Topic 6 - Equations of Value

Question 1

What is an equation of value?

Answer 1

Equates the PV of monies received and monies paid out

Question 2

Give the general formula for an equation of value

Answer 2

PV income = PV outgo OR

PV income –PV outgo = 0

Question 3

How is the premium for an insurance policy calculated?

Answer 3

By equating PV premiums to PV expected benefits and other outgo

Question 4

Derive the Yield equation using discrete cashflows

Answer 4

Consider the following transaction:
∑n,(j=1) a,tj x e^-𝛿𝑡,j = ∑n,(j=1) b,tj x e^-𝛿𝑡,j
a,tj is outgo at time t,j and b,tj is income at time t,j
For constant δ this can be re-written as:
∑n,(j=1) c,tj x e^-𝛿𝑡,j = 0
With c,tj = b,tj - a,tj being the net cashflow at time t,j
Using : e^𝛿 = 1+i
equation can be re-written
∑n,(j=1) c,tj x (1+i)^-𝑡,j = 0 OR
∑n,(j=1) c,tj x v^𝑡,j = 0 (Yield equation)
[Sum of present value of net cashflows]

Question 5

Example
Investor pays £100 now to get £120 in five years time
what is the effective annual interest i pa?

Answer 5

100 = 120v^5
100 = 120 x (1+i)^-5
(1+i)^5 = 1.2
i = 3.71%

Question 6

Example

Investor pays £100 now to get £60 in five years time and another £60 in ten years calculate i

Answer 6

100 = 60v^5 + 60v^10
Using v^5 as your base this becomes a quadratic eqn
v^5 = [−60 ± sqrt(60^2+4x60x100)]/120
i = 2.49% or -188%
i = 2.49%

Question 7

Derive the Yield equation using continuous cashflows

Answer 7

ρ1(t) is rate of outgo and ρ2(t) is rate of income
∫0,∞ ρ(t).e^-𝛿𝑡 dt = 0
With ρ(t)=ρ2(t) - ρ1(t)

Question 8

State the Yield equation using both discrete and continuous cashflows

Answer 8

So when both discrete and continuous cashflows
present the equation of value becomes:
∑n,(j=1) c,tj x e^-𝛿𝑡,j + ∫0,∞ ρ(t).e^-𝛿𝑡 dt = 0 OR
∑n,(j=1) c,tj x (1+i)^-𝑡,j + ∫0,∞ ρ(t).(1+i)^-𝑡,j dt = 0

Question 9

What can equations of value be used for?

Answer 9

To determine an unknown value

Question 10

Consider a security with the following equation of value
P = Ian¬ + Rv^n
• P = Price/PV
• I  = Interest payments (coupons)
• R = Redemption payment

Find P if I = 25, R = 250, i = 8% pa and n = 10

Answer 10

P = 25a10¬ + 250v10
P = 25 x 6.7101 +250 x 1.08^-10
P = £283.55

Question 11

Consider a security with the following equation of value
P = Ian¬ + Rv^n
• P = Price/PV
• I  = Interest payments (coupons)
• R = Redemption payment

Find I if P = 200, R = 250, i = 8% pa and n = 10

Answer 11

P = Ia10¬ + 250v^10
200 = I x 6.7101 +250 x 1.08^-10
I = (200−250x1.08^−10)/6.7101
I = £12.55

Question 12

Consider a security with the following equation of value
P = Ian¬ + Rv^n
• P = Price/PV
• I  = Interest payments (coupons)
• R = Redemption payment

Find R if P = 200, I = 25, i = 8% pa and n = 10

Answer 12

P = 25a10¬ + Rv^10
200 = 25 x 6.7101 + R x 1.08^-10
R = (200−25x6.7101)/1.08^−10
R = £69.62

Question 13

Consider a security with the following equation of value
P = Ian¬ + Rv^n
• P = Price/PV
• I  = Interest payments (coupons)
• R = Redemption payment

Find n if P = 270, I = 25, R = 250 and i = 8% pa

Answer 13

P = 25an¬ + 250v^n
270 = 25 x (1−v^n)/0.08  + 250v^n
270 = 312.5 - 62.5v^n 
v^n = (312.5−270)/62.5 = 0.68 = 1.08^-n
-nln1.08 = ln0.68
Therefore, n = ln0.68/−ln1.08= 5

Question 14

Consider a security with the following equation of value
P = Ian¬ + Rv^n
• P = Price/PV
• I  = Interest payments (coupons)
• R = Redemption payment

Find i if P = 298 I = 25, R = 250 and n = 5

Answer 14

P = 25a5¬ + 250v^5 @ i
298 = 25 x [(1−(1+i)^−5)/i] + 250(1+i)^-5
This is the most difficult type of equation to solve
Use linear interpolation to determine i

P1 = 25a5¬ + 250v^5 @ i1 =5%
P1 = 25 x 4.3295 + 250 x 1.05^-5
P1 = £304.12 
this is higher than £298
so need to increase i to reduce PV
Try i2 = 6%

P2 = 25a5¬ + 250v^5 @ i2 =6%
P2 = 25 x 4.2124 + 250 x 1.06^-5
P2 = £292.12

Now find i
i ≈ 0.05 + (298−304.12)/(292.12−304.12) x (0.06 –0.05)
i ≈ 0.055 = 5.5%
25 x (1−1.055^−5)/0.055 + 250x1.055^-5 = £298

Question 15

Define the linear interpolation formula for i

Answer 15

i ≈ i1 + (P−P1)/(P2 −P1) x (i2 –i1)

Question 16

State what each variable in the linear interpolation formula for i represent

Answer 16

P is present value based on i
P1 is present value based on i1
P2 is present value based on i2

Question 17

When does linear interpolation work best?

Answer 17

Approximation works best if the test i’s are close to the true value e.g. 1% apart

Question 18

Consider a security with the following equation of value
P = Ian¬ + Rv^n
• P = Price/PV
• I  = Interest payments (coupons)
• R = Redemption payment

Find i if P = 216 I = 15, R = 200 and n = 8

Answer 18

P1 = 15a8¬ + 200v^8 @ i1 =6%
P1 = 15 x 6.2098 + 200 x 1.06^-8
P1 = £218.63 
P2 = 15a8¬ + 200v^8 @ i2 =7%
P2 = 15 x 5.9713 + 200 x 1.07^-8
P2 = £205.97

i ≈0.06 + (216−218.63)/(205.97−218.63) x (0.07 –0.06)
i ≈ 0.062 = 6.2%
15 x (1−1.062^−8)/0.062 + 200x1.062^-8 = £216
[Always perform the check at the end to ensure answer is reasonable]

Question 19

If there is uncertainty about the payment of a cashflow, we can allow for this in two different ways:

Answer 19

• Apply a probability to the cashflow at each time

- Use a higher discount rate

Question 20

How would one apply a probability to a cashflow payment and how would this affect/change the Yield equation?

Answer 20

Apply a probability to the cashflow at each time
∑n,(j=1) p,tj x c,tj x (1+i)^-𝑡,j + ∫0,∞ p(t).ρ(t).(1+i)^-𝑡,j dt = 0
p,tj and p(t) represent the probability of a cashflow at t

Where the force of interest is constant and we can say that the probability is itself in the form of a discount factor
∑n,(j=1) c,tj x e^-𝛿𝑡,j x e^-μ𝑡,j + ∫0,∞ ρ(t).e^-𝛿𝑡 x e^-μ𝑡,j dt = 0
𝜇 is a constant force of probability of a cashflow at t

Question 21

Uncertain Payments
Example
You enter a competition with the following payments,
probability of winning and time before you receive the monies

Prize Prob of winning Time before payment
£100 1 in 50 1 day
£500 1 in 1,000 2 days
£2,000 1 in 10,000 1 week

The effective rate of interest is 0.02% per day
Q) Calculate the Expected Present Value

Answer 21

EPV = ∑n,(j=1) p,tj x c,tj x (1+i)^-𝑡,j + ∫0,∞ p(t).ρ(t).(1+i)^-𝑡,j dt
EPV = 100x(1/50)v + 500x(1/1,000)v^2 + 2000x(1/10000)v^7
EPV = 100x(1/50)(1.0002)^-1 + 500x(1/1,000)(1.0002)^-2 + 2000x(1/10,000)(1.0002)^-7 =  £2.70

Question 22

Where the force of interest is constant and we can say that the probability is itself in the form of a discount factor state how we would modify our yield equation for uncertain payments

Answer 22

∑n,(j=1) c,tj x e^-𝛿𝑡,j x e^-μ𝑡,j + ∫0,∞ ρ(t).e^-𝛿𝑡 x e^-μ𝑡,j dt = 0
𝜇 is a constant force of probability of a cashflow at t
Can combine both into single time dependent function

∑n,(j=1) c,tj x e^-𝛿𝑡,j + ∫0,∞ ρ(t).e^-𝛿𝑡 dt = 0
𝛿`= 𝛿+ 𝜇
Revised force of discount is greater than actual force of discount as 𝜇 must be positive to give a probability
between 0 and 1.

Question 23

Question
A pensioner expects to receive a pension of £1,000 annually in arrears from a pension scheme for the next 10 years, however the latest valuation has indicated that there are insufficient funds to meet the pension. Calculate the PV of the pension @ i = 4% pa assuming:
- All payments will be received in full
- Prob of receiving the 1st payment is 0.95, 2nd is 0.90 etc
- Increasing the force of interest by 0.03774

Answer 23

Solution
- All payments will be received in full
1000𝑎10¬ @4%
1000 x 8.1109 = £8,110.90

• Prob of receiving the 1st payment is 0.95, 2nd is 0.90 etc
  1000𝑎10¬ - 50(Ia)10¬ @4%
  1000 x 8.1109 –50 x 41.9922
  £6,011.29

- Increasing the force of interest by 0.03774
ln(1.04) + 0.03774 = 𝛿`= 0.07696
e^𝛿` = 1+i` = 1.08
1000𝑎10¬ @8%
1000 x 6.7101
£6,710.10