Under PPSWOR of size $2$ or $3$, inclusion probability of $i$th unit exceeds that of $j$th unit when $i$th unit has largr size measure than $j$th unit

Question

I was reading PPSWOR (Probability Proportional to Size and WithOut Replacement) sampling and this question suddenly struck me

Is it true that under PPSWOR of size $2$ or $3$, inclusion probability of $i$th unit exceeds that of $j$th unit whenever $i$th unit has a larger size measure than $j$th unit?

I can't decide for sure whether it's true or not. All I could think of are-

Selection probability of unit $i$ is $$p_i=\frac{x_i}{\sum x_i}$$ Since we have a PPSWOR distribution, it remains the same in each draw.

Probability of selecting unit $i$ and $j$ is $p_i\cdot p_j$.

So, possible samples are $(1,2)$, $(1,3)$ and $(2,3)$.

But, I can't proceed any more than this. Can anyone please help?

Also, is there any $n>3$ for which PPSWOR of size $n$ satisfies the given condition?

Also, I am not sure about the tags. Feel free to edit them.

Answer 1

Having larger items being more likely to be sampled overall than smaller items seems obviously true. Here is a demonstration for a sample of size $2$ without replacement, where the weights are $x_1, x_2, x_3, \ldots$ with finite $s=\sum x_i$.

The probability of selecting item $1$ in a sample of size $2$ without replacement is $$p_1= \frac{x_1}{s} + \frac{x_2}{s} \frac{x_1}{s-x_2} + \sum\limits_{k\ge 3} \frac{x_k}{s} \frac{x_1}{s-x_k}= \frac{x_1}{s-x_2} +x_1 \sum\limits_{k\ge 3} \frac{x_k}{s(s-x_k)}$$

The probability of selecting item $2$ in a sample of size $2$ without replacement is $$p_2=\frac{x_2}{s} + \frac{x_1}{s} \frac{x_2}{s-x_1} + \sum\limits_{k\ge 3} \frac{x_k}{s} \frac{x_2}{s-x_k}= \frac{x_2}{s-x_1} +x_2 \sum\limits_{k\ge 3} \frac{x_k}{s(s-x_k)}$$

If $x_1>x_2$ and $s>x_1+x_2$ then

• $\frac{x_1}{s-x_2}>\frac{x_2}{s-x_1}$ since their difference is $(x_1-x_2) \frac{s-x_1-x_2}{(s-x_1)(s-x_2)}$

• $x_1 \sum\limits_{k\ge 3} \frac{x_k}{s(s-x_k)} > x_2 \sum\limits_{k\ge 3} \frac{x_k}{s(s-x_k)}$

so $p_1>p_2$ as expected. You could do something more complicated for larger samples without replacement.

This does not mean that item $1$ is more likely to be selected than item $2$ in the second round of sampling. For example with weights $x_1=90, x_2=9, x_3=1$, you get

• the probability of item $1$ being selected in the first round being $0.9$ and being selected in the second round about $0.0981$ for a total of about $0.9981$,

• the probability of item $2$ being selected in the first round being $0.09$ and being selected in the second round about $0.8109$ for a total of about $0.9009$,

• the probability of item $3$ being selected in the first round being $0.01$ and being selected in the second round about $0.0910$ for a total of about $0.1010$.