W woolley New member Joined Sep 9, 2007 Messages 22 Jan 21, 2009 #1 The original question: draw a graph to show that \(\displaystyle \Phi(a)=1-\Phi(-a)\). i know that this is because of symmetry and symmetry also implies third moment is zero. how do i draw a graph to show this?
The original question: draw a graph to show that \(\displaystyle \Phi(a)=1-\Phi(-a)\). i know that this is because of symmetry and symmetry also implies third moment is zero. how do i draw a graph to show this?
W woolley New member Joined Sep 9, 2007 Messages 22 Jan 21, 2009 #2 third moment is symmetry right? so does third moment of zero mean that the distribution is perfectly symmetric? do i simply draw a bell curve???
third moment is symmetry right? so does third moment of zero mean that the distribution is perfectly symmetric? do i simply draw a bell curve???
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Jan 21, 2009 #3 I'd graph f(a) = Phi(a) + Phi(-a) and show it was constant (1).
