I am working on an assignment regarding stratified sampling.
The assignment hints that in my situation the bandwidth of the strata is proportional to the standard deviation, so something along the lines of:
$$ \sigma_i \propto (b_i-a_i) \implies \sigma_i = K(b_i-a_i) $$
Any hints on how to go about proving this? I have searched online but I haven't found anything.
Ok, I believe I figured it out. I'll post this for anyone interested. Comments greatly appreciated.
I left out a crucial detail. We are working with a discrete uniform distribution.
The distribution has some nice properties, such as:
$$ N_i = (b_i-a_i) \cdot N_{ij} \tag{1}$$ $$\sigma_i = \frac{N_i}{k_i} \tag{2}$$
Where $N_i$ is the total population in the strata and $N_{ij}$ is the number of element for a specific value $j$ that is between $a_i$ and $b_i$. These properties gives:
$$\sigma_i = \frac{N_i}{k_i} \implies \sigma_i = \frac{N_{ij}}{k_i} \cdot (b_i - a_i) \implies \sigma_i = K(b_i - a_i)$$
Then I have to assume that $\frac{N_{ij}}{k_i}$ is equal across all strata. This seems like reasonable assumption since it's a discrete uniform distribution. I have difficulties explaining why tough. If that assumptions holds, then $\frac{N_{ij}}{k_i} = K$.