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# This note has been used to help create the Normal Distribution wiki

Begin with a row from the Pascal triangle, preferably some large exponent , derive the Gaussian Distribution.

Solution

The bell curve is a probability density curve of binary systems. Then the probability at a some displacement from the medium is $P(n, k) = \left( \begin{matrix} n \\ k \end{matrix} \right) {2}^{-n}= \frac{n!}{(\frac{1}{2}n + k)! (\frac{1}{2}n - k)! {2}^{n}}$

Using the Stirling approximation and treating $$k = \frac{\sigma}{2}$$, we have
$P(n, \sigma) \sim {\left(\frac{n}{2\pi} \right)}^{\frac{1}{2}} {\left(\frac{n}{2}\right)}^{n} {\left(\frac{{n}^{2} - {\sigma}^{2}}{4}\right)}^{-\frac{1}{2}(n+1)}{\left(\frac{n + \sigma}{n-\sigma}\right)}^{\frac{-\sigma}{2}}.$

For $$n>>\sigma$$, $$\frac{n + \sigma}{n-\sigma} \sim 1+\frac{2\sigma}{n}$$; hence, for large $$n$$ $P(n, \sigma) \sim {\left(\frac{n}{2\pi} \right)}^{\frac{1}{2}} {\left(1- \frac{{\sigma}^{2}}{{n}^{2}}\right)}^{-\frac{1}{2}(n+1)}{\left(1+\frac{2\sigma}{n}\right)}^{\frac{-\sigma}{2}}.$

Taking the logarithm yields $ln(P(n,\sigma)) \sim \frac{1}{2}ln \left (\frac{2}{\pi n}\right) - \frac{1}{2}(n+1)ln \left (1- \frac{{\sigma}^{2}}{{n}^{2}}\right) - \frac{\sigma}{2}ln \left (1+\frac{2\sigma}{n}\right).$

For small $$x$$, $$ln(1+x) \approx x$$; subsequently, $ln(P(n,\sigma)) \sim \frac{1}{2}ln \left (\frac{2}{\pi n}\right) - \frac{1}{2}(n+1) \left (-\frac{{\sigma}^{2}}{{n}^{2}}\right) - \frac{\sigma}{2} \left (\frac{2\sigma}{n}\right)$ or $ln(P(n,\sigma)) \sim \frac{1}{2}ln \left (\frac{2}{\pi n}\right) + \frac{{\sigma}^{2}}{{n}^{2}} - \frac{{\sigma}^{2}}{2n}.$

Since $$\frac{{\sigma}^{2}}{{n}^{2}}$$ vanishes faster than $$\frac{{\sigma}^{2}}{2n}$$ for very large $$n$$, we arrive at the result:

$P(n, \sigma) = {\left(\frac{2}{\pi n} \right)}^{\frac{1}{2}} {e}^{\frac{-{\sigma}^{2}}{2n}}.$

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
3 years, 4 months ago

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