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 April 16th, 2017, 09:36 AM #1 Member   Joined: Sep 2013 Posts: 83 Thanks: 0 How are the expressions equal? In a paper i recently read , the author claims that: $\displaystyle (1){e^{ - a{{(r + (1/2){\sigma ^2})}^2}\tau /(2\sigma {}^2)}} = (2){e^{ - {1 \over 4}{{(\gamma + 1)}^2}a\tau }}$ For (2) the following are true $\displaystyle \gamma = {{2r} \over {{\sigma ^2}}}$ $\displaystyle \tau = \tau *{{{\sigma ^2}} \over 2}$ I think he has made an error because i cannot see the two expressions being equal? Can someone spot a possible typo?
April 16th, 2017, 10:20 AM   #2
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 Originally Posted by Ku5htr1m In a paper i recently read , the author claims that: $\displaystyle (1){e^{ - a{{(r + (1/2){\sigma ^2})}^2}\tau /(2\sigma {}^2)}} = (2){e^{ - {1 \over 4}{{(\gamma + 1)}^2}a\tau }}$ For (2) the following are true $\displaystyle \gamma = {{2r} \over {{\sigma ^2}}}$ $\displaystyle \tau = \tau *{{{\sigma ^2}} \over 2}$ I think he has made an error because i cannot see the two expressions being equal? Can someone spot a possible typo?
Is this equation correct?

$\displaystyle \tau = \tau *{{{\sigma ^2}} \over 2}$

... if so, $\gamma = r$ ?

 April 16th, 2017, 11:14 AM #3 Member   Joined: Sep 2013 Posts: 83 Thanks: 0 Well, it is $\displaystyle \tilde \tau = \tau *{{\sigma ^2} \over 2}$ But the author drops tilde, I forgot to put tilde on tau in (2). Anyhow, this is only relevant for (2), so if you insert it you get $\displaystyle e^{ - {1 \over 4}(\gamma + 1)^2a*\tau *{{\sigma ^2} \over 2}}$ Last edited by skipjack; April 16th, 2017 at 02:48 PM.
 April 16th, 2017, 02:50 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,962 Thanks: 1606 In that case, the dropped tilde accounts for the confusion and there's nothing else to explain.

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