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Let $V_1$ be the variance of the estimated mean from a stratified random sample of size $n$ with proportional allocation. Assume that the strata sizes are such that the allocations are all integers.Let $V_2$ be the variance of the estimated mean from a simple random sample of size $n$. Show that the ratio $\frac{V1}{V2}$ is independent of $n$.

We define the estimator $\bar y_{st}=\sum_{h}w_h \bar{y}_h$, where $w_h=\frac{N_h}{N}$. We know this is an unbiased estimator of the population mean. For proportional allocation, $n_h=nw_h $

$V(\bar y_{st})=V_1=\frac{1}{n}\sum w_h \sigma^2_h$ (If we assume SRSWR is applied to sample from each strata)

Again, $V_2=V(\bar y)=\frac{\sigma^2}{n}$

Now, $\frac{V_1}{V_2}=\sum w_h \frac{\sigma^2_h}{\sigma^2}=\frac{1}{N}-\sum \frac{w_h(\bar Y_h- \bar Y)^2}{\sigma^2}$ Now, how can I proceed from here?

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$V(\bar{y}_{st}) = V(\sum_h w_h \bar{y}_h) = \sum_hV(w_h\bar{y}_h)$ (by independence of the estimates for different strata). For any non-random $a$ and random variable $X$ with finite variance we have $V(aX) = a^2 V(X)$, so $$V_1 = V(\bar{y}_{st}) = \sum_hw_h^2V(\bar{y}_h) = \sum_hw_h^2\sigma_h^2/n_h,$$ because $V(\bar{y}_h) = \sigma_h^2/n_h$.

We also have $V_2 = \sigma^2/n$.

Hence $$\frac{V_1}{V_2} = \frac{\sum_hw_h^2\sigma_h^2/n_h}{\sigma^2/n}.$$ We have two different formulations for $w_h$: $w_h = n_h/n$ and $w_h = N_h/N$. Using both we have $$\frac{V_1}{V_2} = \frac{\sum_h(N_h/N)(n_h/n)\sigma_h^2/n_h}{\sigma^2/n} = \frac{\sum_h(N_h/N)\sigma_h^2}{\sigma^2},$$ which does not depend on $n$.

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  • $\begingroup$ I have done exactly the same thing. So, is this the correct solution? $\endgroup$ – Legend Killer Dec 23 '18 at 14:41
  • $\begingroup$ I'm not sure I understand your comment. Are you saying what you wrote in your original question is the same as my solution, or that you have done the same thing yourself since posting your original question? Our solutions differ: in your description of the question you have $V_1 = V(\bar{y}_{st}) = 1/n \sum_h w_h \sigma_h^2$, whereas I have put $V_1 = V(\bar{y}_{st}) = \sum_h w_h^2 \sigma_h^2/n_h$, which is not the same, as an example of where our solutions differ. $\endgroup$ – Alex Dec 23 '18 at 17:35
  • $\begingroup$ I agree with your statement that $V_1/V_2 = \sum_h w_h \sigma_h^2 / \sigma^2$, and stating that none of the terms in this sum depend on $n$ because $w_h = N_h/N$ and $\sigma_h, \sigma$ are constants finishes the proof. $\endgroup$ – Alex Dec 23 '18 at 17:41
  • $\begingroup$ Of course @Alex the two expressions of $V(\bar y_{st})$ are same because you did not use $n_h=nw_h$ due to proportional allocation $\endgroup$ – Legend Killer Dec 24 '18 at 3:06
  • $\begingroup$ Quite right, not sure how I missed that $\endgroup$ – Alex Dec 24 '18 at 8:38

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