C. Valuing Forward and Futures Contracts
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RISK MANAGEMENT AND FINANCIAL INSTITUTIONS
EXAMPLE C.1
Consider a six-month futures contract on the S&P 500. The current value of the index
is 1,200, the six-month risk-free rate is 5% per annum, and the average dividend
yield on the S&P 500 over the next six months is expected to be 2% per annum
(both rates continuously compounded). The futures price is 1,200e(0.05−0.02)×0.5 or
1,218.14.
EXAMPLE C.2
The current forward price of a commodity for a contract maturing in nine months
is $550 per ounce. A company has a forward contract to buy 1,000 ounces of the
commodity for a delivery price of $530 in nine months. The nine-month risk-free
rate is 4% per annum continuously compounded. The value of the forward contract
is 1,000 × (550 − 530)e−0.04×9/12 , or $19,409.
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APPENDIX
D
Valuing Swaps
plain vanilla interest rate swap can be valued by assuming that the interest rates
that are realized in the future equal today’s forward interest rates. As an example,
consider an interest rate swap that has 14 months remaining and a notional principal
of $100 million. A fixed rate of 5% per annum is received and LIBOR is paid, with
exchanges taking place every six months. Assume that (a) four months ago, the sixmonth LIBOR rate was 4%, (b) the forward LIBOR interest rate for a six-month
period starting in two months is 4.6%, and (c) the forward LIBOR for a six-month
period starting in eight months is 5.2%. All rates are expressed with semiannual
compounding. Assuming that forward rates are realized, the cash flows on the swap
are as shown in Table D.1. (For example, in eight months the fixed-rate cash flow
received is 0.5 × 0.05 × 100, or $2.5 million; the floating-rate cash flow paid is
0.5 × 0.046 × 100, or 2.3 million.) The value of the swap is the present value of the
net cash flows in the final column.1
An alternative approach (which gives the same valuation) is to assume that the
swap principal of $100 million is paid and received at the end of the life of the swap.
This makes no difference to the value of the swap but allows it to be regarded as
the exchange of interest and principal on a fixed-rate bond for interest and principal
on a floating-rate bond. The fixed-rate bond’s cash flows can be valued in the usual
way. A general rule is that the floating-rate bond is always worth an amount equal
to the principal immediately after an interest payment. In our example, the floating
rate bond is worth $100 million immediately after the payment in two months.
This payment (determined four months ago) is $2 million. The floating rate bond
is therefore worth $102 million immediately before the payment at the two-month
point. The value of the swap today is therefore the present value of the fixed-rate
bond less the present value of a cash flow of $102 million in two months.
A
Currency Swaps
A currency swap can be valued by assuming that exchange rates in the future equal
today’s forward exchange rates. As an example, consider a currency swap in which
4% will be received in GBP and 6% will be paid in USD once a year. The principals
in the two currencies are 10 million USD and 5 million GBP. The swap will last for
1
Note that this is not perfectly accurate because it does not take account of day count
conventions and holiday calendars.
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RISK MANAGEMENT AND FINANCIAL INSTITUTIONS
TABLE D.1
Valuing an Interest Rate Swap by Assuming Forward Rates Are Realized
Time
2 months
8 months
14 months
Fixed Cash
Flow ($ mill.)
Floating Cash
Flow ($ mill.)
Net Cash
Flow ($ mill.)
2.5
2.5
2.5
−2.0
−2.3
−2.6
0.5
0.2
−0.1
TABLE D.2
Valuing a Currency Swap by Assuming Forward Exchange Rates Are
Realized (cash flows in millions)
Time
1
2
3
3
USD Cash
Flow
GBP Cash
Flow
Forward
Exchange Rate
USD Value of
GBP Cash Flow
Net Cash
Flow in USD
−0.6
−0.6
−0.6
−10.0
0.2
0.2
0.2
5.0
1.8000
1.8400
1.8800
1.8800
0.360
0.368
0.376
9.400
−0.240
−0.232
−0.224
−0.600
another three years. The swap cash flows are shown in the second and third columns
of Table D.2. The forward exchange rates are (we assume) those shown in the
fourth column. These are used to convert the GBP cash flows to USD. The final
column shows the net cash flows. The value of the swap is the present value of these
cash flows.
An alternative approach (which gives the same valuation) is to regard the swap
as a long position in a GBP bond and a short position in a USD bond. Each bond
can be valued in its own currency in the usual way and the current exchange rate
can be used to convert the value of the GBP bond from GBP to USD.
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APPENDIX
E
Valuing European Options
T
he Black–Scholes–Merton formulas for valuing European call and put options on
an investment asset that provides no income are
c = S0 N(d1 ) − Ke−r T N(d2 )
and
p = Ke−r T N(−d2 ) − S0 N(−d1 )
where
d1 =
ln(S0 /K) + (r + 2 /2)T
√
T
d2 =
√
ln(S0 /K) + (r − 2 /2)T
= d1 − T
√
T
The function N(x) is the cumulative probability distribution function for a standardized normal distribution (see tables at the end of the book or Excel’s NORMSDIST
function). The variables c and p are the European call and European put price, S0 is
today’s asset price, K is the strike price, r is the continuously compounded risk-free
rate, is the stock price volatility, and T is the time to maturity of the option.
When the underlying asset provides a cash income, the present value of the income during the life of the option should be subtracted from S0 . When the underlying
asset provides a yield at rate q, the formulas become
c = S0 e−qT N(d1 ) − Ke−r T N(d2 )
and
p = Ke−r T N(−d2 ) − S0 e−qT N(−d1 )
where
d1 =
ln(S0 /K) + (r − q + 2 /2)T
√
T
d2 =
√
ln(S0 /K) + (r − q − 2 /2)T
= d1 − T
√
T
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TABLE E.1
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RISK MANAGEMENT AND FINANCIAL INSTITUTIONS
Greek Letters for Options on an Asset That Provides a Yield at Rate q
Greek Letter
Delta
Call Option
e
−qT
N(d1 )
−qT
Gamma
Theta (per yr)
Vega (per %)
Rho (per %)
Put Option
e
−qT
[N(d1 ) − 1]
N (d1 )e
√
S0 T
√
−S0 N (d1 )e−qT /(2 T)
N (d1 )e−qT
√
S0 T
√
−S0 N (d1 )e−qT /(2 T)
+ qS0 N(d1 )e−qT
− qS0 N(−d1 )e−qT
−r T
+ r Ke−r T N(−d2 )
√
S0 T N (d1 )e−qT
100
K Te−r T N(−d2 )
−
100
− r Ke N(d2 )
√
S0 T N (d1 )e−qT
100
K Te−r T N(d2 )
100
Options on a foreign currency can be valued by setting q equal to the foreign risk-free
rate. Options on a futures or forward price can be valued by using these formulas
by setting S0 equal to the current forward or futures price, r = q, and equal to the
volatility of the forward or futures price.
Table E.1 gives formulas for the Greek letters for European options on an asset
that provides income at rate q. N (x) is the standard normal density function:
1
2
N (x) = √ e−x /2
2
EXAMPLE E.1
Consider a six-month European call option on a stock index. The current value of
the index is 1,200, the strike price is 1,250, the risk-free rate is 5%, the dividend
yield on the index is 2%, and the index volatility is 20%. In this case, S0 = 1,200,
K = 1,250, r = 0.05, q = 0.02, = 0.2, and T = 0.5. The value of the option is
53.44, the delta of the option is 0.45, the gamma is 0.0023, the theta is –0.22, the
vega is 3.33, and rho is 2.44. Note that the formula in Table E.1 gives theta per year.
The theta quoted here is per calendar day.
The calculations in this appendix can be done with the DerivaGem software by
selecting Option Type: Black–Scholes European. Option valuation is described more
fully in Hull (2012).1 The implied volatility of an option is defined as the volatility
that causes the Black–Scholes–Merton price of the option to be equal to the market
price (see Section 10.2).
1
See J. C. Hull, Options, Futures, and Other Derivatives, 8th ed. (Upper Saddle River, NJ:
Pearson, 2012).
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APPENDIX
F
Valuing American Options
o value American-style options, we divide the life of the option into n time steps
of length ⌬ t. Suppose that the asset price at the beginning of a step is S. At the
end of the time step it moves up to Su with probability p and down to Sd with
probability 1 − p. For an investment asset that provides no income the values of u,
d, and p are given by
T
u = e
√
⌬t
d=
1
u
p=
a−d
u−d
with
a = er ⌬ t
Figure F.1 shows the tree constructed for valuing a five-month American put
option on a non-dividend-paying stock where the initial stock price is 50, the strike
price is 50, the risk-free rate is 10%, and the volatility is 40%. In this case, there
are five steps so that ⌬ t = 0.08333, u = 1.1224, d = 0.8909, a = 1.0084, and p =
0.5073. The upper number at each node is the stock price and the lower number is
the value of the option.
At the final nodes of the tree, the option price is its intrinsic value. For example,
at node G, the option price is 50 − 35.36 = 14.64. At earlier nodes, we first calculate
a value assuming that the option is held for a further time period of length ⌬ t and
then check to see whether early exercise is optimal. Consider first node E. If the
option is held for a further time period, it will be worth 0.00 if there is an up move
(probability: p) and 5.45 if there is a down move (probability: 1 − p). The expected
value in time ⌬ t is therefore 0.5073 × 0 + 0.4927 × 5.45 or 2.686, and the 2.66
value at node E is calculated by discounting this at the risk-free rate of 10% for
one month. The option should not be exercised at node E as the payoff from early
exercise would be zero. Consider next node A. A calculation similar to that just
given shows that, assuming it is held for a further time period, the option’s value at
node A is 9.90. If exercised, its value is 50 − 39.69 = 10.31. In this case, it should
be exercised and the value of being at node A is 10.31.
Continuing to work back from the end of the tree to the beginning, the value of
the option at the initial node D is found to be 4.49. As the number of steps on the
tree is increased, the accuracy of the option price increases. With 30, 50, and 100
time steps, we get values for the option of 4.263, 4.272, and 4.278, respectively.
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RISK MANAGEMENT AND FINANCIAL INSTITUTIONS
At each node:
Upper value = Underlying asset price
Lower value = Option price
Shading indicates where option is exercised.
Strike price = 50
Discount factor per step = 0.9917
Time step, dt = 0.0833 years, 30.42 days
Growth factor per step, a = 1.0084
Probability of up move, p = 0.5073
Up-step size, u = 1.1224
Down-step size, d = 0.8909
F
62.99
0.64
56.12
D
2.16
C
50.00
50.00
4.49
3.77
44.55
6.96
B
39.69
10.36
89.07
0.00
79.35
0.00
70.70
0.00
70.70
0.00
62.99
0.00
56.12
1.30
44.55
6.38
E
50.00
2.66
A
39.69
10.31
35.36
14.64
56.12
0.00
44.55
5.45
G
35.36
14.64
31.50
18.50
28.07
21.93
Node Time:
0.0000
0.0833
0.1667
0.2500
0.3333
0.4167
FIGURE F.1 Binomial Tree from DerivaGem for American Put on
Non-Dividend-Paying Stock
To calculate delta, we consider the two nodes at time ⌬ t. In our example, as we
move from the lower node to the upper node, the option price changes from 6.96 to
2.16 and the stock price changes from 44.55 to 56.12. The estimate of delta is the
change in the option price divided by the change in the stock price:
Delta =
2.16 − 6.96
= −0.41
56.12 − 44.55
To calculate gamma we consider the three nodes at time 2⌬ t. The delta calculated
from the upper two nodes (C and F) is −0.241. This can be regarded as the delta for
a stock price of (62.99 + 50)/2 = 56.49. The delta calculated from the lower two
nodes (B and C) is −0.639. This can be regarded as the delta for a stock price of
(50 + 39.69)/2 = 44.84. The estimate of gamma is the change in delta divided by
the change in the stock price:
Gamma =
−0.241 − (−0.639)
= 0.034
56.49 − 44.84
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537
Appendix F: Valuing American Options
We estimate theta from nodes D and C as
Theta =
3.77 − 4.49
2 × 0.08333
or –4.30 per year. This is –0.0118 per calendar day. Vega is estimated by increasing
the volatility, constructing a new tree, and observing the effect of the increased
volatility on the option price. Rho is calculated similarly.
When the asset underlying the option provides a yield at rate q, the procedure
is exactly the same except that a = e(r −q)⌬ t instead of er ⌬ t in the equation for p.
When the option is on the forward or futures price, a is set equal to one and the tree
shows the forward or futures price at each node. The calculations we have described
can be done using the DerivaGem software by selecting Option Type: Binomial
American. Binomial trees and other numerical procedures are described more fully in
Hull (2012).1
1
See J. C. Hull, Options, Futures, and Other Derivatives, 8th ed. (Upper Saddle River, NJ:
Pearson, 2012).
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APPENDIX
G
Taylor Series Expansions
onsider a function z = F (x). When a small change ⌬ x is made to x, there is a
corresponding small change ⌬ z in z. A first approximation to the relationship
between ⌬ z and ⌬ x is
C
⌬z =
dz
⌬x
dx
(G.1)
This relationship is exact if z is a linear function of x and approximate in other
situations. A more accurate approximation is
⌬z =
dz
1 d2 z
⌬x+
(⌬ x)2
dx
2 dx2
(G.2)
This relationship is exact if z is a quadratic function of x and approximate in other
situations. By adding more terms in the series, we can increase accuracy. The full
expansion is
⌬z =
1 d2 z
dz
1 d3 z
1 d4 z
2
3
⌬x+
(⌬
x)
+
(⌬
x)
+
(⌬ x)4 + · · ·
dx
2! dx2
3! dx3
4! dx4
EXAMPLE G.1
Consider the function z =
√
x so that
dz
1
=
dx
2x1/2
d2 z
1
= − 3/2
dx2
4x
Suppose that x = 2 and ⌬ x = 0.1 so that ⌬ z =
x=2
dz
= 0.35355
dx
d2 z
= −0.08839
dx2
d3 z
3
=
dx3
8x5/2
√
2.1 −
√
2 = 0.034924. When
d3 z
= 0.06629
dx3
The first order approximation to ⌬ z, given by equation (G.1), is
⌬ z = 0.35355 × 0.1 = 0.035355
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RISK MANAGEMENT AND FINANCIAL INSTITUTIONS
The second order approximation, given by equation (G.2), is
⌬ z = 0.35355 × 0.1 +
1
× (−0.08839) × 0.12 = 0.034913
2
The third order approximation is
⌬ z = 0.35355 × 0.1 +
1
1
× (−0.08839) × 0.12 + × 0.06629 × 0.13 = 0.034924
2
6
It can be seen that the series expansion quickly converges to the correct answer of
0.034924.
Functions of Two Variables
Consider next a function of two variables, z = F (x, y). Suppose that ⌬ x and ⌬ y are
small changes in x and y, respectively, and that ⌬ z is the corresponding small change
in z. In this case, the first order approximation is
⌬z =
∂z
∂z
⌬x+
⌬y
∂x
∂y
(G.3)
The second order approximation is
⌬z =
∂z
∂z
1 ∂2z
1 ∂2z
∂2z
2
2
⌬x+
⌬y +
(⌬ x⌬ y)
(⌬
x)
+
(⌬
y)
+
∂x
∂y
2 ∂ x2
2 ∂ y2
∂ x∂ y
EXAMPLE G.2
Consider the function z =
√
xy so that
y1/2
∂z
=
∂x
2x1/2
∂2z
y1/2
=
−
∂ x2
4x3/2
x1/2
∂z
=
∂y
2y1/2
∂2z
x1/2
=
−
∂ y2
4y3/2
1
∂2z
=
∂ x∂ y
4(xy)1/2
Suppose that x = 2, y = 1, ⌬ x = 0.1, and ⌬ y = 0.1 so that
⌬z =
√
√
2.1 × 1.1 − 2 × 1 = 0.10565
When x = 2 and y = 1
∂z
= 0.35355
∂x
∂2z
= −0.08839
∂ x2
∂z
= 0.70711
∂y
∂2z
= −0.35355
∂ y2
∂2z
= 0.17678
∂ x∂ y
(G.4)