The PDF for the standard normal is: #mathbb P(z) = 1/sqrt(2 pi) e^(- z^2/2)#
It has mean value:
# mu = int_(-oo)^(oo) dz \ z \ mathbb P(z) = 1/sqrt(2 pi) int_(-oo)^(oo) \ dz \ z e^(- z^2/2) #
# =1/sqrt(2 pi) int_(-oo)^(oo) \ d( - e^(- z^2/2)) #
# =1/sqrt(2 pi) [ e^(- z^2/2) ]_(oo)^(-oo) = 0 #
It follows that:
# Var(z) = int_(-oo)^(oo) dz \ (z - mu)^2 mathbb P(z) #
# = 1/sqrt(2 pi) int_(-oo)^(oo) \ dz \ z^2 e^(- z^2/2) #
This time, use IBP:
# Var(z) = - 1/sqrt(2 pi) int_(-oo)^(oo) \ d( e^(- z^2/2)) \ z #
# = - 1/sqrt(2 pi) ( [ z e^(- z^2/2) ]\_(-oo)^(oo) - int_(-oo)^(oo) dz \ e^(- z^2/2) ) #
# = - 1/sqrt(2 pi) ( [ z e^(- z^2/2) ]\_(-oo)^(oo) - int_(-oo)^(oo) dz \ e^(- z^2/2) ) #
Because # [ z e^(- z^2/2) ]\_(-oo)^(oo) = 0#
# = 1/sqrt(2 pi) int_(-oo)^(oo) dz \ e^(- z^2/2) #
This integral is well known. It can be done using a polar sub, but here the result is stated.
# Var(z) = 1/sqrt(2 pi) sqrt(2 pi) = 1 #