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#1
08-29-2006, 04:48 PM
 michael solomon Member Join Date: Apr 2006 Posts: 70
ASM Section 1b-f Q24 - problem with answer

The question: It takes 11.553 years for an initial investment to double at a force of interest (delta). How long will it take for an initial investment to triple at a nominal rate of interest numerically equal to (delta) and convertable twice a year.

Answers: (A) less than 19 years (B) at least 19 but less than 19.5 etc.

I worked it out like Dr. Cherry and got the same answer of 18.58 years, which seems to be (A). But I don't understand: if the interest is only convertible twice a year, then there is no more money in the investment at 18.58 years than at 18.5 years, and since this isn't enough, don't I have to wait until 19 years to triple my investment (though I know at 19 years I would have more than triple)? This would be answer (B). To borrow terminology from Paper P, it seems to me like a disrete "distribution" rather than a continuous one.

#2
08-29-2006, 07:32 PM
 ebony Member SOA Join Date: Mar 2006 Studying for CSP - GH Favorite beer: +Hb!l pnq Posts: 985

Interest is a continuous distribution. I guess there are some step functions for interest, but I haven't been encountering any in this material. Convertible twice a year just helps you to define what the rate of interest is. 18.58 years just defines the investment period.
Hope that helps you.
#3
08-29-2006, 09:19 PM
 atomic Member CAS Join Date: Jul 2006 Posts: 4,088

Quote:
 Originally Posted by michael solomon The question: It takes 11.553 years for an initial investment to double at a force of interest (delta). How long will it take for an initial investment to triple at a nominal rate of interest numerically equal to (delta) and convertable twice a year. Answers: (A) less than 19 years (B) at least 19 but less than 19.5 etc. I worked it out like Dr. Cherry and got the same answer of 18.58 years, which seems to be (A). But I don't understand: if the interest is only convertible twice a year, then there is no more money in the investment at 18.58 years than at 18.5 years, and since this isn't enough, don't I have to wait until 19 years to triple my investment (though I know at 19 years I would have more than triple)? This would be answer (B). To borrow terminology from Paper P, it seems to me like a disrete "distribution" rather than a continuous one. Please help
The question, in my opinion, is imprecise for the reason you stated. The assumption that needs to be made is that interest is paid continuously, and not paid in arrears.

For sake of completeness, the solution is as follows: We must have, for a constant force of interest $\delta$,

$A(11.553) = 2A(0) = A(0) e^{11.553 \delta}$

from which it follows that

$\delta = \frac{\log 2}{11.553} = 0.06.$

Hence the nominal annual interest rate convertible semiannually is

$i^{(2)} = \delta = 0.06$

and we want to find $n$ such that

$A(n) = 3A(0) = A(0) \left(1 + \frac{i^{(2)}}{2}\right)^{2n} = A(0) (1.03)^{2n}$

or equivalently,

$n = \frac{\log 3}{2 \log 1.03} = 18.5835.$

Since

$(1.03)^{37} < 3 < (1.03)^{38}$,

the precise answer would be to say that it takes n = 19 years to triple the investment if interest were paid at the end of each six-month period (i.e. in arrears), and n = 18.5835 years if interest were paid continuously.

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