The Gaussian (Normal) Distribution
Probability Distribution Function
The Normal (Gaussian) distribution probability distribution function is
\[
f\left(x; \mu, \sigma\right) = \frac{1}{\sqrt{2\pi}\sigma}\,e^{-\left(x-\mu\right)^2/2\sigma^2},
\]
normalized to unity,
\[
\int\limits_{-\infty}^{\infty} f\left(x; \mu,\sigma\right)\,dx = 1,
\]
is symmetrically distributed about its mean, \(\mu\), with width \(\sigma\).
Full Width at Half Maximum (FWHM)
The full width at half maximum (FWHM) is the distance between points on a curve at which the function reaches half its maximum value. The FWHM is often used to describe the “width” of a distribution.
For a 1-dimensional Gaussian, it is seen that as the maximum value occurs as \(x = \mu\) (by definition), half of the maximum value is
\[\begin{split}
\begin{align*}
\left.\frac{1}{\sqrt{2\pi}\sigma}\,e^{-\left(x-\mu\right)^2/2\sigma^2}\right|_{x = x \text{ of (max/2)}} &= \frac{1}{2} f\left(x_{\text{max}}\right)\\
&= \frac{1}{2} f\left(\mu\right) = \frac{1}{2} \frac{1}{\sqrt{2\pi}\sigma},
\end{align*}
\end{split}\]
resulting in the equality
\[
e^{-\left(x-\mu\right)^2/2\sigma^2} = \frac{1}{2},
\]
which is (taking the log)
\[
-\frac{\left(x-\mu\right)^2}{2\sigma^2} = -\ln2.
\]
Thus, solving the equality,
\[
\left(x-\mu\right)^2 = 2\sigma^2 \ln2,
\]
yields
\[
x_{\pm} = \pm \sigma \sqrt{2 \ln 2} + \mu.
\]
Thus, the FWHM is
\[\begin{split}
\begin{align*}
\text{FWHM} &= x_{+} - x_{-}\\
&= \left(\sigma \sqrt{2 \ln 2} + \mu\right) - \left(-\sigma \sqrt{2 \ln 2} + \mu\right)\\
&= \boxed{2\sqrt{2\ln2}\sigma}\,.
\end{align*}
\end{split}\]
N.B.: It is seen that the FWHM for a Gaussian is independent of both the normalization constant and the mean, and is only dependent on the standard devaiation of the Gaussian.
Probability and the Error Function
The probability that a Normally distributed random variable will lie in a range of values symmetrically integrated over is given by
\[\begin{split}
\begin{align*}
\text{Pr}\left(\mu - y \leq x \leq \mu + y\right) &= \int\limits_{\mu - y}^{\mu + y}\frac{1}{\sqrt{2\pi}\sigma}\,e^{-\left(x-\mu\right)^2/2\sigma^2}\,dx\\
&= \int\limits_{\mu}^{\mu + y}\frac{2}{\sqrt{2\pi}\sigma}\,e^{-\left(x-\mu\right)^2/2\sigma^2}\,dx.
\end{align*}
\end{split}\]
Making the substitution
\[
t = \frac{\left(x-\mu\right)}{\sqrt{2}\sigma},
\]
then
\[\begin{split}
\begin{align*}
\text{Pr}\left(\mu - y \leq x \leq \mu + y\right) &= \int\limits_{\mu}^{\mu + y}\frac{2}{\sqrt{2\pi}\sigma}\,e^{-\left(x-\mu\right)^2/2\sigma^2}\,dx\\
&= \boxed{\frac{2}{\sqrt{\pi}} \int\limits_{0}^{y/\sqrt{2}\sigma}e^{-t^2}\,dt \equiv \text{erf}\left(\frac{y}{\sqrt{2}\sigma}\right)}\,.
\end{align*}
\end{split}\]
Cumulative Distribution Function (cdf)
For the cumulative distribution function (cdf),
\[
\Phi\left(x\right) = \int\limits_{-\infty}^{x}f\left(t;\mu,\sigma\right)\,dt,
\]
it is seen, noting from the form of the error function, that
\[
\begin{equation*}
\frac{2}{\sqrt{\pi}} \int\limits_{0}^{y/\sqrt{2}\sigma}e^{-t^2}\,dt = \frac{2}{\sqrt{2 \pi}} \int\limits_{0}^{y/\sigma}e^{-t^2/2}\,dt = \text{erf}\left(\frac{y}{\sqrt{2}\sigma}\right),
\end{equation*}
\]
then for the standard Gaussian (\(\mu = 0\), \(\sigma=1\))
\[\begin{split}
\begin{align*}
\Phi\left(x\right) &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{x} e^{-t^2/2}\,dt\\
&= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{0} e^{-t^2/2}\,dt + \frac{1}{\sqrt{2\pi}}\int\limits_{0}^{x} e^{-t^2/2}\,dt\\
&= \frac{1}{2} + \frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)\\
&= \frac{1}{2} \left(1 + \text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)
\end{align*}
\end{split}\]
so it is likewise seen that
\[\begin{split}
\begin{align*}
\text{Pr}\left(\mu - n \sigma \leq x \leq \mu + n\sigma\right) &= \Phi(n) - \Phi(-n)\\
&= \Phi(n) - \left(1-\Phi(n)\right)\\
&= \frac{1}{2} \left(1 + \text{erf}\left(\frac{n}{\sqrt{2}}\right)\right) - \left[1-\frac{1}{2} \left(1 + \text{erf}\left(\frac{n}{\sqrt{2}}\right)\right)\right]\\
&= \text{erf}\left(\frac{n}{\sqrt{2}}\right).
\end{align*}
\end{split}\]
Show this for: generic normal distribution \(f\) with mean \(\mu\) and std \(\sigma\)
It is noted that in the case that \(\left|y\right|=n\sigma\),
\[
\text{Pr}\left(\mu - y \leq x \leq \mu + y\right) = \text{Pr}\left(\mu - n\sigma \leq x \leq \mu + n\sigma\right) = \text{erf}\left(\frac{n}{\sqrt{2}}\right).
\]
So, for \(n=1\),
\[
\text{Pr}\left(\mu - \sigma \leq x \leq \mu + \sigma\right) = \text{erf}\left(\frac{1}{\sqrt{2}}\right)
\]
However, at this point we are at an impass analytically, as the integral of a Gaussian function over a finite range has no analytical solution, and must be evaluated numerically.
\(p\)-values
Two-Tailed \(p\)-value
One-Tailed \(p\)-value