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#1
 Doc Member Join Date: Sep 2009 Posts: 111 Panning Hidden Formulas

Do you think we would need to know formulas in the background used to generate Table 1 modified durations?

For example, to calculate the modified duration of Premiums:

PV(P) = P * d/(1-d). Given that P as stated at the end of section 2 on page 6, both d and P depend on y. Using natural logarithm of PV(P) and taking the derivative with respect to y gives the negative duration. Thus the modified duration for premium is given by

[ (1+k)S + L ] / [ P (1+y)^2 ] + 1 / (1+y-cr) = [ (1.15)(50) + 75 ] / [ (101.19) (1.05)^2 ] + 1 / (1.05 - 0.90)
= 132.5 / 111.562 + 1 / 0.15 = 7.854.

Similar approach for losses and expenses remembering to use PV(L) = 71.429 instead of just L = 75 since each of these are at time =0.

Note this same approach of using the derivative of ln(F) is useful for obtaining a modified duration formula of the franchise value quickly (produces equivalent formula; can be made into formula in text using algebra) with only memorization of F.
#2
 Doc Member Join Date: Sep 2009 Posts: 111 Discovered a shorter form of the current economic value as

C = (1+k)S/(1+y)

once you substitute for P in the equation. Both this version and Panning's are intuitive to remember. However this form confirms that it is the amount surplus grows above the risk free rate.
#3
 Doc Member Join Date: Sep 2009 Posts: 111 Another for total value T = C + F = (1 + k - cr) S / (1 + y - cr) after taking the above expression for C and combining with F on page 7 substituting for d as cr / (1+y).

This interestingly shows detentions impact to the total compared to if it was not considered. Quick formulas if you need time.
#4
 Doc Member Join Date: Sep 2009 Posts: 111 However do not use the duration implied by this formula since the text uses a dollar weighted average for duration of total value between current and franchise values.

Example following Goldfarb's questions in his manual I get total duration = 5 using the formula implied from above whereas using the dollar durations gives 5.53.
#5
 Doc Member Join Date: Sep 2009 Posts: 111 Found out why...

Author assumes durations of assets and liabilities 1.0 and .9523 respectively. Assets defined as

A = P+S-E = S (1+k)/(1+y) + L/(1+y)

so that the duration of assets is

1/(1+y) - 1 / (1+a+by) + 1/(1+y) = 1.0352.

This difference leads to why the text result for duration of total value 5.563 is different from 5 obtained using the formulas above. Note using duration of current value 0.0828 derived from above instead of 1.062 gives (54.76 * 0.0828 + 45.24 * 10.952)/100 = 5.0

I guess this means for the exam I will do it the way shown in the text and not calculate durations for current and total value as above.

Last edited by Doc; 04-30-2016 at 10:53 AM..
#6
 Rojo Habanero Member CAS Join Date: Dec 2010 Favorite beer: Mead actually Posts: 423 I stumbled onto this thread, which I found Doc's solution really interesting. Can anyone show me the process of the differentiation? Calculus isn't my strongest suit, and I would be really thankful if anyone can show me how it's done?
#7
 act_123 Member CAS Join Date: Dec 2013 Posts: 4,984 His stuff seems right but I wouldn't know how to differentiate either.
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#8
 Rojo Habanero Member CAS Join Date: Dec 2010 Favorite beer: Mead actually Posts: 423 Yeah, I (think?) am able to get the duration of L and E based on that formula. It's probably a long shot during exam they asked us to this the alternative method to calculate the duration (and differentiate somehow)... but then it's CAS, long shot happens during exam...

I guess the resulting formula [ (1+k)S + L ] / [ P (1+y)^2 ] + 1 / (1+y-cr) isn't hard to memorize, and to get the duration of L and E you only need to modify it slightly.

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