What is the reason for the one-half in the normal pdf's gaussian (i.e. : why $\exp(-x^{2}/2)$ instead of $\exp(-x^{2})$ )

Question

It doesn't seem to relate to normalization, as the normalizing constant adapts to every possible "upstairs formulation", and in the standard case is $\displaystyle\frac{1}{\sqrt{2\pi}}$. Does it relate to fundamental probabilistic assumptions or to useful functional properties?

Answer 1

It is for normalization, to ensure that the variance is $1$.

Answer 2

and note that since the mean is zero by symmetry, you can calculate the variance quite simply. with parameter $a$ for the distribution, the variance is: $$ v = \frac {I_a(x^2)}{I_a(1)} \\ $$ where $$ I_a(f(x)) = \int_{-\infty}^{\infty} f(x) exp(-ax^2)dx $$ a straightforward integration by parts gives: $$ I_a(x^2) = \frac1{2a}I_a(1) $$ so for $v=1$ we require $a=\frac12$