Actuarial Outpost > MFE SOA Sample #22
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#1
10-04-2008, 04:30 AM
 MarcNguyen Member Join Date: Oct 2007 Posts: 60
SOA Sample #22

You are given:
i) Ito process of short-rate: dr(t) =[0.09-0.5r(t)]dt +0.3dZ
ii) Risk-neutral: dr(t) = [0.15-0.5r(t)]dt + sigma(r(t))dZ~
iii) g(t) is derivative of r(t)
dg(r,t) = m(r,g,t)dt -0.4g(r,t)dZ
Determine m(r,g)?

SOA solution: (r+0.08)g
But I think this solution isn't correct because formula (24.19) in textbook that
so Shapre ratio of short-rate is -0.2 then solution must be g(r-0.08).

am I right?
#2
10-04-2008, 10:04 AM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,262

Quote:
 Originally Posted by MarcNguyen You are given: i) Ito process of short-rate: dr(t) =[0.09-0.5r(t)]dt +0.3dZ ii) Risk-neutral: dr(t) = [0.15-0.5r(t)]dt + sigma(r(t))dZ~ iii) g(t) is derivative of r(t) dg(r,t) = m(r,g,t)dt -0.4g(r,t)dZ Determine m(r,g)? SOA solution: (r+0.08)g But I think this solution isn't correct because formula (24.19) in textbook that is dr =[a(r)- (risk premium)]dt+sigma(r)dZ in stead of dr =[a(r) + (risk premium)]dt+sigma(r)dZ so Shapre ratio of short-rate is -0.2 then solution must be g(r-0.08). am I right?
The errata to McDonald corrected a systematic problem with the sign conventions in much of chapter 24. See

and in particular the corrected pages in the early part of chapter 24

In the revised pages equation (24.19) reads
$dr = $a(r) + \sigma(r) \phi(r, t)$\,dt + \sigma(r) \,d\tilde{Z}$
__________________
The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.
#3
10-04-2008, 05:28 PM
 colby2152 Note Contributor SOA Join Date: Feb 2006 Location: Virginia Studying for FA, GH Core College: PSU '07 Favorite beer: Oskar Blues Old Chub Scotch Ale Posts: 4,176

Quote:
 Originally Posted by MarcNguyen You are given: i) Ito process of short-rate: dr(t) =[0.09-0.5r(t)]dt +0.3dZ ii) Risk-neutral: dr(t) = [0.15-0.5r(t)]dt + sigma(r(t))dZ~ iii) g(t) is derivative of r(t) dg(r,t) = m(r,g,t)dt -0.4g(r,t)dZ Determine m(r,g)? SOA solution: (r+0.08)g But I think this solution isn't correct because formula (24.19) in textbook that is dr =[a(r)- (risk premium)]dt+sigma(r)dZ in stead of dr =[a(r) + (risk premium)]dt+sigma(r)dZ so Shapre ratio of short-rate is -0.2 then solution must be g(r-0.08). am I right?
This is a question I bombed in my only sitting of MFE. What does statement 3 mean?

$g(r,t)=dr(t)$?

If so, then $g(r,t)=[0.15-0.5r(t)]dt + \sigma(r(t))dZ$
#4
10-04-2008, 05:47 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,262

Quote:
 Originally Posted by colby2152 This is a question I bombed in my only sitting of MFE. What does statement 3 mean? $g(r,t)=dr(t)$? If so, then $g(r,t)=[0.15-0.5r(t)]dt + \sigma(r(t))dZ$
No. The actual problem states

Quote:
 (iii) $g(r,t)$ denotes the price of an interest-rate derivative at time t, if the short rate at that time is r.
This allows you to use a Sharpe ratio argument to understand $d g(r(t), t)$.
__________________
The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.
#5
10-05-2008, 01:49 AM
 MarcNguyen Member Join Date: Oct 2007 Posts: 60

in the last paragraph on page 661 (ch.20) and previous (24.19) we have: The risk-neutral process is obtained by subtracting risk premium from the drift.
If risk-neutral of interest rate is
dr = [a(r) + sigma*sharpe ratio]dt + sigma*dZ~
where dZ~ = (sharpe ratio)*dt + DZ
It couldn't equal to dr = a(r)dt + sigma*dZ.
#6
10-05-2008, 03:19 AM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,262

Quote:
 Originally Posted by MarcNguyen It made me confused. in the last paragraph on page 661 (ch.20) and previous (24.19) we have: The risk-neutral process is obtained by subtracting risk premium from the drift. If risk-neutral of interest rate is dr = [a(r) + sigma*sharpe ratio]dt + sigma*dZ~ where dZ~ = (sharpe ratio)*dt + DZ It couldn't equal to dr = a(r)dt + sigma*dZ.
Sigh.

The only difference between saying
$dr = [a(r) + \sigma(r) \phi(r, t)] \,dt + \sigma(r) \,d\tilde{Z}$

and
$dr = [a(r) - \sigma(r) \phi'(r, t)] \,dt + \sigma(r) \,d\tilde{Z}$

is that then $\phi'(r,t) = - \phi(r,t)$, i.e. it's a matter of sign convention. Unfortunately sign conventions can be a pain in the neck to keep straight, and McDonald screwed up in the original version; this meant that a number of other formulas in the chapter were wrong.

Anyway, the way he originally defined things the Sharpe ratio should come out negative, but in that case you need to be careful about signs when computing the Sharpe ratio. In this problem it would become
$\frac{\frac{\mu(r, g(r,t))}{g(r,t)} - r}{-0.4} = -0.2$

and so you still end up with
$\mu(r, g(r, t)) = (r + 0.08) g(r, t)$

That said, you'd be infinitely better off if you never looked at the original version of that part of Chapter 24 again, and just stuck with the errata pages.
__________________
The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.
#7
10-05-2008, 06:27 AM
 MarcNguyen Member Join Date: Oct 2007 Posts: 60

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