# Can I make a proportional-to-size without replacement sample (PPS WOR) self weighted?

Let's say I have 100 schools and each has a different number of students. I want to estimate which % of students are in schools with electricity. Simulation and theory indicate it is more efficient to give a higher probability to schools with a larger number of students.

If I do a proportion-to-size with replacement sample, I am able to get a self weighted sample, where my estimator is a plain average of the sample (no weighted needed). Below is my logic (if you already know this is correct, you can skip to the questions):

Let's say we have:

• A sample selected proportional to the student size of the school (with replacement)
• A sample size 2

Based on this, let's say we got the following sample: 1, 0 (with 1 being the school has electricity); a good estimate of the proportion of students with electricity I think would be:

$$s_i =$$ number of students school i

$$s_t =$$ total number of students

$$\bar{y} =$$ estimate of proportion of students that have electricity $$\bar{y} =\frac{\frac{s_1 \frac{1}{\frac{s_1}{s_t}} + s_2 \frac{0}{\frac{s_2}{s_t}}}{2}}{s_t}$$

which is the same as $$\bar{y} =\frac{\frac{s_t\cdot1 + s_t\cdot0}{2}}{s_t} = \frac{1 + 0}{2} = 0.5$$ Which is the sample average.

Question 1: Does this look right? Basically, the selection probability compensates for the school size, and the average now represents the proportion of students rather than the proportion of schools.

Question 2: Is there a way that I can design my sample so the calculation is a plain average of the sample with sampling without replacement? I have read a few papers, and I haven't found an answer to this question.

Taking a design-based approach where $$t_i$$ represents the total number of students in each school and is considered fixed, $$p_i$$ is the probability of selecting school $$i$$, and $$W_i$$ is a (random) variable that indicates whether a school was selected we can easily show that the weighted sample total is an unbiased estimator for the total.
\begin{aligned} \mathbb{E}(\hat{t}) &= \sum^N_{i = 1} W_i \cdot \frac{t_i}{p_i} \\ &= \sum^N_{i = 1} \mathbb{E}(W_i) \cdot \frac{t_i}{p_i} \\ &= \sum^N_{i = 1} p_i \cdot \frac{t_i}{p_i} \\ & = t \end{aligned}
This holds whether sampling is with replacement or without replacement as long as $$p_i$$ is fixed. Of course, the variance for the two cases is not the same.