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 April 19th, 2017, 06:08 AM #1 Member   Joined: Sep 2013 Posts: 81 Thanks: 0 Can this equation be expressed in this way? I recently read an article and they provide a formula for the first-passage 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: 6,213 Thanks: 491 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: 81 Thanks: 0 so you think its alright _
 April 20th, 2017, 04:11 AM #4 Senior Member   Joined: May 2016 From: USA Posts: 574 Thanks: 248 I must have slipped a cog (or perhaps 20) because I am not following. $a = \dfrac{1}{b} * \{ \ln(c) + d \} = \dfrac{ln(c)}{b} + \dfrac{d}{b} \implies$ $\dfrac{ab - d}{b} = \dfrac{ln(c)}{b} \implies ab - d = ln(c) \implies ln \left (e^{(ab-d)} \right ) = ln(c) \implies$ $e^{(ab-d)} = c \implies c = exp(ab - d).$
April 20th, 2017, 07:27 AM   #5
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 Originally Posted by JeffM1 I must have slipped a cog (or perhaps 20) because I am not following. $a = \dfrac{1}{b} * \{ \ln(c) + d \} = \dfrac{ln(c)}{b} + \dfrac{d}{b} \implies$ $\dfrac{ab - d}{b} = \dfrac{ln(c)}{b} \implies ab - d = ln(c) \implies ln \left (e^{(ab-d)} \right ) = ln(c) \implies$ e^{(ab-d)} = c \implies c = exp(ab - d).$hm... the negative sign in$\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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