# Is there a lower bound to the standard deviation of a Gaussian (Normal) distribution?

A Gaussian or Normal distribution is defined by the probability density function $$f(x \; | \; \mu, \sigma) = \frac{1}{\sigma\sqrt{2\pi} } \; e^{ -\frac{(x-\mu)^2}{2\sigma^2} }$$ with $\sigma$ as the standard deviation and $\mu$ as the mean. For all positive-definite $\sigma$, the probability distribution is normalized; i.e., $$\int_{-\infty}^{+\infty} dx \; f(x \, | \, \mu, \sigma) = 1 , \quad \forall \, \sigma > 0 ~.$$ Since, the range of $f$ is $\mathbb{R}^+$ (positive real numbers), then the normalization implies that $$0 < f(x) \leq 1, \quad \forall \, x \in \mathbb{R} ~.$$ which is valid as $x = \mu$, so that $$f(x = \mu) = \frac{1}{\sigma\sqrt{2\pi} } \leq 1$$ which gives $$\sigma \geq \frac{1}{\sqrt{2\pi}} \approx 0.4 ~;$$ that is a lower bound for the standard deviation!

I cannot see where I have made a mistake in the course of the argument.

• The normalisation does not imply that the distribution is lower than 1 everywhere. The value of $f$ is not a probability, only $f(x)\mathrm dx$ is the probability to be in $(x,\,x+\mathrm dx)$. Since $\mathrm dx$ is infinitesimal, this is always lower than 1. – Tom-Tom Dec 10 '15 at 16:08
• $f(x)$ can be as big as you can imagine. – Zhanxiong Dec 10 '15 at 16:09

The bound $f(x)\leq 1$ is not correct - the probability density function can take arbitrarily large values.
Alternatively, consider the uniform distribution on the interval $[0,1/10]$, it has probability density function $$f(x)=\begin{cases} 10 & x\in\left[0,\frac{1}{10}\right]\\ 0 & \text{otherwise} \end{cases} .$$
• My mistake was a naïve comparison with summations; as if it was $\sum_{i} f(x_i) = 1$. – AlQuemist Dec 10 '15 at 20:26