How do sample size and measurement accuracy affect the measurement of the parameters of a log-normal distribution?

Question

Assume that we have objects (e.g. particles) whose properties (e.g. diameter) follow a log-normal distribution that can be described by a geometric mean $\mu_g$ and a geometric standard deviation $\sigma_g$.

Now consider that we want to measure $\mu_g$ and $\sigma_g$, by sampling $n$ samples from the distribution (e.g. measuring the diameters of $n$ individual particles). Consider further that the utilized measurement technique is imperfect, i.e. we don't receive the true value $A_i$ from a measurement, but instead we receive $A_i'=A_i+\Delta A_i$.

If we define that the errors are random, then a reasonable assumption for $A_i'$ could be that it is normally distributed around $A_i$ with a certain standard deviation $\sigma_{measurement}$.

We then calculate the measured properties of the initial log-normal distribution, based on the imperfect measurements, so we receive

$\mu_g'=\mu_g+\Delta \mu_g= \sqrt[n]{(A_1+\Delta A_1) (A_2+\Delta A_2) \cdots (A_n+\Delta A_n)}$

and

$\sigma_g'=\sigma_g+\Delta \sigma_g=\exp{\sqrt{ \sum_{i=1}^n \left(\ln { {A_i+\Delta A_i} \over \mu_g' } \right)^2 \over n }}$.

Questions:

1. Can we make general statements about $\Delta \mu_g$ and $\Delta \sigma_g$? E.g. $\Delta \mu_g<0$ or $\Delta \sigma_g>0$, or something along these lines?
2. How does the sample size $n$ play into this? Often, we would assume that errors will approach 0 for $n\rightarrow\infty$. Is this also the case in this scenario?

Note: I simulated the process, and obviously I'm willing to share the results, but I'd like to have an unbiased discussion first.

Thanks a lot!

Answer 1

I find this question easier after replacing $A$ by $e^X$ and $\Delta A$ by $Y$:

If $X$ and $Y$ are independent normals and $Y$ has mean $0$, are the moments of $\ln(e^X+Y)$ bigger or smaller than the moments of $X$?

This only makes sense when $Y$ is small by comparison with $e^X$, so that the cases where $e^X+Y$ is negative are negligible -- which is the same as requiring that the errors are small by comparison with the measurements, so that the geometric mean mentioned in the question is meaningful.

With that mild assumption, the answer is:

• $E[\ln(e^X+Y)]<E[X]$
• $Var[\ln(e^X+Y)]>Var[X]$

The claim about means follows from the concavity of $\log$, or in more detail:

\begin{align} (e^x+y)(e^x-y)&<(e^x)^2\\ \frac12[\ln(e^x+y)+\ln(e^x-y)]&<x\\ E[\ln(e^x+Y)]&<x\\ E[\ln(e^X+Y)]&<E[X] \end{align}

The claim about variances follows, at least approximately, from

$$Var[\ln(e^X+Y)]-Var[X]\simeq Var[e^{-X}Y]$$

which is accurate to second order in $Y$.