I am trying different methods for propensity score matching, and I have an issue on matching with replacement.

I use standardized difference to measure the "quality" of my match, and for this specific case where I can have the same control for different pairs I do not get how to compute the mean of any variable for my control group.

I know how many different control observation are used to form the pairs and the number of times each control is used to form a pair. I heard it has something to do with weighted mean, but I do not get how/why.

Example: Let say A is my case group and B the control group, weight the variable. matched sample

count the number of time each control is used to form a pair

The mean of case group a is just (70+60+90)/3 (if I am not mistaken) But what about the mean (and variance) of the control group.

Thank you for your help.

  • $\begingroup$ To be clear, you are asking for help on how to compute a weighted mean and (weighted) variance? $\endgroup$ – Repmat May 31 '17 at 6:43
  • $\begingroup$ I believe when using matching with replacement specific measures have to be taken when estimating the causal effect but also the step when checking for imbalance among the covariates. Among the solutions I heard about is boostraping for the estimaaation of causal effect (the variance), but also weighted mean for the computation needed to check for imbalance. Indeed standardized differences uses the mean and variance (for continuous variables) for both group in the matched sample. $\endgroup$ – FFF Jun 1 '17 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.