# mean and std deviation of a population equal?

Hypothetically, if we have a population of size $n$ whose mean and std deviation are equal, I think with some work we have a constraint that the ratio, (Sum of squared points)/(Sum of points$)^2$ $= \frac{(2n-1)}{n^2}$, which gets small quickly as $n$ gets large. Are there heuristic considerations that might render such a population plausible as an extension of, say, the binomial distribution (as with the Poisson distribution, although that distribution the mean is equal to the variance)?

Does this property (mean = Sqrt[variance] ) suggest anything about the population generally, if that question is not too vague? I have not encountered a population with this property in any texts, but am fairly sure it has been considered...?

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Given any population, you can shift it, affecting the mean but not the variance or standard deviation. In particular, you can always shift it so the mean equals the standard deviation (or the logarithm of the standard deviation, or its cosine, or pretty much anything you want). So I don't think it says a lot about the population. – Gerry Myerson Oct 24 '11 at 0:29

The distributions of exponential type whose variance and mean are related, so that $\operatorname{Var}(X) \sim (\mathbb{E}(X))^p$ for a fixed $p$, are called an index parameter, are known as Tweedie family.

The case you are interested in corresponds to index $p =2$. $\Gamma$-distribution possesses this property (the exponential distribution is a special case of $\Gamma$-distribution).

$\Gamma$-distribution has probability density $f(x) = \frac{1}{\Gamma(\alpha)} x^{\alpha - 1} \beta^{-\alpha} \exp\left(- \frac{x}{\beta}\right) \mathbf{1}_{x > 0}$.

Its mean is $\mu = \alpha \beta$ and variance $\mu_2 = \alpha \beta^2$, hence $\mu_2 = \frac{\mu^2}{\alpha}$. The equality requires $\alpha = 1$, so we recover the exponential distribution as already noted by Michael Hardy.

But we do not have to stay within the exponential family. You can certainly achieve the equality with normal distribution, and with beta distribution, and with many discrete distributions, for instance, the generalized Poisson distribution with $\lambda = \frac{1}{1-\mu}$, both the mean and variance equal $(1 - \mu)^{-2}$.

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One situation in which the mean is equal to the standard deviation is with the exponential distribution whose probability density is $$f(x) = \begin{cases} \frac1\theta e^{-x/\theta} & \text{if }x>0, \\ \\ \\ 0 & \text{if } x < 0. \end{cases}$$ The mean and the standard deviation are both equal to $\theta$.

These are the only distributions that are "memoryless" in the sense that if (capital) $X$ is a random variable that is so distributed, then $$\Pr(X > x + y \mid X > y) = \Pr(X > x)$$ for all positive numbers $x$ and $y$.

How you got $\dfrac{2n-1}{n^2}$ escapes me.

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Concerning the matter that I said escapes me. It is true that equality of the mean and standard deviation entails that $\dfrac{\mathbb{E}(X^2)}{(\mathbb{E}(X))^2} = 2$. I'm guessing you were trying to do something similar to that. – Michael Hardy Oct 24 '11 at 0:36
Dividing by $n-1$ rather than by $n$ in calculating the variance is "Bessel's correction" (see the Wikipedia article of that title), and is done when one is estimating the variance based on an i.i.d. random sample rather than on the whole population. I think what you should get here is $\left(nm^2 - 2\sum_{i=1}^n x_i + \sum_{i=1}^n x_i^2\right)/n = m^2$. There's a missing factor of $n$ in "$nm^2$". Solving for $m$ gets you $\sum_i x_i^2/(2\sum_i x_i) = m$. Dividing the numerator and denominator by $nm$ then gets you $E(X^2)/(2(EX)^2) = 1$. – Michael Hardy Oct 24 '11 at 1:57