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I have a question regarding the use of propensity score in a survival analysis with use of mutliple imputation to handle missing data. The question is of theoretical nature and may well apply to other situations.

I have a data set of n individuals. The aim is to estimate the effect of a treatment on a binary outcome (death). The analysis is based on propensity score; the propensity score is derived by means of logistic regression, which includes 30 predictors variables. Effect estimation is carried out by means of Cox regression (which uses the propensity score in various ways [stratification, covariate adjustments etc]). There are a large number of patients, and on average 2–7% missing for each variable (of which there are 30 included in the prop. score).

Thus, I have a large data set with a substantial amount of missing data (at least in terms of complete cases) which is why I use multiple imputation - 5 complete data sets are imputed. Now the question is what to do with the muliply imputed data sets; which one of the strategies below should I prefer?

1. Calculate one average propensity score for each individual using the 5 separate data sets. That way, each individual will have one propensity score, which is the average from the n complete data sets. Then do the Cox regression..

2. Analyze each separate multiply imputed data set (with Cox regression), and then pool the 5 hazard ratio estimates to one hazard ratio.

The second method appears to be used more often, but is it better/worse?

Any thoughts about this?

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You could simply try both approaches and see which one performs better under cross-validation.

To hazard a guess, my intuition is that #2 might perform better because essentially the propensity scores from pooled imputed data will have a different distribution from the propensity scores of cases that required no imputation.

In symbols, if $X$ is the random variable corresponding to the case you want to impute, $p(X)$ is the propensity score and $h(X, t) = e^{\beta p(X)} h_0(p(X), t)$ is the hazard learned by the Cox regression on propensity score, approach 1 will estimate the hazard as $$e^{\beta E(p(X))} h_0(E(p(X)), t),$$ whereas approach 2 will estimate the hazard as $$E[e^{\beta p(X)}h_0(p(X), t].$$ These agree when $X$ is not random (i.e. does not require imputation), but disagree when $X$ is random. To me, the second one seems better-justified, but I'm not certain that this would work out well in practice.

One way you could test whether this is a problem is by including in your Cox regression an indicator for whether the case required imputation, and see how the effect of this indicator changes between methods 1 and 2.

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  • $\begingroup$ Thanks for the answer. I might have failed to explain that in both cases (1 and 2), the data is "complete" after multiple imputation (as opposed to complete case analysis). You definitely have a good point about trying both and comparing them under cross validation, but I thought there might be some guidelines on this. D'agostino (1998) argues that pooling should be used (stat.ubc.ca/~john/papers/DAgostinoSIM1998.pdf), whereas Hill (2008) votes for averaging of the propensity score (academiccommons.columbia.edu/catalog/ac:129151). $\endgroup$ Mar 15, 2015 at 10:52
  • $\begingroup$ I tried both pooling the separate HRs, and averaging the prop score; it yielded roughly the same estimates. However, it is considerably technically easier to do the prop score averaging, as it allows you to work with just 1 data set instead of 5 or more. $\endgroup$ Mar 15, 2015 at 10:54
  • $\begingroup$ @Ben, in what way do you mean the two approaches agree when X is non-random? I see that the expressions agree, but in the OP's set-up, even without missing values, the two approaches would give different results. $\endgroup$
    – swmo
    Mar 20, 2015 at 9:11
  • $\begingroup$ @swmo "even without missing values, the two approaches would give different results"--really? If there were no missing values, the five imputed datasets would be the same, so the pooled propensity scores would be equal to the original propensity scores, and hence the model trained on the pooled data would be the same as the model trained on an individual dataset. Right? $\endgroup$
    – Ben Kuhn
    Mar 20, 2015 at 16:12
  • $\begingroup$ @Ben, I misread both the question and your answer. I see what you mean. $\endgroup$
    – swmo
    Mar 20, 2015 at 16:45

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