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#11
07-11-2012, 05:44 PM
 hello_actuary Member SOA Join Date: Jul 2012 Posts: 103

Quote:
 Originally Posted by Abraham Weishaus I allow posting exercises if you ask specific questions about them, something you don't understand, and specify where the exercise came from. Your list above seems very unspecific. What exactly didn't you understand about the solutions? Which lines puzzled you? In some cases you want an alternative solution. Why wasn't the solution provided adequate? I admit that occasionally students do find shorter and better solutions than the ones I provide.
Hello, I will ask you questions I had in other chapters later on. I am in Chapter 26 of your ASM MFE 9th edition and have a question with you now. On P521 Quiz 26-2, You did change the model from risk-netural (let's use *)to true probability measure.

d(Z(t)*) = d(Z(t)) + sharpe ratio (dt).

According to P518 formula 26.8, sigma(r) * sharpe ratio = 0.08 * 0.2 = 0.016, so a(r) = -0.01 and the true drift should be -0.01 + 0.016 = 0.006. Would you please explain why did you solve the true drift as -0.01 - 0.016 = -0.026?

Thank you.
#12
10-31-2012, 06:57 PM
 yminev Join Date: Jul 2011 Posts: 12

Quote:
 Originally Posted by hello_actuary Hello, I will ask you questions I had in other chapters later on. I am in Chapter 26 of your ASM MFE 9th edition and have a question with you now. On P521 Quiz 26-2, You did change the model from risk-netural (let's use *)to true probability measure. d(Z(t)*) = d(Z(t)) + sharpe ratio (dt). According to P518 formula 26.8, sigma(r) * sharpe ratio = 0.08 * 0.2 = 0.016, so a(r) = -0.01 and the true drift should be -0.01 + 0.016 = 0.006. Would you please explain why did you solve the true drift as -0.01 - 0.016 = -0.026? Thank you.
Was there an explanation for why in quiz 26.2, the true drift is -0.026 instead of 0.006? I am also confused with this!

Thank you!
#13
02-02-2013, 03:53 PM
 tysox45 Join Date: Mar 2012 Posts: 13

I could be wrong but I think this is why. On 518 he says that the coefficient of dZ(t) will be negative. Since we are given a bond process with a positive coefficient it means the related process for assets had the opposite coefficient which is -.08, and must be used when calculating the true drift, since the Sharpe ratio given was for assets and not bonds. But when calculating the final probability we use the the positive .08 that is given in
dr(t)=-.01dt+.08dZ(hat)(t).

If anybody could confirm this reasoning it would be greatly appreciated, as I was initially confused about this.
#14
02-02-2013, 08:18 PM
 tkt Member CAS SOA Join Date: Jun 2011 Location: Des Moines College: Graduated from Drake University Posts: 145

Quote:
 Originally Posted by hello_actuary Hello, I will ask you questions I had in other chapters later on. I am in Chapter 26 of your ASM MFE 9th edition and have a question with you now. On P521 Quiz 26-2, You did change the model from risk-netural (let's use *)to true probability measure. d(Z(t)*) = d(Z(t)) + sharpe ratio (dt). According to P518 formula 26.8, sigma(r) * sharpe ratio = 0.08 * 0.2 = 0.016, so a(r) = -0.01 and the true drift should be -0.01 + 0.016 = 0.006. Would you please explain why did you solve the true drift as -0.01 - 0.016 = -0.026? Thank you.
Under the true probability measure, the short rate process is:
dr(t) = a(r) dt + sigma(r) dZ

Under the risk-neutral probability measure, the short rate process is:
dr(t) = [a(r) + sigma(r)*phi(r,t)] dt + sigma(r) d(Z_tilde)

In regards to quiz 26-2 in the ASM manual, we are given the short rate process under the risk-neutral probability measure:
dr(t) = -0.01 dt + 0.08 d(Z_tilde)

We are also given that phi(r,t) = 0.2.

So, from the given short rate process, we can deduce that:
sigma(r) = 0.08
[a(r) + sigma(r)*phi(r,t)] = -0.01

Thus, a(r) + 0.08*0.2 = -0.01 and so a(r) = -0.01 - 0.08*0.2 = -0.026.

Hope this helps!
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