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?
$V(\bar{y}_{st}) = V(\sum_h w_h \bar{y}_h) = \sum_hV(w_h\bar{y}_h)$ (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 $n$.