Finance/Actuarial question. Please help!

**cirrus74** · April 9th 2008, 03:58 AM

Any help on this is much appreciated!

**kalagota** · April 9th 2008, 05:01 AM

Originally Posted by cirrus74

Any help on this is much appreciated!

i'll use $s_n$ to denote the accumulated value of the annuity since i dont know the code in LaTex..

so, $s_{10} = \frac{(1+i)^{10} - 1}{i}$

and assuming that this is a compound interest, then we can use

$a(t) = \underbrace{e^{\int_0^t \delta_s \, ds}}_{in \, terms \, of \, force \, of \, interest} = \underbrace{(1+i)^t}_{basic \, formula}$

therefore, the middle part can be simplified (evaluated at $t = 10$ ) as $\ln 2$ (verify) and together with the right hand side, we have $(1+i)^{10} = \ln 2$ or $i = \sqrt[10]{\ln 2} - 1$

therefore, $s_{10} = \frac{(1+i)^{10} - 1}{i} = \frac{\ln 2 - 1}{\sqrt[10]{\ln 2} - 1}$

**colby2152** · April 9th 2008, 05:09 AM

Please do not double post. You question was first answered in the Business Math forum: http://www.mathhelpforum.com/math-he...-function.html

**TKHunny** · April 9th 2008, 05:10 AM

You should have these:

$a(t) = e^{\int_{0}^{t}\delta_{r}dr}$

$a_{angle-n} = \sum_{t=1}^{n}\left(\frac{1}{a(t)}\right)$

$s_{angle-n} = a(n)*a_{angle-n}$

Let's see what you get.

Note: I'm not sure what Kalagota is missing. It doesn't make sense for 10 payments to be accumulated to only 8½, so you get to tell us what you manage to calulate.

**kalagota** · April 9th 2008, 06:22 AM

now i wonder what i missed?

i think, my mistake was that $\ln 2$ thing.. i forgot that the integral was a power of $e$ and thus, $a(t)$ should be equal to $2$ .. this should make sense...

now, let me correct my mistake..

$(1+i)^{10} = 2$ or $i = \sqrt[10]{2} - 1$

thus, $s_{10} = \frac{(1+i)^{10} - 1}{i} = \frac{1}{\sqrt[10]{2} - 1} \approx 13.9327262$

**TKHunny** · April 9th 2008, 06:44 AM

Try again. I'm holding out for 14.5.

**kalagota** · April 9th 2008, 06:55 AM

how you did yours?

is $a(t) = \frac{1}{2}$ ? it can't be i think, and i shall stick with $a(t)=2$ ..

**TKHunny** · April 9th 2008, 07:56 AM

Formulas shown above. I think yours is missing some moving parts.

Start from $s_{10} = (1+i)^{10} + (1+i)^{9} + ... + (1+i)$ , rather than the summed version that assumes regularity of interest for each payment.

With variable interest, $a(10) \neq \left(a(5)\right)^{2}$ .

**cirrus74** · April 9th 2008, 01:43 PM

Originally Posted by TKHunny

Try again. I'm holding out for 14.5.

You've got the right answer. How did you do it? And thank you both so much for helping.

I did as was suggested in the business post and integrated exp(delta(t)dt) with limits of 1 to 20. And got 1.9... which is definitely not the answer.

**TKHunny** · April 9th 2008, 03:17 PM

The integral is good. You will need that.

Look at my first post very carefully. There is a summation in there. If you are trying to do it in one piece, you do not yet have the right idea.

**kalagota** · April 9th 2008, 06:49 PM

did you notice notice that your summation is just the sum of the first n geometric sequence.. i think, you must be the one to check your solution.. or the best way is to post your whole solution in here..

besides, if you will push your 14.5, try to work it backward, and you will derive at $a(t) = 1.71...$ and not 2..

**kalagota** · April 9th 2008, 07:23 PM

maybe i'll put the derivation here..

$a(t) = (1+i)^t$

thus, $a_{angle-n} = \sum_{t=1}^n \frac{1}{a(t)}|_{n=10} = \frac{1}{a(1)} + \frac{1}{a(2)} + ... + \frac{1}{a(10)}$

or $a_{angle-10} = \frac{1}{(1+i)} + \frac{1}{(1+i)^2} + ... + \frac{1}{(1+i)^{10}}$

and $s_{angle-10} = a(10) \, a_{angle-10} = (1+i)^{10}\, a_{angle-10} = (1+i)^9 + (1+i)^8 + ... + (1+i) + 1$

$= \frac{(1+i)^{10} - 1}{(1+i) - 1}$ and just do your substitution..

**cirrus74** · April 9th 2008, 11:31 PM

Thanks, I've got the answer. Do you mind if I ask a question though?

Why can I not just use this? Or at least, the answer doesn't turn out right, but I don't understand why.

$<br /> \int _1 ^{10} e^{\delta t} dt<br />$

exp(-ln(20 - 10)+ ln(20-1)) = exp (ln(19/10)) = 19/10

Did I make a mistake somewhere?

**TKHunny** · April 10th 2008, 09:37 AM

Originally Posted by kalagota

thus, $a_{angle-n} = \sum_{t=1}^n \frac{1}{a(t)}|_{n=10} = \frac{1}{a(1)} + \frac{1}{a(2)} + ... + \frac{1}{a(10)}$

Perfect.

or $a_{angle-10} = \frac{1}{(1+i)} + \frac{1}{(1+i)^2} + ... + \frac{1}{(1+i)^{10}}$

This is where it falls apart. Your interest is static.

Calculate from the integrals:

a(1) = 20/19

a(2) = 10/9

a(3) = 20/17

a(4) = 5/4

It is not a Geometric Sequence.

**TKHunny** · April 10th 2008, 10:06 AM

Another Possible Answer:

As it stands, we have 't' defined as the time from EACH deposit. That may not seem quite right. If we use 't' to mean the time from the beginning of the financial structure, we get:

$s(t) = e^{\int_{t}^{10}\delta^{t}\;dr}$

Then:

$s_{angle-10} = \sum_{0}^{9}s(t) = 15.5$

Very interestingly, we get:

s(0) = 2.00

s(1) = 1.90

s(2) = 1.80

etc...

It IS an arithmetic sequence.

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