@KyleCSN It all depends how the coefficient in Cuped is fit. Using pooled or weighted average from two groups. Explained here in more detail. https://t.co/9j1dQ8LiyV
@gregakespret@ronnyk It is based on this paper:
https://t.co/sF8LYps0bY
It is using Bayes Factor, but use a prior trained with historical experiment data.
@ronnyk@drsimonj 2/2 since mean of covariates for treatment and control (variant and base) are the same, they will be canceled when computing the delta. Therefore in the actual computation we don't compute it at all. The adjusted metric is simply metric-theta*covariate
@deaneckles@ronnyk@drsimonj Also cuped is not exactly regression without interaction with treatment assignment term. We still estimate two theta for treatment and control group, then use a weighted average of the two theta in adjustment. We also normally think about super population PATE.
@deaneckles@ronnyk@drsimonj Yes I’m aware of his paper. Using same theta is finite sample unbiased (at least under Gaussian case) reason is thetahat is function of sample covariance and variance and they are independent of sample mean.
@deaneckles@ronnyk@drsimonj It is an in sample fit but we force the same theta for treatment and control, Vs. use separate theta fit by treatment and control in sample data. The asymptotic difference between this and regression adjustment is negligible for our typical applications.
@lukasvermeer@LizzieEardley Two things are mixed up here. What Lizzie pointed out is easy to tackle with details method. What @lukasvermeer is saying is denominator movement make the metric movement hard to interpret. But when session are hardly moved, using it as denominator leads to more sensitive metrics
@rbelo@ronnyk@deaneckles@seanjtaylor Welch test assumes normality or CLT to kick in. For highly skewed dist, equal size T vs C makes CLT kick in way faster.