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QFI Core Exam Old Advanced Portfolio Management Forum

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  #1  
Old 06-13-2018, 12:52 PM
komorgan komorgan is offline
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Default Paul Wilmott, Chapt. 10

My question pertains to the equation that's shown as:



When I work out the left side of this equation, I get:




So this seems to imply that = 0

Is this correct? If so, why do we assume it's 0?
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Old 06-13-2018, 01:11 PM
Academic Actuary Academic Actuary is offline
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Quote:
Originally Posted by komorgan View Post
My question pertains to the equation that's shown as:



When I work out the left side of this equation, I get:




So this seems to imply that = 0

Is this correct? If so, why do we assume it's 0?
What do the V's represent?
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  #3  
Old 06-13-2018, 01:26 PM
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Zakfischer Zakfischer is offline
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Quote:
Originally Posted by komorgan View Post
My question pertains to the equation that's shown as:



When I work out the left side of this equation, I get:




So this seems to imply that = 0

Is this correct? If so, why do we assume it's 0?
You are close. The integral has an upper and lower bound.

- When plugging in the upper bound of "T" you get 0.

- When plugging in the lower bound of "t0" you get

- Therefore, we get the integral by doing upper bound - lower bound

Here are the details of how we know this:

Important realization: At expiration, the value of an option is its payoff. This is obvious, but important. Because at time T.....we must have Vi = Va. The value at expiration is equal to the payoff, and the volatility parameter used does not matter at expiration! In other words - at expiration, our value (if we use a call as an example) is max(ST-K,0) and the same value regardless of sigma. So at expiration time T, Vi = Va. Therefore, when we plug in at the upper bound T for our integration, that part will cancel out.



The idea is that the integral accumulates the total present value of profit. We see that for the hedging with actual volatility case, the total present value of profit is very simple - it is the difference between the OV with actual volatility and the market price.

You will see this again in QFI-115, so feel free to look at both sections to solidify your understanding. If you are using TIA, you will find it helpful to go to page 39 of the Section 2 DSG. This will likely clear everything up.

Hopefully that helps clarify

Also see here for a similar discussion: http://www.actuarialoutpost.com/actu...d.php?t=328799
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Last edited by Zakfischer; 06-13-2018 at 01:35 PM..
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Old 06-13-2018, 01:59 PM
komorgan komorgan is offline
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Quote:
Originally Posted by Zakfischer View Post
You are close. The integral has an upper and lower bound.

- When plugging in the upper bound of "T" you get 0.

- When plugging in the lower bound of "t0" you get

- Therefore, we get the integral by doing upper bound - lower bound

Here are the details of how we know this:

Important realization: At expiration, the value of an option is its payoff. This is obvious, but important. Because at time T.....we must have Vi = Va. The value at expiration is equal to the payoff, and the volatility parameter used does not matter at expiration! In other words - at expiration, our value (if we use a call as an example) is max(ST-K,0) and the same value regardless of sigma. So at expiration time T, Vi = Va. Therefore, when we plug in at the upper bound T for our integration, that part will cancel out.



The idea is that the integral accumulates the total present value of profit. We see that for the hedging with actual volatility case, the total present value of profit is very simple - it is the difference between the OV with actual volatility and the market price.

You will see this again in QFI-115, so feel free to look at both sections to solidify your understanding. If you are using TIA, you will find it helpful to go to page 39 of the Section 2 DSG. This will likely clear everything up.

Hopefully that helps clarify

Also see here for a similar discussion: http://www.actuarialoutpost.com/actu...d.php?t=328799
Thank you, Zach. That makes sense. I just wish they explained that in the book!
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  #5  
Old 06-14-2018, 10:56 AM
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Zakfischer Zakfischer is offline
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Originally Posted by komorgan View Post
Thank you, Zach. That makes sense. I just wish they explained that in the book!
Glad that cleared things up!
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QFI Core | QFI Advanced | QFI IRM | LRM | ERM
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  #6  
Old 07-02-2018, 03:29 PM
komorgan komorgan is offline
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Sorry, just realized the left side of the first equation I posted should've been:

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