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April 16th, 2017, 09:36 AM   #1
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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?
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April 16th, 2017, 10:20 AM   #2
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Quote:
Originally Posted by Ku5htr1m View Post
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, $\gamma = r$ ?
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April 16th, 2017, 11:14 AM   #3
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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.
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April 16th, 2017, 02:50 PM   #4
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In that case, the dropped tilde accounts for the confusion and there's nothing else to explain.
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