Question

Do Problem 6.4 with f(x) in (8.1).

Short answer

The mean is equal to 0.

The variance is equal to ${\sigma }^2$

Step-by-step solution

Given information

A function that is proportional to$e^{-a{x}^2}$.

Definition of variance

A statistical measurement of the dispersion between values in a data collection is known as variance.

The density of harmonic oscillator

The probability density functions of a harmonic oscillator$f\left(x\right)=c\left(a\right).e^{-a{x}^2}$.

The density function is $g\left(x\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{\frac{-x^2}{\left(2{\sigma }^2\right)}}$

It is the only symmetric function proportional to $e^{-a{x}^2}$ that integrates to 1.

$\begin{array}{rcl}a& =& \frac{1}{2{\sigma }^2}c\left(a\right)\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}\end{array}$

Hence the mean is equal to 0 and the variance is equal to $\begin{array}{rcl}& & {\sigma }^2\end{array}$