What's new

# Hull Risk Management Ch10 - EOC-QUESTIONS 10.21 and 10.5

##### Active Member
Hi,
In Reference to HULLCH10_Question 10.5:

What is Simplified Approach Equation 10.4 the the solution is referring to here ..?

Question 10.5: Suppose that observations on an exchange rate at the end of the past 11 days have been 0.7000, 0.7010, 0.7070, 0.6999, 0.6970, 0.7003, 0.6951, 0.6953, 0.6934, 0.6923, and 0.6922. Estimate the daily volatility using both approaches in Section 10.5.

Answer: The standard formula for calculating standard deviation gives 0.547% per day. The simplified approach in equation (10.4) gives 0.530% per day.

#### Nicole Seaman

Staff member
Subscriber
Hi,
In Reference to QUANT-HULL_CH10_VOLATILITY_PROBLEM-10.5 :-

What is Simplified Approach Equation 10.4 the the solution is referring to here ..?

View attachment 1838

Here is the equation that you are looking for. I'm actually working on an updated version of this document that has not been published yet, and it includes the actual equation instead of just referencing equation 10.4. I hope this helps!

##### Active Member

Here is the equation that you are looking for. I'm actually working on an updated version of this document that has not been published yet, and it includes the actual equation instead of just referencing equation 10.4. I hope this helps!

View attachment 1839
Thanks so much @Nicole Seaman for coming to the rescue so promptly

##### Active Member
Hi,

In the Problem below, how and where did we get Alpha = LN( 1/.98 ) ...? which formula is this ...?
Also, in Part C of this problem, where did we get the part highlighted in yellow ... ...?

Thanks all for any insights on this.

Question 10.21

Suppose that the parameters in a GARCH(1,1) model are α = 0.03, β = 0.95 and ω = 0.000002.

a) What is the long-run average volatility?
b) If the current volatility is 1.5% per day, what is your estimate of the volatility in 20, 40, and 60 days?
c) What volatility should be used to price 20-, 40-, and 60-day options?
d) Suppose that there is an event that increases the volatility from 1.5% per day to 2% per day. Estimate the effect on the volatility in 20, 40, and 60 days.

Estimate by how much the event increases the volatilities used to price 20-, 40-, and 60-day options.

Last edited by a moderator:

#### QuantMan2318

##### Active Member
Subscriber

The solution is confusing because the formula that you have highlighted is not the estimate of alpha but is the estimate of long run Gamma of the equation in $\gamma V_L$, where:

we try to find the estimate of the long run Variance as V(t) as $E({\sigma^{2_{}}}_{(n+t)})$ for the term structure of Variance/Volatility

and the Coefficient $\gamma\;=\;nl(1/(\alpha+\beta))$>>>>>>>>> 1

Why is this so? we know that Gamma = 1 - (alpha +Beta) and hence the logarithmic transformation of the same is nl(1/(alpha+beta)) - this was my understanding of the formula based on rough log equivalence

We know that the estimate of Variance for a future time period is given as:

$E( {\sigma^2}_{(n+t)})\;=\;V_{L_{}}+(\alpha+\beta)^t\;\ast\;({\sigma^2}_n\;-\;V_L)$

Therefore, we substitute equation (1) in the above formula which becomes:

$V(t)\;=\;V_L+e^{-\gamma t}\lbrack V(O)-V_L\rbrack$V

To find the estimate of implied volatility at time T we take the average of the above (Average volatility) which is the integral of the above equation:

$\sigma(T)^2\;=\;252\;\ast\;(V_L\;+\;(1\;-\;e^{-\gamma T})/\gamma T\ast\lbrack V(O)\;-\;V_L\rbrack)$

In other words

You get the Gamma as nl(1/(0.03+0.95)) which is 0.0202

furthermore as (alpha+beta)^t is nothing but e^-gamma *t taking the inverse, you get the above solution

Let me know if the above would help

Thanks
Mani

Last edited:

##### Active Member
Thanks so much @QuantMan2318 - thought I almost understood this - and then hit this hurdle ...
As you rightly pointed out: Given that Gamma = [ 1 - (Alpha + Beta ) ] , How is Gamma = LN [ 1 / (Alpha +Beta ) ] ...?

#### QuantMan2318

##### Active Member
Subscriber

Why is this so? we know that Gamma = 1 - (alpha +Beta) and hence the logarithmic transformation of the same is nl(1/(alpha+beta)) - this was my understanding of the formula based on rough log equivalence

Let me know if the above would help

Thanks
Mani
Thanks so much @QuantMan2318 - thought I almost understood this - and then hit this hurdle ...
As you rightly pointed out: Given that Gamma = [ 1 - (Alpha + Beta ) ] , How is Gamma = LN [ 1 / (Alpha +Beta ) ] ...?

This is just based on log transformation, it is not strictly equal of course, I presumed that the authors assumed a transformation of the original formula to arrive at that (you may refer my first quote above) , @David Harper CFA FRM can correct me if I am wrong.

We know that Alpha + Beta + Gamma = 1

Divide both sides by Alpha + Beta

1/(1-Gamma) = 1/(Alpha + Beta)

Taking log on both sides

-nl(1-Gamma) = nl(1/(Alpha +Beta))

which I can assume is roughly equivalent to or some way related to Gamma - I do not know more than this on why the author took that formula for prediction of Variances other than the fact that it is transformation of some kind

More mathematical members can shed more light here

Thanks

##### Active Member
This is just based on log transformation, it is not strictly equal of course, I presumed that the authors assumed a transformation of the original formula to arrive at that (you may refer my first quote above) , @David Harper CFA FRM can correct me if I am wrong.

We know that Alpha + Beta + Gamma = 1

Divide both sides by Alpha + Beta

1/(1-Gamma) = 1/(Alpha + Beta)

Taking log on both sides

-nl(1-Gamma) = nl(1/(Alpha +Beta))

which I can assume is roughly equivalent to or some way related to Gamma - I do not know more than this on why the author took that formula for prediction of Variances other than the fact that it is transformation of some kind

More mathematical members can shed more light here

Thanks
Thanks again so much @QuantMan2318 - Think I got it this time around
As Gamma is a small no, (Gamma)^2 and Higher Powers of Gamma and so on are negligible and can be approximated to 0.
So by the Taylor Series Expansion of Ln ( 1+ x) approximates to x with the higher powers of x being almost equal to 0.

Again Thanks so much for taking the time to breaking this formula down for me - It was a Huge AHA Moment for me !! THANK YOU

##### Active Member
@QuantMan2318 @David Harper CFA FRM Have one more follow up question on this one..

So, the Variance for a future time period is given by :-
()=+−[()−] -------Equation 1

To find the estimate of implied volatility at time T we take the average of the above (Average volatility) which is the integral of the above equation:
()2=252∗(+(1−−)/∗[()−]) -------Equation 2

Why do we have to further take the Integral of the above Equation 1 and get to the Equation 2 ...? Could we NOT use the Equation 1 itself to find the 20, 40 60 Day Volatilities...?

P.S : Please ignore this last Follow Up Question- Think I was able to wrap my head around it finally.

Last edited:

##### Active Member

I am a bit rusty on Integrals and so have a follow up question
V(t) = V(L )+ e^ -Gamma*T [ V(0) - V(L) ]
So, I/T * Integral [ V(t) ] = 1/T * Integral { V(L )+ e^ -Gamma*T [ V(0) - V(L) ] }

Now, Integral { V(L )+ e^ -Gamma*T [ V(0) - V(L) ] } = Integral { V(L ) . dT } + Integral { e^ -Gamma*T [ V(0) - V(L) ] }

Integral { V(L ) . dT } = V(L ) * T -----(a)
Integral { e^ -Gamma*T [ V(0) - V(L) ] } = [ V(0) - V(L) ] * [ ( -1/ Gamma) * e^ - Gamma*T ]
so, Integral { e^ -Gamma*T [ V(0) - V(L) ] } = { - [ V(0) - V(L) ] / Gamma } * e^ - Gamma*T -----(b)

V(L ) * T { - [ V(0) - V(L) ] / Gamma } * e^ - Gamma*T

I am not able to derive/get the [ V(0) - V(L) ] / Gamma that needs to be added...

Can some one expand the Integration of the V(t) ..I know I am NOT doing the Integral right ...

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @gargi.adhikari I won't have time to work detailed calculus issues four (4) days before the exam, if you think about it (if ever!). Have you plugged this into Mathematica? That's what I do with difficult calculus, it shows step-by-step. Their alpha is really amazing, it's saved me so many times with calculus, in particular integrals, see https://www.wolframalpha.com/

(and if you find the solution, maybe you can share it back and try to be helpful to somebody else?)

Thanks,

Last edited:

#### akrushn2

##### Member
Subscriber
Not able to actually calculate out to the answer on 10.21 c- I keep getting 23.569% This is my calculation. 252*(0.0001 + 1-e^-20.0202/0.0202*20*(0.015^2/252-0.0001))= this is 252*(0.000222139)= 0.055978 sqrt of this is = 23.659%. Where am I going wrong here?

Last edited:

Staff member
Subscriber
Last edited: