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

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