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 QFI Core Exam Old Advanced Portfolio Management Forum

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#1
06-13-2018, 12:52 PM
 komorgan Member Non-Actuary Join Date: Mar 2012 Posts: 109
Paul Wilmott, Chapt. 10

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

$\int_{t_0}^{T}exp(-rt_o)d[exp(-rt)(V^i-V^a)] = V^a - V^i$

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

$V^a - V^i + exp(-r(T-t_0))(V^i - V^a)$

So this seems to imply that $exp(-r(T-t_0))(V^i - V^a)$ = 0

Is this correct? If so, why do we assume it's 0?
#2
06-13-2018, 01:11 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 7,938

Quote:
 Originally Posted by komorgan My question pertains to the equation that's shown as: $\int_{t_0}^{T}exp(-rt_o)d[exp(-rt)(V^i-V^a)] = V^a - V^i$ When I work out the left side of this equation, I get: $V^a - V^i + exp(-r(T-t_0))(V^i - V^a)$ So this seems to imply that $exp(-r(T-t_0))(V^i - V^a)$ = 0 Is this correct? If so, why do we assume it's 0?
What do the V's represent?
#3
06-13-2018, 01:26 PM
 Zakfischer Member SOA Join Date: Jan 2015 Location: Chicago College: Brown University Posts: 269

Quote:
 Originally Posted by komorgan My question pertains to the equation that's shown as: $\int_{t_0}^{T}exp(-rt_o)d[exp(-rt)(V^i-V^a)] = V^a - V^i$ When I work out the left side of this equation, I get: $V^a - V^i + exp(-r(T-t_0))(V^i - V^a)$ So this seems to imply that $exp(-r(T-t_0))(V^i - V^a)$ = 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 $e^{-r(t_{0}-t_{0})} (V^{i}-V^{a}) = V^{i} - V^{a}$

- Therefore, we get the integral by doing upper bound - lower bound $= 0 -( V^{i} - V^{a}) = V^{a} - V^{i}$

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.

$e^{r t_{0}} \displaystyle\int_{t_{0}}^{T} d(e^{-rt} (V^{i}-V^{a})) = 0 - e^{-r(t_{0}-t_{0})} (V^{i}-V^{a}) = V^{a} - V^{i}$

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
__________________
Zachary Fischer, FSA CERA
TIA Instructor

QFI Core | QFI Advanced | QFI IRM | LRM | ERM

Last edited by Zakfischer; 06-13-2018 at 01:35 PM..
#4
06-13-2018, 01:59 PM
 komorgan Member Non-Actuary Join Date: Mar 2012 Posts: 109

Quote:
 Originally Posted by Zakfischer 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 $e^{-r(t_{0}-t_{0})} (V^{i}-V^{a}) = V^{i} - V^{a}$ - Therefore, we get the integral by doing upper bound - lower bound $= 0 -( V^{i} - V^{a}) = V^{a} - V^{i}$ 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. $e^{r t_{0}} \displaystyle\int_{t_{0}}^{T} d(e^{-rt} (V^{i}-V^{a})) = 0 - e^{-r(t_{0}-t_{0})} (V^{i}-V^{a}) = V^{a} - V^{i}$ 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!
#5
06-14-2018, 10:56 AM
 Zakfischer Member SOA Join Date: Jan 2015 Location: Chicago College: Brown University Posts: 269

Quote:
 Originally Posted by komorgan Thank you, Zach. That makes sense. I just wish they explained that in the book!
Glad that cleared things up!
__________________
Zachary Fischer, FSA CERA
TIA Instructor

QFI Core | QFI Advanced | QFI IRM | LRM | ERM
#6
07-02-2018, 03:29 PM
 komorgan Member Non-Actuary Join Date: Mar 2012 Posts: 109

Sorry, just realized the left side of the first equation I posted should've been:

$\int_{t_0}^{T}exp(rt_o)d[exp(-rt)(V^i-V^a)]$

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