A Concrete Example: The Gaussian Case

To understand how a universal scaling form emerges, let us analyze in detail the case of a Gaussian distribution with zero mean and variance $\sigma ^{2}$. Using integration by parts, we can write :

$P^{<}(E)=\int _{-\infty }^{E}{\frac {dx}{\sqrt {2\pi \sigma ^{2}}}}\,e^{-{\frac {x^{2}}{2\sigma ^{2}}}}={\frac {1}{2{\sqrt {\pi }}}}\int _{\frac {E^{2}}{2\sigma ^{2}}}^{\infty }{\frac {dt}{\sqrt {t}}}e^{-t}={\frac {\sigma }{{\sqrt {2\pi }}|E|}}e^{-{\frac {E^{2}}{2\sigma ^{2}}}}-{\frac {1}{4{\sqrt {\pi }}}}\int _{\frac {E^{2}}{2\sigma ^{2}}}^{\infty }{\frac {dt}{t}}e^{-t}$

Hence we derive the following asymptotic expansion for $E\to -\infty$ :

$P^{<}(E)\approx {\frac {\sigma }{{\sqrt {2\pi }}|E|}}e^{-{\frac {E^{2}}{2\sigma ^{2}}}}+O({\frac {e^{-{\frac {E^{2}}{2\sigma ^{2}}}}}{|E|^{2}}})$

It is convenient to introduce the function $A(E)$ defined as

$P^{<}(E)=\exp(A(E))\quad A_{\text{gauss}}(E)=-{\frac {E^{2}}{2\sigma ^{2}}}-\log({\frac {{\sqrt {2\pi }}|E|}{\sigma }})+\ldots$

Using this expansion and the second relation introduced earlier, show that for large $M$, the typical value of the minimum energy is:

$E_{\min }^{\text{typ}}=-\sigma {\sqrt {2\log M}}\left(1-{\frac {1}{4}}{\frac {\log(4\pi \log M)}{\log M}}+\ldots \right)$