# impossible distribution statistics?

I'm currently reviewing an article where authors presented distribution statistics that look erroneous to me. But I'm not able to find a way to ascertain it. The article presented results with a mean of 95% and standard deviation of 25%. Maximum value can't be more than 100% and minimum value can't be less than 0%. I don’t have the sample number. I tried to generate lognormal random numbers with these statistics without success:

# R Code
require("Runuran")
d1 <- urlnorm(n = 1000, meanlog = log(95), sdlog = log(25), lb = 0, ub = 100)


How can I generalize this conclusion whatever the distribution?

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The maximum standard deviation under the constraints that all data lie in $[0,1]$ and the mean is $0.95$ can be computed to be $\sqrt{0.95(1-0.95)}\approx 0.218$ (and is attained when $95\%$ of the data equal $1$ and $5\%$ of them equal $0$). – whuber Jun 13 '14 at 18:54
Of interest as well (expounding on whuber's limit), Variance of a bounded random variable. – Andy W Jun 13 '14 at 19:46
@Andy Thank you for the reference. Generalizing Juho Kokkala, when the essential range of $X$ is in $[0,1]$ and $E[X]=c$, then $\text{Var}(X)=E(X^2)-c^2\le c(1-c)$ because $E(X^2)\le E(X)=c$. Equality can hold iff $X^2=X$ a.e., implying $X$ is Bernoulli. The general result for bounded $X$ follows immediately by rescaling $[0,1]$ to the range of $X$. – whuber Jun 13 '14 at 20:37
Playing with simulation from lognormals isn't a useful way to investigate the properties of something bounded to [0,1], since lognormals don't have such an upper bound. – Glen_b Jun 13 '14 at 23:18

If you are working in the context that there are $n$ observations for which $\hat{\mu} = 0.95$ and $\hat{\sigma} = 0.25$, then I think you are correct. Playing around with sample sizes, the sample standard deviation is strictly less than 0.25 (can get around .22). The largest variance will be when the observations are either $1$'s or $0$'s. In that case, we need $\frac{19}{20}$ of the $n$ observations to be $1$ with the rest $0$, and $\frac{19}{20}\cdot 1 + \frac{1}{20}\cdot 0 = 0.95$.

For the standard deviation to be $0.25$, we need a variance of $0.0625$, and thus we need: \begin{align} &=\sum_n\left(x_i - 0.95\right)^2 = 0.0625\\ &=\frac{1}{n - 1}\left(\frac{19n}{20}.05^2 + \frac{n}{20}.95^2\right) = 0.0625\\ &=\left(\frac{19n}{20}.05^2 + \frac{n}{20}.95^2\right) = 0.0625n - 0.0625\\ &= n\left(.95\cdot{.05}^2 + .05\cdot{.95}^2 - 0.0625\right) = -0.0625\\ &\Rightarrow n = 4.1666666 \end{align} Which is an non-integral number and impossible.

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That's it, thank you. – denisC Jun 13 '14 at 18:21

Without the original context I'm speculating, but I suspect this is an odds vs probability issue?

If event A has a 95% chance of happening then what is 25% more likely to happen than event A? That's the same as calculating something 75% less likely than A' or (1-.95)*(1-.25)= 0.0375 which means that .9625 is 25% more likely than .95 and is 1 standard deviation above your distribution's mean.

Hope this helps.

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Hi, this was not was I was looking for. I should have been more clear. Maybe taking the problem upside will make it more understandable. Is it possible to generate a sample with mean of 95, standard deviation of 25 (sample values has to be positive and can't exceed 100)? But thank you to take time for this. – denisC Jun 13 '14 at 18:04
Oh I see. Your model isn't necessarily binary but has some sort of limit? And somehow the model's mean is 95% of the limit and the standard deviation is 25%? – Eric Jun 13 '14 at 18:12

If you are reviewing for a journal and the data look incorrect, this is something you can raise in your written response. Even if the data are not in error, it is still a problem because it appears to be an error or an incorrect / unconventional way of reporting data. I think you have done due diligence by thinking carefully about the paper and the results, but the burden is ultimately on the authors to make the results clear to the readership.

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