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February 14th, 2017, 07:05 PM   #21
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Denis

I too wondered about how the actuaries cope with it.

For payments covering a period certain, the only issue involves truncation and rounding. The way I persuaded my bank to deal with it was to truncate payments but accrue the amounts for compounding purposes and accumulate them until meaningful.

What is harder to ponder is how to deal with the risk on "lifetime" payments. I am guessing that is dealt with through contractual language about what is meant by "lifetime." It's been over 40 years since I had to deal with the practicalities of computing payments. That was back in the day when many US banks claimed to offer "continuous compounding."
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February 14th, 2017, 07:51 PM   #22
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Continuous compounding!
Close to "false advertising".
6% cpd. continuously : annual = 6.18365465...
6% cpd. daily (365 days): annual = 6.18313106...

I've seen it all...1961 to 2001 with same financial institution.
In 1961, we were using ledger cards and the likes...
Computerized around 1964.
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February 15th, 2017, 05:41 AM   #23
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Assume the first month payment is $P_0$ and it is increased yearly at a rate of $r$. Then if $P_k$ denotes the payment amount after the $k^{th}$ year you have the formula. $p_1 = P_0(1+r), P_2 = P_0(1+r)^2,....P_Y = P_0(1+r)^Y$ where $Y$ is the number of full years for which they receive payments. Set $M_k$ to denote the number of payments taking place in the $k^{th}$ year so that in this case, $M_k = 12$ except potentially when $k = Y$ and $M_Y$ is the number of additional payments in the incomplete year.

Now, it is simple to compute the total payoff after $Y$ years and an additional $M = M_Y$ months:
$$ S = \sum_{k \geq 0} M_kP_k = 12\sum_{0}^{Y-1} P_k + MP_Y.$$

Recalling that $P_k = P_0(1+r)^k$ and recognizing that partial sums of a geometric sequence have the nice formula $(1 + (1+r) + (1+r)^2 + ... + (1+r)^{Y-1}) = \frac{1 - (1+r)^{Y}}{-r}$ we obtain:
$$ S = 12P_0\sum_{0}^{Y-1} (1+r)^k + MP_0(1+r)^Y = 12P_0(\frac{1 - (1+r)^Y}{-r}) + MP_0(1+r)^Y.$$

Finally, if a particular value for $S$ is specified one can factor and solve for $P_0$ explicitly which is
$$ P_0 = \frac{S}{12(\frac{1-(1+r)^Y)}{r}) + M(1+r)^Y}.$$

This formula works up to algebra mistakes but if I've made some you should be able to see the idea and fix it appropriately. Hope it helps.
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February 15th, 2017, 06:08 AM   #24
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After thinking for a moment there is a more elegant method which might be a little less accurate. This probably doesn't actually make a difference in actual payouts since the final payment can always be adjusted slightly. The idea is to image a continuous payout schedule so that $S(x)$ is the total amount paid after $x$ years. Expressing $S$ explicitly it is clear it should aggregate payments so it is appropriate to write
$$ S(x) = \int_0^x f(t) \ dt$$
where $f(t)$ is the payment value for an infinitesimal length of time. In this case, we are prescribing a continuously varying payment so we notice that $f(t) = 12P_0(1+r)^t$.

To put this into the context of the solution above, you should notice that the sum computed above would be a Riemann sum approximation of the integral I need to compute here. In that sense, we are simplifying a tedious summation by replacing it with a reasonable integral. That said, we can easily compute $S$ explicitly:
$$ S(x) = \int_0^x f(t) \ dt = 12P_0 \int_0^x (1+r)^t \ dt = \frac{12P_0}{\ln (1+r)}((1+r)^x - 1))
$$
which is much nicer to deal with and still highly accurate. It also deals with the partial payment problem quite naturally since it is defined continuously. For example, if you want to make payments for 10 years 5 months as in your example, you simply compute this integral for $x = 10 + 5/12$. Moreover, this would deal with additional days just as easily.

As before, if $S$ is specified we can determine the appropriate starting payments by solving for $P_0$ which is in this case
$$P_0 = \frac{S\ln (1+r)}{12((1+r)^x -1)}. $$
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February 15th, 2017, 10:41 AM   #25
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Quote:
Originally Posted by SDK View Post
As before, if $S$ is specified we can determine the appropriate starting payments by solving for $P_0$ which is in this case
$$P_0 = \frac{S\ln (1+r)}{12((1+r)^x -1)}. $$
Your equation yields 1440.647466... as the starting payment
instead of the required 1454.63404..., close to 14 dollars short.

This will result in 198,076.96 being disbursed instead of the
required 200,000.00, a difference of ~1923 dollars.

The insurance company will sure hire you

I'm 99% sure this is due to the way the way the "5/12" is handled
in your formula, as it discounts the remaining 7/12.
In my formula, I stop the process after 120 months, then
apply the 2% to the resulting 10th year payment.
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February 18th, 2017, 03:54 AM   #26
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Quote:
Originally Posted by Denis View Post
Your equation yields 1440.647466... as the starting payment
instead of the required 1454.63404..., close to 14 dollars short.

This will result in 198,076.96 being disbursed instead of the
required 200,000.00, a difference of ~1923 dollars.

The insurance company will sure hire you

I'm 99% sure this is due to the way the way the "5/12" is handled
in your formula, as it discounts the remaining 7/12.
In my formula, I stop the process after 120 months, then
apply the 2% to the resulting 10th year payment.
As I mentioned, this will lead to a slight error in the payment schedule. This requires the final payment to be corrected by a small amount. $1923.00 seems like a small amount to be concerned with in a 10 year window. It is worth pointing out that the error is not due to the remainder payments. In fact, the biggest reason one would use this sort of reverse quadrature approximation is because of how naturally it handles the approximation of step functions. The error is due to using a particularly crude quadrature rule. If more precision is required, one could apply this same approximation method used here but obtain more accuracy using a different quadrature rule (in this example I have chosen the integrand by assuming the exact sum was its approximation by right endpoints).

Alternatively, one could see the preceding post where I have given an explicit formula for the exact initial payment which utilized no approximations. Though this formula is not as nice and for more complicated problems the derivation done there would become very tedious.
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Last edited by SDK; February 18th, 2017 at 03:58 AM.
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February 18th, 2017, 06:29 AM   #27
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Quote:
Originally Posted by SDK View Post
Finally, if a particular value for $S$ is specified one can factor and solve for $P_0$ explicitly which is
$$ P_0 = \frac{S}{12(\frac{1-(1+r)^Y)}{r}) + M(1+r)^Y}.$$
Sorry Mr. SDK; missed your explicit formula.
Note: a half-bracket is missing in numerator.

With s=200000, r=.02, y=10 and m=5,
your formula yields 1596.147... instead
of the required 1454.634...
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February 18th, 2017, 07:17 AM   #28
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Getting back to practicalities, an approximating method must be sensitive to the amount involved. You have to be able to produce a statement that proves the contract has been honored every statement period, but truncation or rounding error is inherent in any approximation method. So if I were still doing this sort of thing for real, I would be worrying about the fractional pennies on the largest contract we write. Litigation and reputational risk can be killers. Just ask Wells Fargo.
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February 18th, 2017, 07:22 AM   #29
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Not only that Jeff: there's also explaining it to the customer!
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February 18th, 2017, 07:38 AM   #30
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Quote:
Originally Posted by Denis View Post
Not only that Jeff: there's also explaining it to the customer!
Exactly. In addition, you have to be able to explain it to the IT folks, management, reporters, and the board. Some of your audience will be a bit queasy with any multiplication that requires decimals.

I guarantee that I spent more time finding explanations of solutions than I ever spent finding solutions.
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