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I came across this statement and can't decide if it's true or false.

Statement: In a stratified random sampling without replacement, with proportional allocation to the population size, the sample average is unbiased estimator to the population average.

I think it's true but not really sure how to prove it.

Thanks in advance,

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Let's say you are trying to find the average value of some characteristic $X$ for the entire population (i.e., $\mu_X$). If we assume the population is much larger than your sample then we can assume that the probability of selecting someone with a particular characteristic value $X=x$ does not change. If the population is small , then the bias of the samples will not be affected (not to be mistaken with a small sample from a population...where there can be bias), but the precision of the estimator will be different from the "large population" estimate. Since we are focused on the bias, I will assume a large population for the sake of simplicity. The proof for small populations requires dealing with entire samples (since each unique sample of size N has the same probability of being drawn, we can treat these groups as independent but not the individual sample values) as opposed to point-by-point arguments, so there are a lot more factorials and such involved to get to the same conclusion.

OK, preliminaries aside: Let's say you've stratified your population of size $K$ into M groups $G_i$, where the number of people in each group is $n_i$ and its mean for characteristic $X$ is $E(X|G_i)=\mu_i$.

Let's compare the sampling properties of a simple random sample average ($\bar X$) as an estimate of the population average to the one using proportional sampling of the stratified population $\bar X_s$ for a given sample total size $N$. For simple random sampling:

$\bar X = \frac{1}{N}\sum\limits_{i=1}^N x_i \rightarrow E(\bar X)=\frac{1}{N}\sum\limits_{i=1}^N E(x_i)$. Since we are doing simple random sampling without replacement (with a large population assumption $K\gg N$), each $x_i$ has the same statistical properties. In particular, the same expected value, $E(x_i)=\mu_X$. Therefore, $\frac{1}{N}\sum\limits_{i=1}^N E(x_i) =\frac{1}{N}\sum\limits_{i=1}^N \mu_X = \frac{N\mu_X}{N} = \mu_X$.

However, simple random sampling also implies that $P(x_i \in G_j) = \frac{n_j}{K}$ (again, assuming $K\gg N$). Therefore, we can re-write the calculation of the expected value in terms of the M $G_j$:

$E(\bar X)= \frac{1}{N}\sum\limits_{i=1}^N\sum\limits_{j=1}^M P(x_i \in G_j)E(x_i|G_j)=\frac{1}{N}\sum\limits_{i=1}^N\sum\limits_{j=1}^M \frac{n_j}{K}\mu_j=\frac{1}{N}\sum\limits_{i=1}^N\sum\limits_{j=1}^M\left(\frac{1}{K}\sum\limits_{x_l \in G_j} x_l\right)=$

$\frac{1}{N} \sum\limits_{i=1}^N \left(\frac{1}{K}\sum\limits_{x_l \in \left\{\bigcup \limits_{k=1}^M G_k\right\}} x_l\right)=\frac{1}{N}\sum\limits_{i=1}^N\mu_X = \mu_X$

Now, $\sum\limits_{j=1}^M \frac{n_j}{K}\mu_j$ is an interesting sum, because it is a weighted average of the mean value of each strata, where the weight is the size of the strata relative to the entire population. Compare this to the stratified sampling scheme, where you select a total sample size of $N$, but you select $N_i$ people from group $G_i$, where $N_i = \frac{n_i}{K}N$. In this case, $\bar X_s = \frac{1}{N}\sum\limits_{i=1}^M \sum\limits_{x_j\in G_i} x_j \rightarrow E(\bar X_s) =\frac{1}{N}\sum\limits_{i=1}^M \sum\limits_{x_j\in G_i} E(x_j|G_i) =\sum\limits_{i=1}^M \frac{N_i}{N}\mu_i=\sum\limits_{i=1}^M \frac{N\frac{n_i}{K}}{N}\mu_i=\sum\limits_{i=1}^M \frac{n_i}{K}\mu_i = \mu_X$, which is the same as what we derived under simple random sampling.

Hence, you will get an unbiased estimate of the population mean by proportional stratified sampling. However, the benefit of stratification is that it ensures that each individual sample is more representative (i.e., you cannot get a sample consisting of only individuals from a particular sub-group, $G_i$) so your sampling error is generally reduced (since you are using more of your available knowledge to select the samples).

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