sampling variance.

The method is not a simple average of cohort estimates unless their variances happen to be equal. It is also not a claim that the observed estimates must be identical: sampling variation can make estimates differ even when the shared-effect model is appropriate. The derivation below shows exactly which estimate this method produces and how its uncertainty follows from the model.

Definition: Suppose cohort $i$ reports an effect estimate $\hat{\beta}_i$ and its standard error $SE_i$, for $i=1,\ldots,K$. In an inverse-variance fixed-effect meta-analysis, the cohort estimates are modeled as independent normal observations with a common true effect $\beta$: $$ \hat{\beta}_i \sim N(\beta, SE_i^2). $$ The reported $SE_i$ values are treated as known for this calculation. The fixed-effect pooled estimate is the inverse-variance weighted estimate of the shared effect.

This model is an approximation appropriate when the cohort-level estimates have approximately normal sampling distributions. The effect estimates must refer to the same allele contrast, phenotype, and effect scale; otherwise, they are not estimates of the same parameter. Independence is also required for the likelihood and variance formulas used here. In practice, the standard errors are estimated from the cohort data, but are treated as fixed inputs to the usual meta-analysis calculation.

Deriving the Pooled Estimate

For each cohort, let $w_i=1/SE_i^2$ denote its inverse-variance weight. Under the model, the density of $\hat{\beta}_i$ is a normal density centered at $\beta$, with variance $SE_i^2$. Independence means the joint likelihood is the product of these cohort densities. Taking its logarithm gives a sum:

\ell(\beta) = -\frac{1}{2}\sum_{i=1}^{K} \left[ \log(2\pi SE_i^2) +\frac{(\hat{\beta}_i-\beta)^2}{SE_i^2} \right].

The first term inside the sum does not depend on $\beta$. To find the likelihood maximum, differentiate the remaining expression with respect to $\beta$, set the derivative equal to zero, and solve:

\frac{d\ell(\beta)}{d\beta} =\sum_{i=1}^{K}\frac{\hat{\beta}_i-\beta}{SE_i^2} =\sum_{i=1}^{K}w_i(\hat{\beta}_i-\beta)=0.

\left(\sum_{i=1}^{K}w_i\right)\beta =\sum_{i=1}^{K}w_i\hat{\beta}_i \quad\Longrightarrow\quad \hat{\beta}_{FE} =\frac{\sum_{i=1}^{K}w_i\hat{\beta}_i} {\sum_{i=1}^{K}w_i}.

This stationary point is a maximum, not a minimum: the second derivative of the log likelihood is $-\sum_i w_i$, which is strictly negative because every standard error is positive. Thus the inverse-variance weighted estimate is the maximum-likelihood estimate of the shared effect under the stated model.

Theorem: Under the independent shared-effect normal model, the inverse-variance fixed-effect estimate is $$ \hat{\beta}_{FE} =\frac{\sum_{i=1}^{K}\hat{\beta}_i/SE_i^2} {\sum_{i=1}^{K}1/SE_i^2}. $$ Each cohort's contribution is proportional to its precision, $1/SE_i^2$. A cohort with half the standard error has four times the weight.

Deriving the Standard Error

The pooled estimate is a weighted sum of independent estimates. Since the weights are fixed by the reported standard errors, its sampling variance is the sum of the weighted variances. Substituting $w_i=1/SE_i^2$ gives:

\operatorname{Var}(\hat{\beta}_{FE}) =\operatorname{Var}\left( \frac{\sum_i w_i\hat{\beta}_i}{\sum_i w_i} \right) =\frac{\sum_i w_i^2 SE_i^2}{(\sum_i w_i)^2} =\frac{\sum_i w_i}{(\sum_i w_i)^2} =\frac{1}{\sum_i w_i}.

The third equality uses $w_i^2SE_i^2=w_i$. Taking the square root yields the pooled standard error. The same result follows from the curvature of the normal log likelihood: its negative second derivative is the total precision $\sum_i w_i$.

SE(\hat{\beta}_{FE}) =\sqrt{\frac{1}{\sum_{i=1}^{K}1/SE_i^2}}.

If the normal approximation is suitable, a 95% confidence interval is $\hat{\beta}_{FE}\pm 1.96SE(\hat{\beta}_{FE})$. For testing $H_0:\beta=0$, the Wald statistic is $Z=\hat{\beta}_{FE}/SE(\hat{\beta}_{FE})$, compared with the standard normal distribution under the null. The estimate and interval retain the units of the cohort effect scale—for example, trait standard deviations per effect allele or log odds per effect allele.

Worked Example: Pooling Three Cohort Estimates

Suppose three hypothetical, independent cohorts report effects on the same scale, in trait standard deviations per effect allele. Their estimates and standard errors are $0.08$ and $0.04$, $0.12$ and $0.06$, and $0.02$ and $0.03$, respectively. Assume their estimates are approximately normal and share one true effect.

Calculate the weights. The inverse-variance weights are $1/SE_i^2$:

w_1=\frac{1}{0.04^2}=625,\qquad w_2=\frac{1}{0.06^2}=277.7778,\qquad w_3=\frac{1}{0.03^2}=1111.1111.

The total weight is $625+277.7778+1111.1111=2013.8889$. The weighted numerator is $625(0.08)+277.7778(0.12)+1111.1111(0.02)=105.5556$, using the displayed rounded weights. Therefore:

\hat{\beta}_{FE} =\frac{105.5556}{2013.8889} \approx 0.0524138.

Using the unrounded values gives the same result to the displayed precision. The pooled standard error is:

SE(\hat{\beta}_{FE}) =\sqrt{\frac{1}{2013.8889}} \approx 0.0222835.

A 95% confidence interval is $0.0524138\pm1.96(0.0222835)$, or approximately $0.00874$ to $0.09609$ trait standard deviations per effect allele. Under the shared-effect model, this interval is compatible with a positive effect and excludes zero. The Wald statistic for $H_0:\beta=0$ is $0.0524138/0.0222835\approx2.352$, corresponding to a two-sided standard normal p-value of about $0.0187$. This is evidence against the zero-effect null under the model; it is not the probability that the null is true.

The third cohort has the smallest standard error and hence the largest weight. Its estimate contributes most to the pooled result, but the other estimates still contribute according to their precision.

The Likelihood Interpretation

The weighted estimate is not only an arithmetic rule. It is the value of the shared effect that makes the observed cohort estimates most likely under the model. The likelihood penalizes values of $\beta$ according to the squared distance between each cohort estimate and that value, scaled by that cohort's variance. A precise cohort has a larger penalty for a given discrepancy, because its estimate is expected to vary less.

Completing the square makes this interpretation explicit. The weighted sum of squared deviations from a candidate $\beta$ decomposes into a part that does not depend on $\beta$ and a part centered at $\hat{\beta}_{FE}$:

\sum_i w_i(\hat{\beta}_i-\beta)^2 =\sum_i w_i(\hat{\beta}_i-\hat{\beta}_{FE})^2 +\left(\sum_i w_i\right)(\beta-\hat{\beta}_{FE})^2.

To verify the identity, write $\hat{\beta}_i-\beta=(\hat{\beta}_i-\hat{\beta}_{FE})+(\hat{\beta}_{FE}-\beta)$, square, and sum with weights. The cross-product is zero because $\sum_iw_i(\hat{\beta}_i-\hat{\beta}_{FE})=0$, which is precisely the weighted-mean equation. The second term in the decomposition is minimized when $\beta=\hat{\beta}_{FE}$, confirming the likelihood maximum.

Worked Example: Seeing Precision in the Likelihood

Consider two hypothetical independent cohorts estimating the same effect in trait standard deviations per effect allele. Their estimates are $0.10$ with standard error $0.05$, and $-0.02$ with standard error $0.10$. The respective weights are $400$ and $100$, so:

\hat{\beta}_{FE} =\frac{400(0.10)+100(-0.02)}{400+100} =\frac{38}{500} =0.076, \qquad SE(\hat{\beta}_{FE})=\sqrt{\frac{1}{500}}\approx0.04472.

The first cohort's estimate is more precise and receives four times the weight. To compare likelihood at $\beta=0$ with its maximum, use the weighted squared deviations. At zero, they sum to $400(0.10^2)+100((-0.02)^2)=4.04$. At the pooled estimate $0.076$, they sum to $400(0.10-0.076)^2+100(-0.02-0.076)^2=0.2304+0.9216=1.152$. The log-likelihood difference is therefore:

\ell(0)-\ell(0.076) =-\frac{1}{2}(4.04-1.152) =-1.444.

Thus the likelihood at zero is $\exp(-1.444)\approx0.236$ times the maximum likelihood, with the same normalizing constants in both likelihoods. This comparison illustrates the model's relative support for parameter values; it is not a posterior probability that the effect equals a particular value.

Worked Example: Checking the Pooled Likelihood Maximum

Take three hypothetical independent cohort estimates on the same scale: $0.10$ with standard error $0.05$, $0.04$ with standard error $0.10$, and $0.07$ with standard error $0.05$. Their weights are $400$, $100$, and $400$. The weighted estimate and its standard error are:

\hat{\beta}_{FE} =\frac{400(0.10)+100(0.04)+400(0.07)}{900} =\frac{72}{900}=0.08, \qquad SE(\hat{\beta}_{FE})=\sqrt{\frac{1}{900}}=0.03333.

At $\beta=0.08$, the weighted squared deviations total $400(0.02^2)+100((-0.04)^2)+400((-0.01)^2)=0.16+0.16+0.04=0.36$. At $\beta=0$, the total is $400(0.10^2)+100(0.04^2)+400(0.07^2)=4+0.16+1.96=6.12$. Hence:

\ell(0)-\ell(0.08) =-\frac{1}{2}(6.12-0.36) =-2.88, \qquad \frac{L(0)}{L(0.08)}=\exp(-2.88)\approx0.0561.

The shared-effect likelihood is maximized at $0.08$, as the derivation predicts. The likelihood ratio here compares two specified parameter values under the model; it does not measure the probability that the shared-effect assumption is correct.

Assumptions, Scope, and a Common Pitfall

The fixed-effect calculation is most directly justified when the estimates are aligned to the same estimand and can be treated as independent, approximately normal measurements of one shared effect. Its standard error reflects sampling uncertainty under that model. It does not incorporate extra uncertainty from effects genuinely differing across cohorts. If the shared-effect assumption is not scientifically appropriate, the pooled result may not summarize the cohort effects in a useful way.

Independence matters to the variance derivation: the variance of a sum would include covariance terms if cohorts shared participants or otherwise had correlated estimates. The formula $1/\sum_iw_i$ would then not generally apply. Likewise, using an effect estimate on a different scale or with the opposite allele orientation would break the common-parameter interpretation. Harmonization and phenotype alignment are prerequisites, not adjustments supplied by the weighting formula.

Pitfall: treating the pooled estimate as an unweighted mean. Averaging the three estimates in the first example gives $(0.08+0.12+0.02)/3=0.0733$, not the inverse-variance estimate $0.05241$. The arithmetic mean ignores that the cohort standard errors differ; the fixed-effect likelihood leads to inverse-variance weights.
Pitfall: reading fixed effect as identical observed effects. The model assumes one shared underlying parameter, not equal estimates in every sample. Different estimates are expected through sampling variation.
Pitfall: interpreting a confidence interval or likelihood as a probability for the parameter. A frequentist confidence interval procedure has 95% coverage under repeated sampling when its assumptions hold. The interval from one analysis is not a 95% probability statement about the fixed parameter.

Key takeaway: Under independent normal sampling around one shared effect, inverse-variance weights arise by maximizing the joint likelihood. The pooled estimate is the weighted mean, and its standard error is the reciprocal square root of total precision.

Check Your Understanding

Use the shared-effect model and show the calculation steps where appropriate.

Two cohorts report standard errors of $0.08$ and $0.04$. What is the ratio of their inverse-variance weights, and which cohort receives greater weight?
State the assumptions needed to use the independent normal likelihood for cohort effect estimates.
Explain why differentiating the log likelihood gives a weighted mean rather than an ordinary mean.
If the total inverse-variance weight is $625$, what is the pooled standard error?
Why does the standard error formula $1/\sqrt{\sum_i w_i}$ need modification when cohort estimates are correlated?