I'm using a probability proportional to size method to draw a sample of clustered units. The distribution of the size variable is positively skewed, giving the largest units an inclusion probability greater than 1 (which doesn't make sense as a probability).

I'm using the UPtille function from the sampling package to sample units. The function should return a 0 or a 1 to indicate inclusion in the sample. Using my inclusion probabilities - where a few are > 1 - I get a vector with other numbers too.

What do I do about these inclusion probabilities that are greater than 1, and what is the implication of having units that are basically automatically included in the sample because of how big they are?


1 Answer 1


The usual approach in this situation is to set inclusion probabilities to $1$ when $\pi_i\geq 1$ and to repeat the computation for the remaining units. You do this iteratively until there are no more inclusion probabilities greater than $1$.

inclusionprobabilities function in the sampling package does this for you.

Units with $\pi_i=1$ are called self-selecting units, and the implication is that they will be always present in your sample, and will not contribute to variance.

Be careful if self-selecting units are too many.

  • $\begingroup$ Thanks for pointing me towards inclusionprobabilities - but if calculating by hand, you would calculate the inclusion probabilities for all other units as if the large units (with p > 1) weren't there, and then assign them with probability of 1? Can you please expand on 'will not contribute to variance' with respect to self-selecting units? $\endgroup$ Feb 10, 2018 at 1:18
  • $\begingroup$ Yes, if any $\pi_i \geq 1$, you set them to $1$, then ignore them and repeat computations (at this step you multiply by $n-k$, where $k$ is the number of self-selecting units). If $\pi_i = 1$, then unit $i$ will be always in your sample, no matter how many samples you draw, so unit $i$ is not a source of variability for the HT estimator. To see this numerically, you may try to put $\pi_i = 1 $ in the Horvitz-Thompson variance (in this case $\pi_{ij}=\pi_j$ ), you'll see that for any couple $(i, j)$ the contribution to variance is $0$. $\endgroup$
    – Roberto
    Feb 10, 2018 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.