# Derive househould weights from a uniformly distributed person sample

The Swiss Public-Use Sample of the national census is a 5% sample drawn from the entire census survey. According to the documentation, the persons are sampled uniformly without replacement. Persons are the "primary sampling unit", the selection probability and person weights are all equal: $\pi = 0.05, w = 1/\pi = 20$.

For each person there are household-level attributes as well. The sample is not uniformly distributed anymore when considering the households: Larger households have a bigger sampling probability than smaller households. In order to carry ot analyses at the household level (with variables such as "number of children in the household"), the records must be reweighted.

Intuition suggests that each person record $p$ should be weighted with $w_p / k_p$ (with $k_p$ being the number of persons in the household). This should hold exactly if the persons were drawn with replacement, and at least approximately in the "without replacement" scheme. (The sample size is by five orders of magnitude larger than the maximum household size.) However, I find it difficult to provide a formal argument for this.

Is my intuition correct? How would a proof or a probabilistic argument look like? Can you suggest a textbook that treats similar examples, perhaps even this?

• Is your intent for the data to have one row per household? If so, what are you intending to do when there is more than one person in the sample from a household? Somehow you will have to consolidate the two (or more) points of individual data into an overall estimate of the "household"'s value for each variable. I'm presuming any such analysis will only be on variables that make sense to think of at a household level. – Peter Ellis Apr 8 '12 at 7:04
• @PeterEllis: For analyses at the person level, I can use $w_p$ as weights. So you're right: I want to analyze at the household level, with variables such as "number of children in the household". I'm not concerned too much about households occuring twice or more often in the sample, as I hope that the reweighting will fix this. After all, what's the difference between drawing twice from the same five-children household and drawing from two different five-children households, at least for the analysis? Or am I too optimistic? – krlmlr Apr 8 '12 at 7:10

I think you should conceptualise your new sample as a weighted sample with replacement of households from the original population of households. Then the weighting scheme comes naturally according to standard principles - weights are proportional to the inverse of the probability of a household being chosen in the sample.

• Thank you. This seems to support my intuition. I have found inverse probability weighting on Wikipedia, and it seems that the Horvitz-Thompson estimator is a special case of applying i.p.w. to estimate the mean. I still need to figure out if it matters for my analysis that the household-level weights are now according to a SRSWR strategy. – krlmlr Apr 8 '12 at 7:51

Sampling without replacement is full of various stumbling blocks, often not very intuitive. A concatenation of two samples drawn WITH replacement is a sample with replacement again; a concatenation of two WOR samples is not a WOR sample, for instance.

If you have an SRSWOR sample of size $n$ from the population of size $N$ (American survey statisticians hate the European ones who can just use the population registers!), then the probability of selecting a group of size $k$ (e.g., a household of this size) is

$${\rm Prob}[\mbox{at least 1 member of the group is in the sample}] =$$ $$1 - {\rm Prob}[\mbox{no member of the group is in the sample}] =$$ $$1 - {k \choose 0}{N-k \choose n}\biggl/{N \choose n} = 1 - \frac{(N-k)!(N-n)!}{(N-k-n)!N!}$$ $$\sim 1 - \frac{(N-k)^{N-k+1/2} (N-n)^{N-n+1/2}}{(N-k-n)^{N-k-n+1/2}N^{N+1/2}}$$ $$= 1 - \frac{(1-k/N)^{N-k+1/2} (1-n/N)^{N-n+1/2}}{(1-(k+n)/N)^{N-k-n+1/2}}$$ $$\sim 1 - \frac{\exp(-k)(1-k/N)^{-k+1/2} \exp(-n)(1-n/N)^{-n+1/2}}{\exp(-k-n)(1-(k+n)/N)^{-k-n+1/2}}$$ $$\sim 1 - \frac{(1+k(k-1/2)/N)(1+n(n-1/2)/N)}{1+(k+n)(k+n-1/2)/N}$$ $$\sim 1 - \biggl[1+\frac{k(k-1/2)}N +\frac{n(n-1/2)}N - \frac{(k+n)(k+n-1/2)}N\biggr] \sim 2kn/N$$ where the first approximation uses Stirling's approximation for the factorial, the second one, the exponential limit (I cheated with the power 1/2 though), and the subsequent ones, the Taylor series expansion of $(1+x)^a$ near $x=0$ that ignores the terms $o(1/N)$. I don't know where the factor of 2 is creeping in, it seems totally out of whack there. The first order approximation to the probability of selection should instead be $kn/N=kf$ where where $f=0.05$ is the (individual) sampling fraction. I'd appreciate it if somebody could point out where I made a mistake. A good sampling book should give correct derivation.

A useful approximation is Poisson sampling with the rate $kn/N$. Under this approximation, the probability of having zero hits is $\exp(-kn/N)$, and the probability of at least one hit is thus $1-\exp(-kn/N) = 1 - \exp(-kf)$. So instead of $w(k)=1/(kf)$, as you initially suggested, you'd want to have $w(k)=\bigl[1-\exp(-kf)\bigr]^{-1}$; for the households of size 10, the difference between the two numbers will be notable.

• I did a quick plot of $1/(kf)$ against $1/(1 - \exp(-kf)$ using a free plotting service, indeed both are similar but the second term is slightly larger than the first. -- In the second paragraph, you mention a sample with replacement. However, my starting point is a person sample without replacement. Also, I don't remove duplicates from the household-level data. Could you please explain the initial assumption and the meaning of the resulting weights? – krlmlr Apr 9 '12 at 19:17
• On your second comment: that was a typo, I am talking about SRSWOR, as the formulas suggest (or at least I thought they did :)). – StasK Apr 10 '12 at 1:53
• Thanks for the clarification. But how does your derivation account for households that might be drawn twice or more often in the SRSWOR person sample? – krlmlr Apr 10 '12 at 6:30
• On the most recent one: by using the ratios of the number of combinations, rather than the products of probabilities that would be valid for sampling with replacement. – StasK Apr 11 '12 at 9:25
• Because you asked: I have added my version of the derivation as a separate answer, please feel free to copy it and use it in your answer if you think it is correct. – krlmlr Apr 12 '12 at 6:53

Following up Peter's answer, here is my solution to the problem.

A given household of size $k$ must be reweighted using $w = 1/E[K]$ with $K$ being a random variable that gives the number of persons sampled for the household. The variable $K_i$ is approximately distributed like $k$ independent Bernoulli trials with success probability $\pi$ each, its expecation is $E[k] = k\pi$. My initial intuition was correct! This is even more surprising given that the sampling probabilities of a household follow a hypergeometric distribution which can be approximated by a bionomial distribution, and the expected value of this distribution is beautiful again. (Even more: The mean = expected value of the hypergeometric distribution is also $k\pi$ in our case -- ho need to approximate at all! This also means that it doesn't matter if the person sample was taken with or without replacement.)

The result is given in Cochran (1977), Chapter 9A. (Not Chapter 11, the results there are similar but the underlying sampling scheme is different.) There, they call "clusters" what are "households" in our case. However, this book derives the weights directly, without taking expectation values of membership counts. These expected values are mentioned in a 2009 lecture by James R. Chromy, Some Generalizations of the Horvitz-Thompson Estimator. (So, nothing really new...)

Following up Stas' answer, here is an alternative approach to deriving the sampling probaility for selecting no member of a group of size $k$:

$${\rm Prob}[\mbox{at least 1 member of the group is in the sample}] = \\ 1 - {\rm Prob}[\mbox{no member of the group is in the sample}] = 1 - \pi_0 = \\ 1 - {k \choose 0}{N-k \choose n}\biggl/{N \choose n} = 1 - \frac{(N-k)!(N-n)!}{(N-k-n)!N!} \\ \pi_0 = \frac{(N-k)!}{N!}\frac{(N-n)!}{(N-k-n)!} = \prod_{i=1}^{k}\frac{i+N-k-n}{i+N-k} \\ \approx \left(1 - \frac{n}{N}\right)^k \\ 1 - \pi_0 \approx 1 - \left(1 - \frac{n}{N}\right)^k$$

The approximation uses $k \ll n$ and $\frac{n}{N - k} \approx \frac{n}{N}$.

• This is much more transparent than my derivation. – StasK Apr 12 '12 at 19:43
• I'd challenge this in a sense that you need to use the probabilities of selection to reweight, not the average number of times the household will be selected. It is the inverse probability that figures in Horvitz-Thompson, and that's what needed to achieve unbiasedness of this estimator. – StasK Apr 13 '12 at 13:22
• I didn't mention it in my reply, but I conducted a simulation experiment for that purpose, and from this experiment I know that only $w = 1/(k\pi)$ correctly estimates the population total. And this is, I'm sure not only by accident, the expected number of chosen household members. I would interpret this as a generalization of inverse probability weighting in the case of sapmling with replacement. If not -- which weights would you suggest otherwise? I tried also $1/(1-\pi_0)$, but this constantly overestimated counts of the larger households in my simulation. – krlmlr Apr 13 '12 at 14:26
• @StasK: Please note the comment above and the edits to the answer. – krlmlr Apr 13 '12 at 14:34