Variance derivation for estimator

Question

I am working to try to determine the variance for the following specific situation:

I have a seed trap where I collect and weight a total amount of seeds, I call this seed biomass. I measure the seed biomass with no error. I then proceed to take a random sample of n seeds and divide the total seed biomass by the mean mass of these n seeds in order to approximate the number of seeds in a trap (N). So the aim is to have the variance of the number of seeds where m is potentially normally or lognormally distributed. Eventually I would like see how the variance changes with different n. This is as far as I got to and somehow does not seem correct. Thank you

$$ N = \frac {SBiomass}{\frac{\sum^n_{1}m_{i}}{n}} $$

$$ var(N)=n^2\times SBiomas^2\times var(\frac{1}{\sum^n_{1}m_{i}})$$

$$ var(N)=n^2\times SBiomas^2\times var(\frac{1}{\sum^n_{1}s_{i}})=n^2\times SBiomass^2\times [{n} \times var (\frac{1}{m})] = SBiomass^2\times n^3 \times var(\frac{1}{m}) $$

Answer 1

$\operatorname{var}\left(\frac{1}{\sum_{i=1}^n m_i}\right)$ is not trivial to calculate, and variance cannot be distributed over the reciprocal sum, i.e. this will not be equal to $n\operatorname{var}(1/m)$.

If $m_i$ is assumed to be normal, the sum will be normal (assuming independent $m_i$). But, variance of reciprocal normal is not defined.

If $m_i$ is assumed to be lognormal, the sum is not lognormal but it approximates to lognormal in many cases. The variance of the sum is already $n\sigma_m^2$ and the reciprocal of lognormal is still a lognormal; so that you can use the formulas for this distribution.

So, the answer may depend heavily on the distribution of $m$.

Answer 2

This is an example of an inference about an unknown population size. The method you have proposed for estimating the population size is to use a ratio estimator, taking the ratio of the population mean weight divided by the sample mean weight. Unfortunately, even before you get to calculating the variance, ratio estimators of this kind tend to be biased upward (i.e., they tend to overestimate the quantity of interest) except under extremely fortuitous assumptions. Correcting this bias requires some knowledge of the distribution of the underlying weights, and this then leads to a more complicated adjusted estimator, where both the mean and variance are complex.

Rather than using the proposed ratio estimator, you might consider using more standard methods for inference of an unknown population size, such as the mark-recapture method. What you do here is take a random sample of $n$ seeds and "mark" them (e.g., with a black felt-tip pen), then mix the sample of seeds back into the population, give it all a good shake to randomise them and then resample a new sample. Count the total number of seeds in the new sample and the number of "marked" seeds, and then use this to make an inference about the population size. The advantage of this method over the method you are proposing is that the distribution of the "marks" is known, and so it is much easier to get good inferences about the population size.