Suppose we have a population 1,2,3,4,5 and a sample selected 1,2,3. when using WR we know 1/N equals 1/5. The prob of 1,2,3 for being selected is 1/5. But in case of WOR how it is 1/N? As the definition of SRS says, each unit has an equal prob of being selected and in WOR, the unit is not returned, so wont it be 1/5 for 1, 1/4 for 2 and 1/3 for 3, So how does the definition hold?
I hope you get it, I couldn't explain clearly due to confusion
I assume WR = replacement and WOR = without replacement.
Simple random sampling assumes every group of members in the current sample space has an equal chance of being selected.
When you make multiple selections, the sample space will reduce for WOR and stay the same for WR. For each selection, the probability will be 1/(N - k) for WOR where k previous selections have been made.
I meant to say, for Simple Random Sampling, all groups of the same size have an equal chance of being selected.
A particular unit can be selected in any of the n draws. its selection probability at ith draw conditional on the event that it is not selected in previous i-1 draws is 1/(N-i+1)*{(N-i+1)/N}=1/N for each i=1....N. Summing over i, the probability that ith unit is selected in the sample is n/N which is same for each unit.
Probability of selection will remain the same at each draw.
Furthermore through basic concept of SRS, every possible sample has an equal chance of being selected as a sample from all possible samples (Cochran, 1977).
Please have a look at the linked references (obtained via the first Google page...). The formulas are semi-trivial to reestablish...
https://www.ma.utexas.edu/users/parker/sampling/repl.htm
https://www.ma.utexas.edu/users/parker/sampling/woreplshort.htm
https://en.wikipedia.org/wiki/Sampling_%28statistics%29
It is simple: example for the second selected item.
Imagine you have a group of N items.
In the first draw, you see clear that the probability that an item is in the sample is 1/N.
You also see clear that conditioned to the fact that a different item than item “a” has been drawn in the first draw the probability that item “a” is drawn in the second draw is 1/(N-1).
The key is “conditioned to”.
Now Bayes rule. In the first draw, any (whatever) of the others N-1 items could be drawn with probability 1/N. So the probability of drawing item “a” in the second extraction conditioned to any of the other items has been drawn in the first extraction is, by bayes rule, (N-1)/N * 1/(N-1) = 1/N.
Nadia has explained through words but Jose M.Pavia has given a legitimate theoretical proof.As Gertsbakh says there's no need for further answers.
One problem I have reading Jose's demonstration, 1/N is the unconditional probability.
It appears that in his last statement that 1/N is conditional, it is not.
Agreed with Nadia Saeed, these theorems about SRS with replacement and without replacement are mentioned and proved in Cochran 1977
Basically there are two types of sampling designs: ordered and unordered.
this is the problem related with ordered sampling design in which we consider the position of selection of units in sample, for example:
we have a population of size 3 as {1,2,3} and our objective is to select a sample of size 2 using SRSWOR then the possible samples are
s1=(1,2)
s2=(2,1)
s3=(1,3)
s4=(3,1)
s5=(2,3)
s6=(3,2)
In all possible samples, each unit of population appear 2 times at first draw and 2 times at second draw i.e. units have same chance to select at both draws.
Dear Hamna,
Selecting the units with such equal probability in each extraction of unit for the sample. I recommend you a Spanish book:
Mirás, Julio (1985). Elementos de Muestreo para Poblaciones Finitas. Instituto Nacional de Estadística. Madrid.
This book has been reedited more recently.
Dear Hamna!
We use two conditions for SRS (both in sampling WR or WOR)
1. The probability of selecting an element from the population remains constant.
2. The prob, of selecting a random sample of size 'n' remains constant.
Consider a case with population size N=5 and sample size n=3 { The population units are supposed to be 1, 2, 3, 4, 5 and the sample units are 1, 2,3 (specifically)}
The case of SRS with replacement is clear to you, now consider SRS without replacement.
For condition 1, the probability of selecting an element (e.g. 1 ) is 1/5 at first draw.
At second draw it is (4/5) * (1/4) i.e
(prob. of selecting any unit other than 1 at first draw) * (prob. of selecting 1 at second draw) = (4/5) * (1/4)= 1/5
Means that if one is not selected first time, any of (2, 3, 4, 5) is selected first, so the prob is 4/5
and now element 1 is selected at second, so the prob. is 1/4 (as sampling is done without replacement, so only 4 units left of which 1 is favorable).
Similarly prob. of selecting 1 at third draw is (4/5)*(3/4)*(1/3) = 1/5
In this way probability remains the same for each element at every draw, whether it is SRS WR or SRS WOR.
I hope it will help you. Regards
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