Lecture 3: Hedging with Minimum Cost, Multi-period Discounting, Bonds Economics of Finance
School of Economics, UNSW
43 Good Weather
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(3×2) 20 28 Bad Weather (1×2)
Consider following. Markets are incomplete: number of states is higher then the number of linearly independent securities.
Q = 20 28 Fair Weather pS = 19 35
Bond Stock
Bond Stock
Suppose that an investor asks an investment firm to create a product with the following payment:
40 Good Weather c = 30 Fair Weather (3×1) 20 Bad Weather
Questions: How to do it? What should the firm charge?
Hedging in Incomplete Market
• The Problem: No matter how many bonds and stocks are chosen, the payments in the ”Fair Weather” state and the ”Bad Weather” state will be the same!
• Suppose the firm will select 40 in ’GW’ and 30 in ’FW’ or ’BW’ to cover all outflows. We can do it by:
20 43 Good W. 40 Good W. Q= c=
20 28 Fair or Bad W. (2×1) 30 Fair or Bad W. Bond Stock
−1 20 43−1 40 0.56667 Bond n=Q ·c= 20 28 30 = 0.66667 Stock
0.56667 p=pS·n= 19 35 0.66667 =34.10
Different Scenarios
• It costs 34.10 PA to purchase a portfolio, n, that would cover all future outflows:
• The firm will receive 10 in ’BW’ state;
• Investor will be assured to receive all the promised
• Since there is extra 10 BA, the firm will be happy to sell the product for 34.10PA.
New product on the market
What if the new product is offered for 32PA?
20 43 40 Good W.
Q = 20 28 30 FairW. pS =
(3×3) 20 28 20 Bad W. (1×3) Bond Stock Product
The atomic security prices can be found:
20 43 40−1 patom=pS ·Q−1 =19 35 3220 28 30
= 0.56 0.18 0.21 Good W. Fair W . Bad W.
19 35 32 Bond Stock Product
Completing the market
The new product, linearly independent of the existing ones, completes the market.
Perfect hedging can be achieved when market is complete:
20 43 40−1 40 n=Q−1·c = 20 28 30 30
20 28 20 20
0 Bond = 0 Stock
Hedging at Minimum Cost
Recall our original example. Market is incomplete:
Fair Weather Bad Weather
43 Good Weather Fair Weather
40 Good Weather
Bond Stock
pS = 19 35
Bond Stock
Our objective is to construct the cheapest portfolio, n, that will deliver as least as much as c in every state of nature.
Bad Weather
Linear programming
• Our problem is to select a portfolio, n, to minimize its cost, pS · n, subject to obtaining no less than the required state-contingent payment, c.
• A constrained optimization problem we have to solve is given by
min pS · n subject to Q · n ≥ c. n
• Note: we use the sign ≥ to indicate that every element of vector Q · n is no less than the corresponding element of vector c.
• We are facing a linear programming problem, or simply the problem of finding a vector that minimizes a linear function subject to linear constraints.
Hedging at Minimum Cost
Matlab functions provide us with several tools for solving linear programming problems. They solve this general problem (I use their notation):
min f′ · x subject to A · x ≤ b. x
• linprog(f, A, b) in Matlab.
Linear programming
We will ask this function to perform the simplest task:
• In our context, the role of x played by portfolio vector, n. Note f is assumed as a column vector. The role of f′ is performed by row vector pS.
• this function deals only with inequalities of the form: A · x ≤ b. Let’s multiply both sides of our constraint, Q · n ≥ c, by −1 to obtain −Q · n ≤ −c
• Hence, the role is A is played by −Q, while the role of b is played by −c.
Matlab: Hedging at Minimum Cost
Enter the data in MATLAB’s command prompt:
>> Q = [20 43; 20 28; 20 28];
>> c = [40 30 20]′;
>> ps = [19 35];
Use the linear programming function linprog >> n = linprog(ps′,-Q,-c)
The result of running these commands is:
n= 0.5667 0.6667
The price of the portfolio is:
>> p = ps*n p = 34.1000
Wrapping up
Hedging involves fully covering contingent payments/liabilities and offsetting risks
• With complete market, this involves replicating desired payments/liabilities;
• With incomplete market, perfect hedging is can not be achieved;
• The ideal hedging then involves hedging with minimum cost;
• This can be done using Matlab ”linprog” function;
• Completing the market will generally reduce deadweight loss associated with incomplete hedging.
The Discount Factor
Definition: The discount factor (for a certain date) represents the present value of a payment of one unit to be made with certainty at the specified future date.
The tree example:
PV =p·q= (0.285+0.665) ·20=0.95·20=19
sum of the atomic prices
• The discount factor for a date in question equals to the sum of appropriate atomic prices (prices of basic atomic securities)
The Discount Factor
E.g. df(1) = 0.95
pG = 0.285
Good weather
0.95 apples
pB = 0.665
Bad weather
Constant Interest Rate
Assume interest rate, i, is constant for now. df(1) is the present value (at time 0) of 1 for certain in one period. Hence
df(1) · (1 + i) = 1.
df(1)= 1 . 1+i
df(2) is the present value (at time 0) of 1 for certain in two periods. If interest rate is constant, df (2) · (1 + i)2 = 1, or
To generalise,
1 . (1+i)2
1 . (1+i)n
The present value of an asset that produces a cash flow, C1, a year from now is
PV=df(1)×C1= 1 ×C1 1+i
i = 0.0526 or 5.26 percent and C1 = $100. Then PV= 1 ×C1= 100 =$95.00
1 + i 1.0526
The P V of a cash flow two years from now (assuming i
constant)is: PV =df(2)×C = 1 ×C . SayC =$100: 2 (1+i)2 2 2
PV = 100 =90.2554. (1.0526)2
Multi-period (Variable) Discount Factors
Definition: A nominal discount factor, df(t), is the present value of one unit of currency to be paid with certainty at time t. Notation:
• Discount factor {1×periods}:
df=df(1) df(2) df(3)
• Vector of cash flows known to be certain {periods×1}:
cf (1) cf = cf(2)
Discounted present value of the cash flows: pv = df · cf
Example: Coupon bonds with different maturities
• The Payment Matrix {periods×bonds}: B1 B2 B3
103 4 3Year1 Q = 0 104 3 Year 2 0 0 103 Year 3
• The Price Vector {1×bonds}:
p = 100 101 98
Discount factors
The Question: What are the discount factors for Year 1, Year 2, Year 3?
• The price of each bond should equal its discounted present value:
100 = df (1) · 103
101 = df (1) · 4 + df (2) · 104
98 = df (1) · 3 + df (2) · 3 + df (3) · 103
Inferring the discount function
• In matrix notation: p = df · Q
• If Q−1 exists, the discount function can be inferred as
df = p · Q−1;
• Since Q−1 exists, the discount function is
103 4 3−1 df =100 101 980 104 3
(1×Years) 0 0 103 = 0.9708 0.9338 0.8959
• Any desired set of future certain payments over the next three years can be valued using this discount function.
• Let c = 300 200 103
n=Q−1c= 0 0
100′, then the replicating portfolio is
Replicating bond portfolio
• To find a portfolio of the three bonds that will replicate a desired set of certain cash flows we can use
Q·n=c (Years×Bonds) (Bonds×1) (Years×1)
3 −1 300 2.8107 B1
3 200=1.8951 B2 103 100 0.97087 B3
Multi-period Interest Rates
Investment grows from V (0) to V (t) in t periods, i(t) is the default-free interest rate for time t:
V(0)(1+i(t))t =V(t)
Definition: The ratio of the ending value to the beginning
value, V (t)/V (0), is termed the (t-period) value relative.
• One dollar will grow to 1/df(t) dollars with certainty by
time t, hence
(1+i(t)) =V(0)=df(t); i(t)= df(t) −1.
tV(t)1 11t • Call i(t) multi-period interest rate, or, yield.
Yield curve
The following Matlab code plot a term structure of interest rate given discount factors:
>>df = [0.94 0.88 0.82]
df = 0.9400 0.8800 0.8200
>> vr = 1./df
vr = 1.0638 1.1364 1.2195 >> i = vr.^(1./[1:3])-1
i = 0.0638 0.0660 0.0684 >>plot([1:3],i)
Bond Yields (Yield to maturity)
Definition: Yield-to-maturity (YTM) is a constant interest rate that makes the present value of all the bond’s payments equal its price.
Example: A bond is selling for $97.84 and provides a certain vector of cash flows:
6 Year 1 cf = 6 Year 2 106 Year 3
Yield-to-maturity, y, must satisfy
6 + 6 + 106 =97.84 1+y (1+y)2 (1+y)3
Computing Bond Yields
• Problem: We need to solve a non-linear equation!
• Solution: Use numerical techniques and tools, e.g. Octave
(Matlab) fsolve function
• fsolve function looks for a solution to equation f(y) = 0
f(y)= 6 + 6 + 106 −97.84 1+y (1+y)2 (1+y)3
Matlab code
y0=0.06; % Coupon rate is an initial guess y=
Optimization terminated: first-order optimality is
less than options.TolFun.
6 Year 1 cf = 6 Year 2 106 Year 3
• The market discount function is:
df = 0.94 0.88 0.82 (1×Years)
• We have all types of measurements about bond’s return (yield, YTM, df, etc.) till now;
• In some applications we are interested in how sensitive is the bond value towards market interest rate;
• Consider a vector of certain cash flows associated with a
Periodical Values and Weights
The present value of each year’s cash flow:
v =df(1)·cf(1) df(2)·cf(2) df(3)·cf(3)
= 5.64 5.28 86.92
w(t) is the fraction of the bond’s present value paid in year t:
w = df (1)·cf (1) (1×Years) df ·cf
df (2)·cf (2) df (3)·cf (3)
= 5.64/97.84
= 0.0576 0.0540 0.8884
df ·cf 86.92/97.84
5.28/97.84
The bond’s duration is the average waiting time for (coupon) payments:
D = t · w(t) = 1 · w(1) + 2 · w(2) + 3 · w(3) t=1
= 1 · 0.0576 + 2 · 0.0540 + 3 · 0.8884 = 2.8308
Duration using Bond yield
Duration of a Bond is often calculated using yield-to-maturity.
3 3 cf(t)/(1+y)t
D = t · w(t) = t · 3 cf (t)/ (1 + y)t
t=1 t=1 t=1 3 cf (t)/ (1 + y)t
= t· Pbond t=1
In our example: y = 0.0682, Pbond = 97.84 therefore 6 6 106
D = 1 · 1.0682 + 2 · (1.0682)2 + 3 · (1.0682)3 /97.84 = 2.8315
Modified Duration
Let v(t) = cf(t)/(1 + y)t and note
dv(t) =−t·cf(t)·(1+y)−t−1 ⇒
dv(t)=−t·v(t)· dy ⇒
(1+y) N N dy
dv(t) = − t · v(t) · (1 + y) ⇒ t=1 t=1
N dv(t) N v(t) dy =−t··⇒ t=1 v t=1 v (1+y)
dv =−md·dy v
Modified Duration
• Modified Duration, md: md = D ; (1+y)
• It measures the (negative) relative change in the value of the bond per marginal change in its own yield-to-maturity.
• Or, in short, the interest rate risk of the bond.
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