1. General Principles of Pension Valuation
Basis for valuation of pension liabilities has similar actuarial mathematics as developed for life insurance
Both create a liability
The defined benefit pension plan is clearly a liability of the plan sponsor – they have made the promise to provide benefits – need to support with appropriate assets
Life insurance benefits are clearly a liability of the insurance company – policy reserves are the measure of that liability on the insurance company’s balance sheet - they have made the promise to provide benefits – need to support with appropriate assets

2. General Principles of Pension Valuation
Mathematics often treated separately
Pension laws have grown to develop specific education standards for actuaries who sign off on pension valuation
United States Pension Law requires and Enrolled Actuary (EA designation) see
Licensed by a Joint Board of the Department of Treasury & Department of Labor

3. General Principles of Pension Valuation
Other reasons why the mathematics is often treated separately
Number of participants in a pension plan typically less than similar insureds in a life insurance portfolio
As we saw… Can lead to different variances of benefit estimates
Discount rates used for pension valuation have more discretion
Life Insurance valuation interest rate set by the Standard Valuation Law (“SVL”) adopted by the NAIC

4. General Principles of Pension Valuation
Other reasons why the mathematics is often treated separately
Benefit design in pension plans much broader and wider than benefit designs in life insurance plans
Benefits a function of many variables (service, salary, factors, etc) while life insurance benefit is often fixed
“Gains and/or losses” due to experience being “better or worse” than what the actuarial valuation assumptions can be a big part of a sponsor’s income statement
Life insurance statutory valuation done typically on a very conservative basis so that most “bad scenarios” already planned for

5. General Principles of Pension Valuation
Main concepts of pension valuation
Accrued Liability (or Actuarial Liability) – how much being set aside for future benefits – Balance Sheet
Normal Cost – how much to pay in each year to fund benefits – Income Statement
Unfunded Liability – a comparison of the assets and liabilities – Balance Sheet
Actuarial Gain / Loss - how much was made or lost due to actual activity being different than expected – Income Statement

6. General Principles of Pension Valuation
Actuarial Liability calculations will vary depending mainly on…
whether the liability only looks at past service – often called the traditional methods or accrued method – or whether the liability looks at potential future service – often called a projected method
whether the liability is calculated separately for each participant with specific data – often called the “individual method” – or whether blanket assumptions are made – often call the “aggregate method”

7. Actuarial Liability
General description of an “individual accrued” liability:
In words….
Accrued Benefit that will be paid annually from retirement age forward, times
Factor that pays the Accrued Benefit every year in a monthly annuity, times
Actuarial discount (which takes into account interest, mortality, and completing service) from the retirement age back to the current attained age

8. Actuarial Liability
General description of an “individual accrued” liability:
In mathematical symbols…
ALx = Bx • är(12) • (D(T)r / D(T)x)
where Bx = annual pension benefit that has accrued to age x
(D(T)r / D(T)x) represents the discounting process from the retirement age back to the current age, involving interest and all decrements

9. Actuarial Liability ALx = Bx • är(12) • (D(T)r / D(T)x)
Concrete Example – COUNTRY Financial DB Plan
Normal Retirement Age = 65
Annual Retirement Benefit
1. Calculate A = Average Annual Earnings during five highest consecutive years of income out the ten years immediately preceding retirement
2. Calculate B = Min (Years of Service, 32)
3. Annual Retirement Benefit = 1.65% x A x B
What is the Actuarial Liability under an Individual Accrued Method (“Traditional Unit Credit”) for an employee age 40 with 15 years of service where earnings have been constant at $50,000 the last 15 years?
ALx = Bx • är(12) • (D(T)r / D(T)x)
ALx = [1.65% (50000) min(15,32)] • ä65(12) • (D(T)65 / D(T)40)
ALx = [12,375] • ä65(12) • (D(T)65 / D(T)40)

10. Commutation Functions
A long time ago, in an actuary’s office, far, far away…. An actuary got tired of writing out all the different summations to indicate insurance and annuity benefits
So…commutation functions were invented
Basic idea: take the most common terms in insurance and annuities and create shorthand for them

11. Commutation Functions
Today’s dilemma: The latest version of the Actuarial Mathematics textbook used to teach actuarial principles does not have any reference to commutation functions
But…you can’t avoid them in other texts or more importantly, insurance laws and regulations
And…Defined Benefit Pension Funding formulas make use of them
So…we’ll review briefly

12. Commutation Functions
Commutation Function Initial Definitions
qx = probability of dying during a year
Mortality tables all are written in terms on qx’s
px = probability of surviving during a year
px = 1 – qx
lx = number of people out of original group surviving to age x = p0 * p1 * p2 * …. * px-1
vx = 1 / (1 + i)x = interest discount for x years

13. Commutation Functions
Natural extensions of these definitions
ly / lx = probability of surviving from age x to age y = (y-x)px
Often, y is described in relative terms to x, such as y = x +n, and therefore the notation becomes lx+n / lx = npx

14. Commutation Functions
The next steps in definitions
Actuaries realized that a lot of what they needed to do involved discounting future payments for mortality and interest
“How much do I need to invest today at a given interest rate and assuming a given mortality to provide a benefit down the road?”
Thus… the birth of the Dx function
Dx = vx lx

15. Commutation Functions
The next steps in definitions
More appropriately, the discounting needed to be done from a future attained age to a known attained age today
“How much do I need to invest today at a given interest rate and assuming a given mortality to provide a benefit at known point in time down the road given that the beneficiary has already survived to a known age?”
Answer? Simple. Divide one Dx function into another
Dz / Dx = vz lz / vx lx = vz-x * lz / lx = vz-x *(z-x)px
Read this as “discounted with interest for (z-x) years, times probability of surviving from x to z”
Example….

16. Commutation Functions
The next steps in definitions
More appropriately, the discounting needed to be done from a future attained age to a known attained age today
“How much do I need to invest today at a given interest rate and assuming a given mortality to provide a benefit at known point in time down the road given that the beneficiary has already survived to a known age?”
Answer? Simple. Divide one Dx function into another
Dz / Dx = vz lz / vx lx = vz-x * lz / lx = vz-x *(z-x)px
Read this as “discounted with interest for (z-x) years, times probability of surviving from x to z”
Example….

17. Commutation Functions

18. Commutation Functions
Breaking down the lz / lx part so it makes logical sense….
Recall that lx = number of people out of original group surviving to age x = p0 * p1 * p2 * …. * px-1
lz / lx = (p0 * p1 * p2 * …. * px-1 * px * px+1 * …. * pz-1) / (p0 * p1 * p2 * …. * px-1)
What happens? All the terms from 0 to x-1 fall out
So… lz / lx = px * px+1 * …. * pz-1 = (z-x)px

19. Commutation Functions
Same thing but with hard and fast numbers!!!
Let z = 65, x = 35
For example, discounting a retirement benefit from age 65 for someone currently age 35
D65 / D35 = v65 l65 / v35 l35 = v * l65 / l35 = v30 * 30p35
Read this as “discounted with interest for 30 years, times probability of surviving from 35 to 65”

20. Commutation Functions
Take a look at a general D65 / Dx
What do you see happening at very young employee ages, say age 25 – 35?
Even as x nears 65, does D65 / Dx get close to 1?
And this would just consider mortality – not including employees leaving the company

21. Commutation Functions
General notation for multiple decrements
In pension valuation you have several decrements to the employee population
Employees die
Employees leave the company
So it is common to distinguish the “mortality only” Dx from the “total decrement” D(T)x
(T) denotes that all forces of decrement are being considered

22. Commutation Functions
In pension valuation you have several decrements to the employee population
Employees die
Employees leave the company
An example with multiple decrements…
Assume 1000 people in a cohort, mortality rate of 10% throughout each year, and 5% of the surviving members leave the cohort at the end of the year

23. Commutation Functions
Expansion on Dx
After inventing Dx, actuaries realized that often times they needed to discount a whole series of future payments
So Nx came next
Nx = Dx + Dx+1 + Dx+2 + … (goes on until the end of the mortality table)
We can prove that an annuity due (where payments begin immediately) = Nx / Dx

24. Commutation Functions
For defined benefit pension funding, Dx and Nx are used quite frequently
You are constantly talking about discounting future benefits to the current point in time
Most pension funding notation, and especially that which is used on the current SOA exams, are written in this format

25. Commutation Functions
Monthly versus Annual
Often times, pension benefits are talked about in terms of monthly payments
So annuities due are often the monthly versions of the more common annual notation
Example:
äx is used to describe an annuity of 1 beginning at age x and paid annually
äx(12) is used to describe an annuity of 1/12 beginning at age x and paid monthly

26. Commutation Functions
Combining äx and Dx
Remember, we proved that äx = Nx / Dx
And often times, we are talking about discounting an annuity stream from retirement age, r, to current age, x
So the factor…. Dr / Dx times är … often enters into the discussion
Well… (Dr * är) / Dx = Nr / Dx … so that is a common way to write it
Or perhaps most frequently … N(12)r / D(T)x

27. Increasing Cost Individual Cost Methods
Individual cost methods calculate the cost that a company should pay to fund a pension benefit by looking at each participant one at a time
Increasing cost means that costs typically increase as participants age and are not levelized across the earnings period
Each year, the “slice” of the benefit that is earned is funded in that year

28. Traditional Unit Credit (TUC) Method
Liability under TUC is the value of the accrued pension benefit as of the valuation date
Also called Unit Credit Cost Method (by ERISA) and Accrued Benefit Cost Method
No consideration is given to what salary increases or years of service might occur in the future – strictly look at what has happened before the valuation date

29. Traditional Unit Credit (TUC) Method
Actuarial Liability for each participant:
ALx = Bx • (D(T)r / D(T)x) • är(12)
where Bx = annual pension benefit that has accrued to age x
Perhaps we can read this as “Accrued Benefit, times discount from r back to x, times annuity forward from r”

30. Traditional Unit Credit (TUC) Method
Normal Cost for each participant:
NCx = bx • (D(T)r / D(T)x) • är(12)
where bx = annual pension benefit earned during employment at age x
Perhaps we can read this as “earned benefit this year, times discount from r back to x, times annuity forward from r”

31. Traditional Unit Credit (TUC) Method
Need to do some examples….
Pull out your handy Commutation Function sheet

32. Traditional Unit Credit (TUC) Method
There’s a clear relationship between Bx and the bx’s
bx’s are the slices of the accrued total benefit, Bx
Many times bx is constant across all years
Recall in some plans, benefits are a fixed amount per year of service
So Bx changes every year, but by the same amount all the time
NCx = bx • (D(T)r / D(T)x) • är(12)… and next year…
NCx+1 = bx+1 • (D(T)r / D(T)x+1) • är(12)
But if bx = bx+1, then the only thing different is the discount factor

33. Traditional Unit Credit (TUC) Method
An interesting twist on this…
If bx = bx+1, then take a look at NCx+1 / NCx
NCx+1 / NCx = [bx+1 • (D(T)r / D(T)x+1) • är(12)] / [bx • (D(T)r / D(T)x) • är(12)]
NCx+1 / NCx = D(T)x / D(T)x+1 = vx lx / vx+1 lx+1
NCx+1 / NCx = 1 / (v • px)
That boiled down to something pretty simple…which means that exam constructors can make a problem look really difficult and have it be solved fairly quickly!