Pooling homogenous studies vs. using meta-analysis/bayesian I am working with a student who has collected about 300 participants for his thesis. After collecting 100 participants, he analyzed the data. I had just recently read of a "mini-meta-analysis" strategy where someone collects pilot data, computes the estimates of interest, then performs a larger study and aggregates the estimates later using meta-analysis (Cumming, 2014). I suggested the student do something similar: run the analysis an additional time and use meta-analysis to pool the estimates. 
The student asked what would be the advantage of using a meta-analysis relative to just analyzing all 300 participants. I couldn't provide a good answer (unless there was a lot of heterogeneity between the samples, which there isn't). 
So, now to my question: if we have the raw data from 2+ homogenous studies, is there an advantage to computing the estimates twice then meta-analyzing them vs. computing them once?
My first impressions are:


*

*The first study allows you to "calibrate" the model, then cross-validate in the second model. However, in this case, his statistical analysis was strictly confirmatory (and no "calibration" was required, aside from computing parameter estimates). 

*Having two samples allows us to do empirical bayesian analysis (we can use the parameters estimated in sample 1 as priors in sample 2 and take advantage of the strengths of bayesian statistics). 
Is there another reason to favor one strategy over another?

Cumming, G. (2014). The new statistics: Why and how. Psychological science, 25(1), 7-29.
 A: I will address frequentist meta-analysis, for which the answer is: The two approaches will give asymptotically equivalent point estimates if (1) you are using an effect size measure satisfying a property I will give below; and (2) the samples are homogenous not only in true effect size, but also in within-study variance. 
Let $\mathbf{X}$ be your entire sample of size 300, and let $\mathbf{X}_1$ and $\mathbf{X}_2$ be the first and second parts of this sample (of sizes 100 and 200). 
Let $\widehat{y}_i$ with $i \in {1,2}$ be point estimates from each sample $\mathbf{X}_1$ and $\mathbf{X}_2$. They have within-study variances $\sigma_i^2$, assumed fixed and known (the usual assumption in meta-analysis), and assumed equal by homogeneity. Let $\widehat{y}_R$ be the pooled point estimate from a random-effects meta-analysis, and let $\tau^2$ be the estimated heterogeneity (the variance of the true effects). Suppose you're meta-analyzing a statistic $g(\cdot)$. 
Let's check whether the meta-analytically pooled point estimate matches the simple estimate pooling all the data (call it the aggregate-data estimate). 
$$\eqalign{
\widehat{y}_R &:= \frac{ \sum_{i=1}^2 \frac{1}{\tau^2 +\sigma^2_i }\widehat{y}_i }{\sum_{i=1}^2\frac{1}{\tau^2 +\sigma^2_i }} \\
&\to \frac{ \sum_{i=1}^2 \frac{1}{\sigma^2_i }\widehat{y}_i }{\sum_{i=1}^2\frac{1}{\sigma^2_i }} \tag{$\tau^2 \to 0$, the truth}\\
&= \frac{ \frac{1}{\sigma^2}\widehat{y}_1 + \frac{1}{\sigma^2}\widehat{y}_2}{\frac{2}{\sigma^2}} \\
&= \frac{ \widehat{y}_1 + \widehat{y}_2}{2}
}$$
(The penultimate line comes from assuming heterogeneity within-study variances.) 
Now, this last expression is a simple average of the two samples' point estimates. So that means that if you choose a test statistic such that:
$$\frac{ g(\mathbf{X}_1) + g(\mathbf{X}_2) }{2} = g(\mathbf{X}) \tag{*}$$
then the meta-analytic estimate will be asymptotically equivalent to your aggregrate-data estimate. $(*)$ holds, for instance, if $g(\cdot)$ is the sample mean, but not if $g(\cdot)$ is the odds ratio. (However, note that you wouldn't want to meta-analyze untransformed odds ratios anyway since they do not fulfill the often-used normality assumption.) 
Here is a code example to illustrate the equivalence when the statistic of interest is the sample mean. The inference also appears to be quite similar. 
n1 = 100
n2 = 300 - n1
theta = 2  # true mean
sigw = 1  # common within-study variance

# generate whole dataset
X = rnorm( n1 + n2, mean = theta, sd = sigw )

# split into 2 subsamples
X1 = X[1:n1]
X2 = X[1:n2]

# get point estimates and SEs for each subsample
ests = c( mean(X1), mean(X2) )
ses = c( sd(X1) / sqrt(n1),  sd(X2) / sqrt(n2) )

# meta-analyze them
library(metafor)
ES = escalc( measure = "MD", yi = ests, sei = ses )
m = rma.uni(ES, method = "REML")

# compare point estimate to aggregate analysis
m$b; mean(X)

# compare inference to aggregate analysis
sqrt(m$vb); sd(X) / sqrt(n1 + n2)

