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April 19th, 2017, 06:08 AM   #1
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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 ?
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April 19th, 2017, 04:51 PM   #2
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σ seems to be all over the place. Otherwise it is O.K.
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April 20th, 2017, 01:11 AM   #3
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so you think its alright _
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April 20th, 2017, 04:11 AM   #4
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I must have slipped a cog (or perhaps 20) because I am not following.

$a = \dfrac{1}{b} * \{ \ln(c) + d \} = \dfrac{ln(c)}{b} + \dfrac{d}{b} \implies$

$\dfrac{ab - d}{b} = \dfrac{ln(c)}{b} \implies ab - d = ln(c) \implies ln \left (e^{(ab-d)} \right ) = ln(c) \implies$

$e^{(ab-d)} = c \implies c = exp(ab - d).$
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April 20th, 2017, 07:27 AM   #5
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Quote:
Originally Posted by JeffM1 View Post
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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