Inclusion probabilities for a survey with unequal probability of selection

I am attempting to run a simulation in R that requires me to take a survey where the probability of selection for each individual is not equal.

I have a factor covariate $x$ which has levels $x_1, x_2, ..., x_5$ that is known for every member of the population before sampling, and I wish to give individuals a probability of selection based off their $x$, so that individuals with $x=x_1$ have the lowest chance of selection, and individuals with $x=x_5$ have the highest. So let's suppose $x_1 =1, x_2=2,..., x_5=5$

Suppose I therefore let $\pi_i = \frac{x_i}{\sum_{i=1}^N x_i}$ where $\pi_i$ is the chance of selection for person $i$.

With this in mind, I need to derive the probability of being included in a sample of size $n$ from the population of size $N$. In simple random sampling, this is just $\frac{n}{N}$, but I'm unsure how to derive it in this instance.

Likewise, for the purpose of variance estimation, I need to derive the joint inclusion probability for two individuals $i$ and $j$, the probability both are included in the sample. Again, in simple random sampling this is just $\frac{n(n-1)}{N(N-1)}$, but I'm not sure what it is here.

The reason I need the inclusion probabilities is so that I can find the sample weights, which are needed for estimation of my total $Y$ and its variance.

• Do you select your sample with or without replacement? Jun 1 '17 at 21:43
• You don't have enough information. You need to know the distribution of $x$ in the population. Jul 9 '18 at 14:58

Suppose we take a sample of size $n$, and let's define $I^{(i)}_j$ as the indicator of whether or not individual $i$ is chosen on the $j^\text{th}$ selection (it equals one if the person is selected and otherwise it equals zero). Then

\begin{align} P( \text{individual $i$ is included in the sample}) &= P \left ( \sum_{j=1}^{n} I_j^{(i)} = 1 \right ) \\ &= \text{E} \left ( \sum_{j=1}^{n} I_j^{(i)} \right ) \\ &= \sum_{j=1}^{n} \text{E} \left ( I_j^{(i)} \right ) \\ &= n \pi_i . \end{align}

We can check that when we're sampling uniformly and $\pi_i = 1 / N$ we do in fact recover the original formula $n / N$. A similar technique can be used to find the probability that individuals $i$ and $j$ are both included in the sample

\begin{align} P(\text{individuals $i$ and $j$ are in the sample}) &= P \left [ \left ( \sum_{k=1}^{n} I_k^{(i)} \right ) \left ( \sum_{l=1}^{n} I_l^{(j)} \right ) = 1 \right ] \\ &= \text{E} \left [ \left ( \sum_{k=1}^{n} I_k^{(i)} \right ) \left ( \sum_{l=1}^{n} I_l^{(j)} \right ) \right ] \\ &= \text{E} \left ( \sum_{k=1}^{n} \sum_{l=1}^{n} I_k^{(i)} I_l^{(j)} \right ) \\ &= \sum_{k=1}^{n} \sum_{l=1}^{n} \text{E} ( I_k^{(i)} I_l^{(j)} ) . \end{align}

Now $I_k^{(i)} I_l^{(j)} = 0$ whenever $k = l$ (we can't select both individuals at the same time) so we're left with the remaining $2 \binom{n}{2} = n (n - 1)$ terms which by symmetry are all equal. The sum above therefore becomes

\begin{align} n (n - 1) \text{E}( I_1^{(i)} I_2^{(j)} ) &= n (n - 1) P(I_1^{(i)} = 1 \cap I_2^{(j)} = 1) \\ &= n (n - 1) P(I_1^{(i)} = 1) P(I_2^{(j)} = 1 \mid I_1^{(i)} = 1) \\ &= n (n - 1) \pi_i \frac{x_j}{\sum_{k \neq i} x_k} . \end{align}

Again we can compare this with the original formula where we have uniform sampling by setting $x_i = 1$ for $1 \leq i \leq N$ and our new formula does reduce to $\frac{n (n - 1)}{N (N - 1)}$.

Converting my comment to an answer.

You don't have enough information to answer the question specifically. Rather, it depends on the distribution of $x$ in the sampling frame (which we may assume to be the whole population).

Trivially, if you have a target sample of $n=$(say)$300$. That means that that sample will be apportioned 5:4:3:2:1 (15 parts) to members with $x=5$, $x=4$, $\ldots$, and $x=1$ respectively. (so 100, 80, 60, 40, 20 respectively). Once you know the distribution of $x$ in the population, you just calculate the SRS sampling probability 5 times. For instance, if $X=1$ has 1,000 people in the population, then the sampling probability is $20/1000 = 0.005$ and so on and so forth.

That is the answer to the problem along with a description of the missing information.