My Math Forum Expectation proof

 January 30th, 2017, 12:03 PM #1 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Expectation proof $\displaystyle E[e^x-1] = \Phi (\frac{u +t}{\sqrt(t)}) e^{u+t/2} - \Phi (\frac{u}{\sqrt(t)})$ where $\displaystyle \Phi$ deontes the cumulative distribution function of the N(0,1) Normal distribution i.e $\displaystyle \Phi(x) = (2\pi)^{-0.5} \int_{-\infty}^{x} e^{\frac{-s^2}{2} } ds$ prove that: $\displaystyle E[e^x -1 ] - E[1 - e^x] = e^{u+t/2} - 1$ Any tips on how to prove this. Every time I try, I seem to just end up with $\displaystyle 2E[e^x -1]$
 January 30th, 2017, 05:26 PM #2 Global Moderator   Joined: May 2007 Posts: 6,216 Thanks: 493 Something is wrong in your statement. It looks like you are doubling the original expression. Thanks from 123qwerty
 January 31st, 2017, 04:38 AM #3 Senior Member   Joined: Feb 2015 From: london Posts: 121 Thanks: 0 Which statement do you think is wrong?
January 31st, 2017, 07:36 AM   #4
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Quote:
 Originally Posted by calypso Which statement do you think is wrong?
The last seems slightly odd: $E(e^X - 1) - E(1 - e^X) = E(e^X - 1) + E(e^X - 1) = 2E(e^X - 1)$.

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