Daily Archives: Mon, 22 June 2015

Gaussian distribution

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.3.

Probably the most common continuous probability density is the gaussian distribution, specified by

\displaystyle  \rho\left(x\right)=Ae^{-\lambda\left(x-a\right)^{2}} \ \ \ \ \ (1)

First, we need to normalize the distribution by finding {A}. That is, we must have

\displaystyle  A\int_{-\infty}^{\infty}e^{-\lambda\left(x-a\right)^{2}}dx=1 \ \ \ \ \ (2)

The gaussian integral is very common, and the result is that

\displaystyle   \int_{-\infty}^{\infty}e^{-\lambda\left(x-a\right)^{2}}dx \displaystyle  = \displaystyle  \sqrt{\frac{\pi}{\lambda}}\ \ \ \ \ (3)
\displaystyle  A \displaystyle  = \displaystyle  \sqrt{\frac{\lambda}{\pi}} \ \ \ \ \ (4)

Although there is no closed form indefinite integral, the definite integral can be found by a cute trick.

\displaystyle   \left[\int_{-\infty}^{\infty}e^{-x^{2}}dx\right]^{2} \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}e^{-x^{2}}dx\int_{-\infty}^{\infty}e^{-y^{2}}dy\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}-y^{2}}dx\;dy \ \ \ \ \ (6)

We can now transform to polar coordinates using

\displaystyle   r^{2} \displaystyle  = \displaystyle  x^{2}+y^{2}\ \ \ \ \ (7)
\displaystyle  dx\;dy \displaystyle  = \displaystyle  r\;dr\;d\theta\ \ \ \ \ (8)
\displaystyle  \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}-y^{2}}dx\;dy \displaystyle  = \displaystyle  \int_{0}^{\infty}\int_{0}^{2\pi}re^{-r^{2}}d\theta\;dr\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  2\pi\int_{0}^{\infty}re^{-r^{2}}dr\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  -\pi\left.e^{-r^{2}}\right|_{0}^{\infty}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \pi \ \ \ \ \ (12)

Therefore

\displaystyle  \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi} \ \ \ \ \ (13)

Using Maple (or integration by parts) we can work out the average and variance.

\displaystyle   \left\langle x\right\rangle \displaystyle  = \displaystyle  \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{\infty}xe^{-\lambda\left(x-a\right)^{2}}dx=a\ \ \ \ \ (14)
\displaystyle  \left\langle x^{2}\right\rangle \displaystyle  = \displaystyle  \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{\infty}x^{2}e^{-\lambda\left(x-a\right)^{2}}dx=\frac{1}{2\lambda}+a^{2}\ \ \ \ \ (15)
\displaystyle  \sigma^{2} \displaystyle  = \displaystyle  \left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}=\frac{1}{2\lambda} \ \ \ \ \ (16)

The distribution has the standard bell shape. Here’s a plot for {\lambda=2} and {a=1}: