{\displaystyle \int _{-\infty }^{\infty }(x-\langle f\rangle )^{2}f(x)\,\mathrm {d} x={{{\sqrt {-\pi \mu }}\left(2\mu -\kappa ^{2}\right)} \over {4\mu ^{3}}}\,e^{-1+\langle f\rangle \kappa +\nu -{{\kappa ^{2}} \over {4\mu }}}=\sigma ^{2}}