# Are sampling designs characterized by their first and second-order inclusion probabilities?

If you have two sampling designs $p_1(\cdot)$ and $p_2(\cdot)$ (defined on the same set of samples $\mathcal{S}$ from the same population $\mathcal{U}$) such that they induced the same inclusion probabilities of first and second-order, then $p_1(s) = p_2(s)$ for all sample $s$ in $\mathcal{S}$. Is this sentence true?

• No, it's not. There are standard examples of three random variables being dependent but pairwise independent: search our site to find them. The emphasis on first- and second-order probabilities is that these determine the expectations and variances of important sampling distributions for many of these sampling designs.
– whuber
Jun 1, 2017 at 22:01
• You might be asking about sampling from finite populations with stratification, clusters, unequal probabilities, etc. I think this conclusion is not true in general but I don't have a counterexample. It might be true for many commonly used designs. I can't think why this is interesting or useful. Jun 1, 2017 at 22:27
• Maybe this is a counterexample: Suppose $p_1$ sample the individual with probability 0.1, 0.2 and 0.3 one by one. I devise the $p_2$ as following: Grouping the individuals into 3 groups according to their selection probabilities. Then sampling the group according to that probabilities. Then $p_1$ and $p_2$ have the same inclusion probability, but $p_1(s) \ne p_2(s)$ for some $s$. Jun 1, 2017 at 23:35
• Thank you for your comments. @whuber I can't see how those counterexamples are related to sampling designs. Can you explain a little more please? Jun 2, 2017 at 5:39
• @David Smith you're right, my question is in the context of sampling from a finite population and I think about it because I read in a book (Cassel, Sarndal and Wretman (1977) Foundations of inference in survey sampling)that first- and second-order inclusion probabilities characterized simple random sampling, but I can't find a proof. Jun 2, 2017 at 5:40

The assertion is not necessarily true, as the following counterexample demonstrates.

Let $\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}$ be a population of four elements. Consider two sampling plans.

1. Select $\omega_4$ no matter what. Independently select $\omega_1,\omega_2,\omega_3$, each with a chance of $1/2$ of being included.

There are eight possible samples, each with a chance of $1/8$. A convenient way to write them is with the binary inclusion vector $s=(s_1,s_2,s_3,s_4)$ where $s_i=1$ when $\omega_i$ is in $s$ and $s_i=0$ otherwise. The possible samples (all of which are equally likely) can be denoted $$\mathcal{S}_1=\{0001,0011,0101,1001,0111,1011,1101,1111\}.$$

2. The second sampling plan draws one element uniformly from this set of possible samples: $$\mathcal{S}_2=\{0001,0111,1011,1101\}.$$

Regardless of the plan, write $\pi_i$ for the chance that $\omega_i$ is in a sample and $\pi_{ij}$ for the chance that $\omega_i$ and $\omega_j$ are both in a sample. As you can readily check, these plans were constructed so that $\pi_i=1/2$ for $i=1,2,3$ and $\pi_4=1$. Moreover, $\pi_{ij}=1/4$ for $i\lt j\lt 4$ and $\pi_{i4}=1/2$ in both plans. Thus, both have identical first and second order inclusion probabilities, yet they are not the same: sampling plan $(1)$ can result in four samples that are impossible in sampling plan $(2)$.

• You're welcome. Did you notice the connection with pairwise independent but not independent variables? The random variables $s_1,s_2,s_3$ have that property in the second sampling plan. That's how this example was constructed. ($\omega_4$ was thrown in only to stave off objections concerning a sampling plan that could produce samples with zero elements.)
– whuber
Jun 2, 2017 at 14:50
• I need to think it over, I was having breakfast while I read your answer. Jun 2, 2017 at 15:03