2. Why Meta-analysis? Part I.
Suppose you have two metrics, A and B.
• Metric A is significant at the 5% level in 5% of all
experiments.
• Metric B is significant at the 5% level in 40% of all
experiments.
All else equal, which should you trust more?
3. Why Meta-analysis? Part I.
Metric A suggests no experiment ever has any effect. All
noise! Our best guess of the true effect = zero.
Metric B suggests some experiments have an effect. Our
best guess of the true effect = estimated effect x 0.82 (if
everything is normal and mean zero).
4. Why Meta-analysis? Part II.
Imagine this is the histogram
of estimated effect sizes from
historical experiments.
You see a 2% lift in your new
experiment. What should you
infer?
5. Why Meta-analysis? Part II.
Each observed effect, y, is the sum of
• the true treatment effect, t
• sampling error, e
If y is very large, likely in part due to a large draw of e. So
should adjust y downwards ("shrink") to offset this, and get a
better estimate of t.
6. Why Meta-analysis? Part III.
Consider a test which improves metric A ("comments") but
degrades a related metric B ("posts").
What, if anything, can we conclude from this?
Can the movement in B give us more information about the
movement in A?
7. Why Meta-analysis? Part III.
Silly example:
• the true lift in each experiment is t ~iid F, equal for both
metrics
• observed values are yA = t + eA, yB = t + eB, for
independent eA, eB
• metric B's contains extra information about metric A
More generally, might have some joint distribution of (tA, tB,
yA, yB), estimated on past experiments.
8. Conclusion
Tech companies run lots of experiments, but they often fall
into a small number of experiment types.
Ignoring the information in past, highly related experiments
is leaving a lot on the table!
Our paper on experiment splitting develops some of these
issues.
9. Appendix
Where does the 0.82 number come from?
Consider the model where
• the true effect t ~ N(0, vt)
• the sampling error e ~ N(0, ve)
• the observed outcome is y = t + e
Can show
• E(t | y) = y x vt/(vt + ve).
• If the fraction of rejections at the 5% level is p, then
vt/(vt + ve) = 1 - (Φ-1(p/2)/1.96)2.