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April 19th, 2017, 06:08 AM  #1 
Member Joined: Sep 2013 Posts: 83 Thanks: 0  Can this equation be expressed in this way?
I recently read an article and they provide a formula for the firstpassage probability as $$Z = {1 \over \sigma }\left[ {\log S/{S_c} + (r  {1 \over 2}{\sigma ^2})t} \right]$$ ${{S_c}}$ value of the stock at time t , $r$ ror on the stock, $\sigma $ standard deviation. Can this be expressed as $$Z = \exp \left\{ {\sigma X(t) + (\sigma r  {1 \over 2}{\sigma ^2})t} \right\}$$ where $$X(t) = \log S/Sc$$ is this the same way just different way to put it ? 
April 19th, 2017, 04:51 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,510 Thanks: 584 
σ seems to be all over the place. Otherwise it is O.K.

April 20th, 2017, 01:11 AM  #3 
Member Joined: Sep 2013 Posts: 83 Thanks: 0 
so you think its alright _

April 20th, 2017, 04:11 AM  #4 
Senior Member Joined: May 2016 From: USA Posts: 1,027 Thanks: 419 
I must have slipped a cog (or perhaps 20) because I am not following. $a = \dfrac{1}{b} * \{ \ln(c) + d \} = \dfrac{ln(c)}{b} + \dfrac{d}{b} \implies$ $\dfrac{ab  d}{b} = \dfrac{ln(c)}{b} \implies ab  d = ln(c) \implies ln \left (e^{(abd)} \right ) = ln(c) \implies$ $e^{(abd)} = c \implies c = exp(ab  d).$ 
April 20th, 2017, 07:27 AM  #5  
Member Joined: Sep 2013 Posts: 83 Thanks: 0  Quote:
hm... the negative sign in $\displaystyle c = exp(ab  d)$ is troubling, should it not be + ? Last edited by Ku5htr1m; April 20th, 2017 at 07:31 AM.  

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