Topic 6 - Equations of Value Flashcards
What is an equation of value?
Equates the PV of monies received and monies paid out
Give the general formula for an equation of value
PV income = PV outgo OR
PV income –PV outgo = 0
How is the premium for an insurance policy calculated?
By equating PV premiums to PV expected benefits and other outgo
Derive the Yield equation using discrete cashflows
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]
Example
Investor pays £100 now to get £120 in five years time
what is the effective annual interest i pa?
100 = 120v^5 100 = 120 x (1+i)^-5 (1+i)^5 = 1.2 i = 3.71%
Example
Investor pays £100 now to get £60 in five years time and another £60 in ten years calculate i
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%
Derive the Yield equation using continuous cashflows
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)
State the Yield equation using both discrete and continuous cashflows
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
What can equations of value be used for?
To determine an unknown value
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
P = 25a10¬ + 250v10 P = 25 x 6.7101 +250 x 1.08^-10 P = £283.55
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
P = Ia10¬ + 250v^10 200 = I x 6.7101 +250 x 1.08^-10 I = (200−250x1.08^−10)/6.7101 I = £12.55
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
P = 25a10¬ + Rv^10 200 = 25 x 6.7101 + R x 1.08^-10 R = (200−25x6.7101)/1.08^−10 R = £69.62
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
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
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
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
Define the linear interpolation formula for i
i ≈ i1 + (P−P1)/(P2 −P1) x (i2 –i1)
State what each variable in the linear interpolation formula for i represent
P is present value based on i
P1 is present value based on i1
P2 is present value based on i2
When does linear interpolation work best?
Approximation works best if the test i’s are close to the true value e.g. 1% apart
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
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]
If there is uncertainty about the payment of a cashflow, we can allow for this in two different ways:
- Apply a probability to the cashflow at each time
- Use a higher discount rate
How would one apply a probability to a cashflow payment and how would this affect/change the Yield equation?
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
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
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
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
∑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
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
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
