# Related:

Trying to understand the derivation of expectation of sample mean $E(\bar x)= \mu$ where $\mu$ is the mean of the population

I've been trying to understand $\mathrm E(x_i)= \mu$ where $x_i$ is the element at the $i^{\textrm {th}}$ draw from a population of size $N$ for making a sample of size $n$ by SRSWOR.

I read between the lines of my book again; here is the excerpt:

Let $x_1,x_2,\ldots,x_n$ represent a simple random sample from the population $X_1,X_2,\ldots,X_n$. In SRSWR, the probability of selection of any particular member of the population at any drawing remains a constant $1/N$; because before any drawing the population contains all the $N$ members. It may be shown that this result is also true in SRSWOR, although the population size varies at each stage of the selection. Thus, the probability of obtaining the population member $X_k$(suppose) at the $i$-th drawing is a constant $1/N$ both in SRSWR and SRSWOR; i.e., $$\mathrm P(x_i= X_k)= \frac{1}{N}$$

While I'm sure of this for SRSWR, I'm not getting this for SRSWOR.

How can the probability of drawing a certain element from the population at a certain draw be $1/N$ for even the author says 'the population size varies at each stage of the selection'?

If after each turn, the population size is not the same, hoe can still the probability of drawing certain element be $1/N$ when the sampling is done without replacement?

Can anyone please explain me the case for SRSWOR?

• It is illustrative to note that $P(X_2 = x) = \sum_{y=1}^n p(X_2 = x | X_1 = y) p(X_1 = y)$ by the law of total probability. Mar 21, 2016 at 15:13
• @AdamO: Is the book saying right?
– user74724
Mar 21, 2016 at 15:22
• Yes it is true that $P(x_i = X_k) = 1/N$, mathematically you would use induction to prove the general case. However, the intuition is that, "It's equally likely that all the other previous draws could be anything else." Mar 21, 2016 at 15:24
• @AdamO: Suppose, you pick up $X_k$ at the first turn; now wouldn't it be wrong to say $\mathrm P(x_2= X_k)= 1/N\;?$ For you've not replaced the element before the second turn, isn't it?
– user74724
Mar 21, 2016 at 15:27
• Yes but if the previous draws do not select that element on the $i$-th turn, you are more likely to draw that element on the $i$-th turn than with SRSWR. Think about the $N$-th draw, the $N-1$ possible cumulative permutations of previous draws which do draw the $k$th element have probability $(N-1)/N$, but the 1 possible draw that does include the $k$-th element has 100% probability of doing so, leading to probability $1/N$ for a successful draw. Mar 21, 2016 at 15:32

By way of induction and the total law of probability it holds that:

$$P(X_2 = k) = \sum_{i=1}^n P(X_2 = k | X_1 = i) P(X_1 = i)$$

and supposing it holds for any $j$ that $P(X_j = k) = 1/N$ then

$$P(X_{j+1} = k) = \sum_{i=1} ^ N P(X_{j+1} = k | X_j = i) P(X_j = i)$$.

I think the comments I've contributed on the post contribute more of the intuition behind this finding.

• I don't think this is correct, because the poster asked about "How can the probability of drawing a certain element from the population at a certain draw be 1/N" in SRSWOR. This is not the same thing as the overall probability of selecting an element, which is the probability of selection at at some draw, which is indeed 1/N. Apr 5, 2016 at 19:30

Let $$A_l^{\textrm {j}}$$ denotes an event that a particular unit $$X_j$$ is not selected at the $$l^{\textrm {th}}$$ draw. The probability of selecting, say, $$k^{\textrm {th}}$$ unit at $$i^{\textrm {th}}$$ draw is: (taking $$A_i^{\textrm {k}}$$= $$A_i$$)

\begin{align} P (x_i = X_k) &= P(A_1 \cap A_2 \cap A_3 \cap A_4 .... A_{i-1} \cap \overline{A_i})\\ &= P(A_1)P(A_2|A_1)P(A_3|A_1A_2).....P(\overline{A_i}|A_1A_2...A_{i-1}) \\ &= (1-\frac{1}{N})(1-\frac{1}{N-1})...(1-\frac{1}{N-i+2})\frac{1}{N-i+1}\\ &= (\frac{N-1}{N})(\frac{N-2}{N-1})...(\frac{N-i+1}{N-i+2})\frac{1}{N-i+1}\\ &= \frac{1}{N} \end{align}