Translations:Analog prediction error/25/es

${\begin{aligned}\int \limits _{0}^{\infty }{{\left[{\tilde {g}}\left(t{\rm {|1}}\right)\right]}^{2}}dt=\int \limits _{0}^{\infty }{\rm {8}}e^{-2\left({\rm {t+1}}\right)dt}{\rm {=4}}\int \limits _{2}^{\infty }{e^{-u}}du{\rm {=4}}e^{-2}.\end{aligned}}$

(202)