Given $n$ numbers, where the value of each number is different, denoted as $v_1, v_2, ..., v_n$, and the probability of selecting each number is $p_1, p_2, ..., p_n$, respectively.
Now if I select $K$ numbers based on the given probabilities, where $K \leq n$, what is the expectation of the sum of those $K$ numbers? Note that the selection is without replacement, so that the $K$ numbers cannot involve duplicate numbers. I understand that if the selection is with replacement, the expectation of the sum of the $K$ numbers equals $K \times E(V)$, where $$E(V) = v_1 \times p_1 + v_2 \times p_2 + ... + v_n \times p_n.$$
Furthermore, what about the expectation of the variance of those $K$ numbers?
I am a CS PhD student who is working on a big data problem, and I don't have any statistics background. I expect that someone can give me a formula as the answer. However, if the answer is too complicated to be described by a formula or intensive computation has to be involved, an approximate answer is totally acceptable.
You can assume $n$ here is quite large, and the probability can vary a lot. In practice, the values of those probabilities come from a query log, which records a series of aggregation queries. The point is that the frequency of each number involved in the queries can be quite skew, i.e., some are rarely queried, while some are queried very frequently. You can assume the probability distribution is normal distribution, zipf distribution or any other reasonable alternatives.
The value distribution is only a contiguous subset of any possible distribution. In other words, if you have a histogram that represents a certain distribution, the all the numbers involved in this problem are the numbers all within a single bucket.
In terms of the value of K, you may assume it is always less than the number of frequently queried elements.