Mean, squared mean and standard deviations of a gaussian

[sebastian616]

Homework Statement

find <x>, <[tex]x^{2}[/tex]> and [tex]\sigma[/tex] of the probability distribution;
[tex]\rho(x)[/tex] = [tex]\sqrt{\frac{\lambda}{\pi}}e[/tex][tex]^{-\lambda(x-a)^{2}}[/tex]

Homework Equations

As shown above

The Attempt at a Solution

I know that I can simply take the mean value (a) and the standard deviation(1/2[tex]\sigma[/tex]) from the fact that this equation is a gaussian, however wanted to know if there was a more rigorous method of calculating these figures than just saying what it is from the gaussian distribution.

 

[mighty2000]

Thank you for your question. it is important to approach problems in a rigorous and systematic manner. In this case, there is indeed a more rigorous method for calculating <x>, <x^{2}>, and \sigma for the given probability distribution.

Firstly, let's define the probability distribution function (PDF) for the given equation:

\rho(x) = \sqrt{\frac{\lambda}{\pi}}e^{-\lambda(x-a)^{2}}

We can rewrite this as:

\rho(x) = \frac{1}{\sqrt{\pi}}e^{-\frac{(x-a)^{2}}{\sigma^{2}}}

where \sigma = \frac{1}{\sqrt{2\lambda}}.

Now, to find <x>, we can use the formula:

<x> = \int_{-\infty}^{\infty}x\rho(x)dx

Substituting in our PDF, we get:

<x> = \int_{-\infty}^{\infty}x\frac{1}{\sqrt{\pi}}e^{-\frac{(x-a)^{2}}{\sigma^{2}}}dx

This integral can be solved using the substitution u = \frac{x-a}{\sigma}, which gives us:

<x> = \frac{a}{\sqrt{\pi}} + \sigma\int_{-\infty}^{\infty}ue^{-u^{2}}du

Using the fact that \int_{-\infty}^{\infty}ue^{-u^{2}}du = 0, we get:

<x> = \frac{a}{\sqrt{\pi}}

Similarly, to find <x^{2}>, we can use the formula:

<x^{2}> = \int_{-\infty}^{\infty}x^{2}\rho(x)dx

Substituting in our PDF, we get:

<x^{2}> = \int_{-\infty}^{\infty}x^{2}\frac{1}{\sqrt{\pi}}e^{-\frac{(x-a)^{2}}{\sigma^{2}}}dx

Using the same substitution as before, we get:

<x^{2}> = \frac{a^{2}}{\sqrt{\pi}} + \sigma^{2}\int_{-\infty}^{\infty}u^{2}e^{-u^{2