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 March 4th, 2018, 09:18 AM #1 Newbie   Joined: Sep 2017 From: xxxx Posts: 8 Thanks: 0 ito lemma Assuming a stock with alpha 0.14, delta 0.01 and volatility of 0.48, how do I derive the price process of a derivative with value V = S^(1/1)e^(0.4)*(1-t) [Equation 1], by using Ito lemma (equation 2)? Ito lemma (equation 2): Where Vs = first derivative of [equation 1] Vt is the first derivative with respect to time of [equation 1] Vss is the second derivative of [equation 1] I thought that the first derivative is: e^(2/5)s and the second one is zero. But I did not take the time component into account since I don't know how do that. At last, one has to substitute the S in equation 2 (ito lemma) with the inverse of V = S^(1/1)e^(0.4)*(1-t) (equation 1). I also don't know what the inverse is and how to substitute. Could someone please help me? I would like to know how to take the derivatives of the first and second order and with respect to time and how to substitute the inverse of equation 1 into equation 2 to replace S? I would be very thankful!!
 March 5th, 2018, 01:32 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 In "equation 1" surely you do not mean $\displaystyle S^{(1/1)}$. What was that supposed to be?
March 5th, 2018, 02:39 AM   #3
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 Originally Posted by Country Boy In "equation 1" surely you do not mean $\displaystyle S^{(1/1)}$. What was that supposed to be?
S^1/1 --> will be just S in this case.
Regarding my original question, valuing a derivative means a contract based on the price of a stock e.g. call option. The other 'derivatives' are derivatives in math (1st and 2nd order).

Last edited by skipjack; March 5th, 2018 at 07:32 AM.

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