You can ignore this context but I think it adds a little interest to the question..
In finance pricing information is often proprietary and firms do not want other firms to know their price, but of course they want to know everyone else's prices. It is potentially valuable information, and useful for bench marking. A compromise is often struck where by a middle man collects all the prices from all firms and then releases some anonymous statistics to all contributors which they can utilise with respect to their own prices. Some financial markets are very illiquid and pricing them is difficult so there can be a lot of variation in the results.
Question
Assume that there is an underlying random variable, $X \sim \mathcal{N}(\mu, \sigma^2)$, (which represents in this case the market-price of a financial instrument measured by each firm).
We are interested in finding optimal estimators, $\hat{\mu}$ and $\hat{\sigma}$, given the following statistics about samples taken from $X$ (i.e. trying to estimate the distribution given the anonymised data from the middle man):
- The mean of the samples from $X = \bar{x}$
- The range of the samples from $X = \max{s_i} - \min{s_i}$
- The number or samples is $n$, i.e. $s_1, ... s_n$
Thoughts
I'm going to assume $\bar{x} =\hat{\mu} $ is the optimal estimator here and that it is unbiased - fairly easy to show.
With some intensive computational statistics found that for the standard normal distribution ($Z \sim \mathcal{N}(0,1)$):
$$ E[max(Z) - min (Z)] = [1.69, 2.06, 2.32, 2.53, 2.70]$$ for $n=[3,4,5,6,7]$
I reverse engineer the estimator $\hat{\sigma} = \frac{\max{s_i} - \min{s_i}}{t_n}$, where $t_n$ is the value from the above lookup table.
Is this appropriate? Is it also an unbiased estimator?
For $\mu$, this is indeed the optimal estimator (in the UMVU sense). This is even the case if you have all the information in the sample, so it is also the case with limited information.
For $\sigma$, this is indeed unbiased. Letting $Z$ be a sample of iid standard normals, we have $$\mathbb{E}[\hat{\sigma}] = \frac{\mathbb{E}[\max(X) - \min(X)]}{t_n} = \frac{\sigma\mathbb{E}[\max(Z)-\min(Z)]}{t_n} = \sigma,$$ since $X$ and $\sigma Z$ have the same (joint) distribution, so $(\max(X), \min(X))$ has the same (joint) distribution as $\sigma(\max(Z), \min(Z)).$
It's also consistent, since $\text{Var}(\hat{\sigma})\to 0,$ using for example this: https://stats.stackexchange.com/questions/229073/variance-of-maximum-of-gaussian-random-variables.