SOA Sample #22

[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?

[jraven]

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

http://www.kellogg.northwestern.edu/...ypos2e_01.html

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

http://www.kellogg.northwestern.edu/...tm/p780-88.pdf

In the revised pages equation (24.19) reads

[colby2152]

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?

?

If so, then

[jraven]

Originally Posted by colby2152

This is a question I bombed in my only sitting of MFE. What does statement 3 mean?

?

If so, then

No. The actual problem states

(iii) 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 .

[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.

[jraven]

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

and

is that then , 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

and so you still end up with

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.

[MarcNguyen]

Thanks for your explian!