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#1
11-03-2008, 03:19 PM
 jvaldez84 Member Join Date: Feb 2007 Location: Houston, TX Studying for C College: Texas A&M Alumni Posts: 1,053
SOA Sample Q #21

I apologize if this question has been asked on this board as I was unable to find it. Can anyone explain this to me?
#2
11-03-2008, 10:52 PM
 jhp8 Member Join Date: Jan 2007 Posts: 278

Quote:
 Originally Posted by jvaldez84 I apologize if this question has been asked on this board as I was unable to find it. Can anyone explain this to me?
Here is how I do this particular one, but this may not be a good technique; please chime in with your opinions

Since the time periods of equal length (7 to 9 vs 11 to 13); they are directly proportional:

.04/.05 * (.06) = .048
#3
11-03-2008, 10:57 PM
 Eagle.K.Y Member Join Date: Sep 2006 Posts: 198

in CIR, Sharpe and sigma are both depending on sqrt(r), and alpha(r) = Sharpe * sigma + r, that is, alpha(r) is based on r. so alpha(0.04)/0.04 = alpha(0.05)/0.05
#4
11-04-2008, 12:23 AM
 ashley_n_h Member Join Date: Apr 2007 Posts: 115

It's discussed here:
http://www.actuarialoutpost.com/actu...39#post3261739
towards the end.

This is the formula I use for it:

$\frac {\phi(r_1)}{\sqrt(r_1)} = \frac {\phi(r_2)}{\sqrt(r_2)}$

Just make sure that when you calculate $\phi$ you use $\sigma \sqrt(r)$
#5
11-04-2008, 08:31 AM
 jvaldez84 Member Join Date: Feb 2007 Location: Houston, TX Studying for C College: Texas A&M Alumni Posts: 1,053

Ashley_n_h, I do not understand how this formula can be used to solve problem 21.
#6
11-04-2008, 11:23 AM
 ashley_n_h Member Join Date: Apr 2007 Posts: 115

So for CIR, the sharpe ratio is this:

$\phi = \frac {\alpha - r}{q}= \frac{\alpha-r}{B\sigma\sqrt(r)}$

Since 9-7=13-11 =2, The B's for both $\phi(0.04)$ and $\phi(0.05)$ are equal. Also, $\sigma$ is the same. So if you use the equation I posted above we have:

$\frac {\alpha(0.04) - 0.04}{B \sigma \sqrt(0.04)} * \frac{1}{\sqrt(0.04)} = \frac {\alpha(0.05) - 0.05}{B\sigma\sqrt(0.05)} * \frac{1}{\sqrt(0.05)}$

Which simplifies to:
$\frac{\alpha(0.04) - 0.04}{0.04} = \frac{\alpha(0.05) - 0.05}{0.05}$

plug in $\alpha (0.05)=0.06$ and you get $\alpha(0.04) = 0.048$
#7
11-04-2008, 11:37 AM
 jvaldez84 Member Join Date: Feb 2007 Location: Houston, TX Studying for C College: Texas A&M Alumni Posts: 1,053

Might be another stupid question but I thought CIR was a function of r.
#8
11-04-2008, 11:40 AM
 ashley_n_h Member Join Date: Apr 2007 Posts: 115

Quote:
 Originally Posted by jvaldez84 Might be another stupid question but I thought CIR was a function of r.
It is...
#9
11-04-2008, 11:48 AM
 jvaldez84 Member Join Date: Feb 2007 Location: Houston, TX Studying for C College: Texas A&M Alumni Posts: 1,053

Nevermind I got it.

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