Consider the Gaussian distribution

$\mathbf{\text{ρ}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{\lambda }{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}}$

where A, a, and λ are positive real constants. (Look up any integrals you need.)

(a) Use Equation 1.16 to determine A.

(b) Find〈x〉,〈x²〉,and σ.

(c) Sketch the graph of ρ(x).

The required integrals are given by:

The given Gaussian probability distribution is assumed to be valid over the whole line

i.e.

${\mathrm{\rho }}_{\left(\mathrm{x}\right)}={\mathrm{Ae}}^{-\mathrm{\lambda }{(\mathrm{x}-\mathrm{a})}^2}$ $-\mathrm{\infty }\le \mathrm{x}\le \mathrm{\infty }$

Calculating for A by normalizing the function,

$\begin{array}{c}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{\rho }}_{\left(\mathrm{x}\right)}\mathrm{dx}=1\\ \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{Ae}}^{-\mathrm{\lambda }{(\mathrm{x}-\mathrm{a})}^2}\mathrm{dx}=1\end{array}$

Assuming, $\mathrm{y}=\mathrm{x}-\mathrm{a}$

Hence, the equation becomes

$\mathrm{A}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{\lambda y}}^2}\mathrm{dy}=1$

$\mathrm{A}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{\frac{-{\mathrm{y}}^2}{{(1/\sqrt{\mathrm{\lambda }})}^2}}\mathrm{dy}=1$

This integral can be taken from $0\mathrm{to}\mathrm{\infty }$, multiplying with a factor 2, since it’s an even function of y.

$2\mathrm{A}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{\frac{-{\mathrm{y}}^2}{{(1/\sqrt{\mathrm{\lambda }})}^2}}\mathrm{dy}=1$

Now, from equation 1, putting $\mathrm{n}=0$

Thus, the value of A is $\sqrt{\frac{\mathrm{\lambda }}{\mathrm{\pi }}}$.

Hence, the Normalised probability distribution comes out to be

${\mathrm{\rho }}_{\left(\mathrm{x}\right)}=\sqrt{\frac{\mathrm{\lambda }}{\mathrm{\pi }}}{\mathrm{e}}^{-\mathrm{\lambda }{(\mathrm{x}-\mathrm{a})}^2}$ $-\mathrm{\infty }\le \mathrm{x}\le \mathrm{\infty }$

First, we solve for $⟨{\mathrm{x}}^2⟩$

role="math" localid="1655379313238"

Since, the integral of an odd function over a symmetric interval is zero

i.e. $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{ye}}^{-{\mathrm{\lambda y}}^2}\mathrm{dy}=0$

Therefore,

Thus, value of $⟨x⟩$ is $\mathrm{a}$.

Now, for $⟨{\mathrm{x}}^2⟩$

Thus, value of $⟨{\mathrm{x}}^2⟩$ is $\frac{1}{2\mathrm{\lambda }}+{\mathrm{a}}^2$.

The standard deviation can be calculated as,

$\begin{array}{l}\mathrm{\sigma }=\sqrt{⟨{\mathrm{x}}^2⟩-{⟨\mathrm{x}⟩}^2}\\ \mathrm{\sigma }=\sqrt{(\frac{1}{2\mathrm{\lambda }}+{\mathrm{a}}^2)-{\mathrm{a}}^2}\\ \mathrm{\sigma }=\sqrt{\frac{1}{2\mathrm{\lambda }}}\end{array}$

Thus, value of σ is $\sqrt{\frac{1}{2\mathrm{\lambda }}}$.

Figure (1)

Here,