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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.

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    $\begingroup$ 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. $\endgroup$ – Tom-Tom Dec 10 '15 at 16:08
  • $\begingroup$ $f(x)$ can be as big as you can imagine. $\endgroup$ – Zhanxiong Dec 10 '15 at 16:09
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The bound $f(x)\leq 1$ is not correct - the probability density function can take arbitrarily large values.

One way to see this is by doing what you did, and then concluding that one of your assumptions must be incorrect since the standard deviation of the normal distribution can be any positive number.

Alternatively, consider the uniform distribution on the interval $[0,1/10]$, it has probability density function $$f(x)=\begin{cases} 10 & x\in\left[0,\frac{1}{10}\right]\\ 0 & \text{otherwise} \end{cases} .$$

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  • $\begingroup$ My mistake was a naïve comparison with summations; as if it was $\sum_{i} f(x_i) = 1$. $\endgroup$ – AlQuemist Dec 10 '15 at 20:26

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