Recursion Formulas: Practice Problems and Solutions

Two basic kinds of recursion formulas will be used on Exam 3L.

The backward recursion formula:

u(x) = c(x) + d(x)*u(x + 1)

The forward recursion formula:

u(x + 1) = -c(x)/d(x) + u(x)/d(x)

What are the functions u, c, and d?

u(x) is the function/life table characteristic being evaluated.

c(x) and d(x) are functions relating u(x) to u(x + 1). They are typically given, along with a starting value of u(x) or u(x +1). Or u(x) and u(x + 1) are given, and one is asked to find c(x) and d(x).

Here, we will do some fairly simple practice problems with recursion formulas, whose main objective will be to get students to memorize the formulas.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. p. 73.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L20-1. It is known that u(x + 1) = 3x³, d(x) = 3/x, c(x) = 5x². Find u(x) using a recursion formula.

Solution S3L20-1. We use a backward recursion formula:

u(x) = c(x) + d(x)*u(x + 1) = 5x² + (3/x)3x³ = 5x² + 9x² = u(x) = 14x².

Problem S3L20-2. It is known that c(x) = 15x², and d(x) = 3. You know that u(16) = 35. Find u(17) using a recursion formula.

Solution S3L20-2. We use a forward recursion formula:

u(x + 1) = -c(x)/d(x) + u(x)/d(x) = -15x²/3 + u(x)/3. We know that x = 16 and u(x) = 35. Thus, u(17) = -15*17²/3 + 35/3 = u(17) = -1433.3333333333.

Problem S3L20-3. You are trying to establish a recursion formula between _tp_x and _t+1p_x. Find c(x) and d(x) for this formula, satisfying the requirement that at least one of these functions must have a "p" in it.

Solution S3L20-3. We know that _tp_x = s(x + t)/s(x) and _t+1p_x = s(x + t + 1)/s(x)

Now we can relate _tp_x to _t+1p_x via a backward recursion formula:

_tp_x = _t+1p_x[s(x + t)/s(x + t + 1)]

We know that s(x + t + 1)/s(x + t) = p_x+t,

so [s(x + t)/s(x + t + 1)] = 1/p_x+t

Thus, _tp_x = [1/p_x+t]_t+1p_x

_tp_x = 0 + [1/p_x+t]_t+1p_x.

This follows the format of the backward recursion formula u(x) = c(x) + d(x)*u(x + 1).

Thus, c(x) = 0 and d(x) = [1/p_x+t].

Problem S3L20-4. You are trying to establish a recursion formula between a(x) and a(x+1) for a population of aquatic aardvarks. It is known that (L_x - l_x+1) = (7/8)(L_x+1 - l_x+2) for all x and (l_x - l_x+1) = (3/4)(l_x+1 - l_x+2) for all x. Find c(x) and d(x) for this formula. Your answers should be expressed in terms of the functions "L," "l," and a(x). (If you need to, feel free to refresh yourself on the formula for L_x in Section 13 and the formula for a(x) from Section 14.)

Solution S3L20-4.

We know from Section 14 that a(x) = (L_x - l_x+1)/(l_x - l_x+1).

Likewise, a(x+1) = (L_x+1 - l_x+2)/(l_x+1 - l_x+2)

Now we can relate a(x) to a(x+1):

a(x) = a(x+1)[(l_x+1 - l_x+2)/(L_x+1 - l_x+2)][(L_x - l_x+1)/(l_x - l_x+1)]

a(x) = a(x+1)[(l_x+1 - l_x+2)/(l_x - l_x+1)][(L_x - l_x+1)/(L_x+1 - l_x+2)]

We know that (L_x - l_x+1) = (7/8)(L_x+1 - l_x+2) and (l_x - l_x+1) = (3/4)(l_x+1 - l_x+2).

So [(l_x+1 - l_x+2)/(l_x - l_x+1)] = 4/3 and [(L_x - l_x+1)/(L_x+1 - l_x+2)] = 7/8

Thus, a(x) = a(x+1)(4/3)(7/8)

a(x) = (7/6)a(x+1)

Hence, using the format of the backward recursion formula, c(x) = 0 and d(x) = 7/6.

Problem S3L20-5. You are trying to establish a recursion formula between x³ and 2(x+1)³. Find c(x) in this formula, given that u(x) = x³ and u(x+1) = 2(x+1)³, and d(x) = 2.

Solution S3L20-5. We use the format of a backward recursion formula:

u(x) = c(x) + d(x)*u(x + 1)

x³ = c(x) + d(x)*2(x+1)³

x³ = c(x) + d(x)*2(x+1)³

x³ = c(x) + d(x)(2x³ + 6x² + 6x + 2)

x³ = c(x) + 2(2x³ + 6x² + 6x + 2)

c(x) = x³ - (4x³ + 12x² + 12x + 4)

c(x) = - 3x³ - 12x² - 12x - 4

See other sections of The Actuary's Free Study Guide for Exam 3L.