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I read the question and the answers regarding the idea of the mean being equal to the standard deviation, in which the mean can always be adjusted to equal anything. But, if the data is restricted to only positive values, then it seems the normal distribution would not work, while the exponential distribution and Poisson would work. Any ideas?

Rick Powers
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  • Please reference the question you are talking about -- I've no idea where to look for the reference. – gt6989b Jul 12 '12 at 18:55
  • I didn't follow. What is the question? – Inquest Jul 12 '12 at 18:59
  • Possibly referring to http://math.stackexchange.com/questions/75244/mean-and-std-deviation-of-a-population-equal – Henry Jul 12 '12 at 19:28
  • The Poisson distribution has mean equal to variance, which is not the standard deviation unless both are $1$ (or $0$) – Henry Jul 12 '12 at 19:30

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There are many other examples.

The simplest non-degenerate one is $1$ or $0$ with equal probabilities which has mean and standard deviation $\frac12$.

Another is to take the value $\sqrt{p(1-p)} - p$ with probability $1-p$ and the value $\sqrt{p(1-p)} - p+1$ with probability $p$, for $p \le \frac12$.

You could devise many more. A $\chi^2$ distribution with two degrees of freedom. Or if $X$ has a $\chi^2$ distribution with $1$ degree of freedom then $X+\sqrt{2}-1$.

You can multiply any example by any positive number $k$ and retain the property.

Henry
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