I'm currently doing my homework for my stats class and I have a question for one of my exercises:

Consider a population P of size N. We define a new sampling scheme in the following way: we first select a sample SA using a simple random sampling without replacement of size n1. Then, we sample SB in P outside of SA according to a simple random sampling without replacement of size n2. We obtain in a such a way the final sample S as S = SA ∪ SB.

Obtain the probability P(S = s) for a realization s (probability mass function). Is it equivalent to a simple random sampling with sample size n = n1 + n2?

My questions is:

I don't really understand the question and don't know what to answer. What kind of PDF am I supposed to describe? What does he mean by P(S=s) for some realization s?

In my understanding the question is asking what the probability is that the sample will contain a certain random variable. Can someone maybe give me a more concrete example so I can work it out?


Unless I am interpreting wrong, here is my translation of the problem

In a population of size $P$, take a simple random sample of size $n_1$ without replacement. Next, without putting anything from this first sample back, take a second simple random sample of size $n_2$. Let our sample $S$ be the union of these two smaller samples. Is this the same as taking one large simple random sample of size $n_1 + n_2$ from the original population of size $P$ ?

A realization is, put simply, an observed value

For $\mathbb{P}(S=s)$, remember what $S$ (big s) represents. It is defined as the union of two disjoint samples, so it would be a set of objects. $s$ (little s) could be any particular set of objects from the population.

  • $\begingroup$ thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function? $\endgroup$ – Nicola Zaugg Nov 25 '18 at 12:04
  • $\begingroup$ We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}\choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}\choose{n_1+n_2}$ $\endgroup$ – WaveX Nov 25 '18 at 16:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.