Options and Other Derivatives
The One-Period Model
The previous chapter introduced the following two methods:
• Replicate the option payoffs with known securities, and calculate the
price of the replicating portfolio,
• Use state price vectors to discount the option’s cash flows.
Rather than using the state-price vector to discount the cash flows, now we
do what’s referred in the book to as normalizing the cash flows, and then
discount them back using a risk-neutral probability. Normalizing the cash
flows means dividing the cash flows of one security by the cash flows of
another security. The risk-neutral probabilities are the probabilities that
equate the discounted normalized cash flows of the security to its normalized
price.
The following are known securities:
105
S1 :
100
105
S2 :
110
100
95
When S2 is normalized, it becomes:
S2 :
110/105
100/100
95/105
Copyright © 2006 by Krzysztof Ostaszewski
We solve for the risk-neutral probability of an “up” move, q, by discounting
the normalized cash flows of S2 .
100  110
95 
2
, thus q = .
= q⋅
+ (1 − q) ⋅
100  105
105 
3
The system is arbitrage-free if and only if risk-neutral probabilities (between
zero and one) exist. Knowing the risk-neutral probabilities allows us to
calculate, for example, the price of a call option on S2 with strike price 104.
We note that the option pays 6 in the “up” scenario and 0 in the “down”
scenario. Its normalized cash flows are:
c
100
6
105
0
105
where c is the call price. We discount its cash flows with interest and the
risk-neutral probabilities to get its price:
c
  2  6   2  
=
+ 1 − ( 0)
100   3  105   3 
and c = 3.81.
Copyright © 2006 by Krzysztof Ostaszewski
The Multiperiod Model
This is the example in the book illustrating the use of arbitrage-free pricing
in a multiperiod model. Here are the payoffs of two securities in all four
possible states, in a two-period model:
S1 (1, )
S2 (1, )
105
105
100
100
1
21
3
4
110
110
95
95
S1 (2, )
110
100
95
100
S2 ( 2, )
120
100
95
90
We have the following evolution of S1 :
110
105
100
100
95
100
100
Let us normalize S2 by dividing its cash flows by those of S1 at each node:
120/110
110/105
100
100
100/100
95/95
95/100
90/100
We can use the normalized cash flows to determine the risk-neutral “up”
probabilities at each node (they need not the same at each node). For the
upper node at time 1 (denote this probability by q(1,1)):
110
120
100
11
= q(1,1)
+ (1 − q (1,1 ))
⇒ q(1,1) =
105
110
100
21
Copyright © 2006 by Krzysztof Ostaszewski
For the lower node at time 1, q(1,0):
95
95
90
1
= q(1,0) + (1 − q (1,0))
⇒ q(1,0) =
100
95
100
2
For the node at time 0, q(0,0):
100
110
95
21
= q( 0,0 )
+ (1 − q (0,0 ))
⇒ q( 0,0 ) =
100
105
100
41
We obtain unique solutions, and all of the risk-neutral probabilities are
between zero and one, indicating that this is an arbitrage-free model. Now,
we can use the risk-neutral probabilities we just found to calculate the
Arrow-Debreu securities prices. This is useful, because all securities are
linear combinations of Arrow-Debreu securities, and price is a linear
operator. For 1 , we must go up at the first node, and then up again at the
second node. So the risk-neutral probability of getting to state 1 at time 2, in
our risk-neutral notation, is:
Q(
1
) = q( 0,0 )q (1,1 ) =
The remaining probabilities are:
Q(
Q(
Q(
21 11 11
⋅ =
41 21 41
21  11  10
⋅ 1−
=
2
41  21 41
21 1 10
)3 = (1 − q (0,0 ))q (1,0) =  1 −  ⋅ =
41 2 41
21
1
10
= (1 − q( 0,0 ))(1 − q(1,1 )) = 1 −  ⋅  1 −  =
4)
 41  2  41
) = q( 0,0 )((1 − q )(1,1 )) =
To calculate the prices of the Arrow-Debreu securities we use the
normalized cash flows and discount utilizing the risk-neutral probabilities.
For the Arrow-Debreu security for state 1 at time 2, its price is denoted by
(2, 1) . It will have normalized payoffs:
1/110
1/105
0/100
0/95
(2, 1)
0/100
0/100
Copyright © 2006 by Krzysztof Ostaszewski
Therefore:
(2, )
1
100
= Q(
1
)
1
⇒
110
(2, ) = 
1
11   100 10
= .
41  110 41
The remaining Arrow-Debreu security values are:
(2, ) = Q
( )
1
⇒
100
(2, )
1
⇒
95
1
⇒
100
2
100
2
3
100
(2, )
4
100
= Q(
3
)
= Q(
4
)
(2, ) = 
2
(2, ) = 
3
10   100  10
= ,
41  100  41
10   100 200
=
,
41  95  779
(2, ) = 
4
10   100 10
= .
41  100 41
Knowing these values we can establish the price of any European option.
The book asks us to price the value of a European call option on Asset 2
with a strike price of 100 and expiring at time 2. Such an option will have a
payoff of 20 in state 1 at time 2 and zero in all other states. Therefore the
price of such an option is
20
(2, ) = 20
1
10  200
=
= 4.878049
41
41
Another example in the book prices a derivative with a payoff equal to
2
(S2 ( 2)) . Squaring the time 2 values for S2 , we obtain the following payoffs
for this derivative:
14,400
10,000
9,025
8,100
1
2
3
4
The price of this derivative is:
14,400
( 2, ) +10,000 (2, ) + 9,025 (2, ) + 8,100 (2, ) =
= 10,243.90
1
Copyright © 2006 by Krzysztof Ostaszewski
2
3
4
The Binomial Option Pricing Model
In this model, in each time period, the asset price either goes up by a factor
of u or down by a factor of d. Each time period is h years (if a year is a unit
of time, which is common, as interest rates are quoted as annual rates) long,
and there are N total time periods (T = Nh). The risk-free force of interest is
rh
r. To make this model arbitrage-free, we must assume: u > e > d . The
e rh − d
probability of an “up” move in the asset price is q =
(this is exactly
u−d
1+i − d
the same formula as q =
proved previously, except for the use of
u−d
the force of interest and possibly fractional time t).
European Call and Put Options
Let m be the minimum number of upward (downward) moves such that the
call (put, respectively) option is in the money. Then m is the smallest
integer such that
 K 
ln  N

 d S (0 )
m>
u
ln  
 d
European call/put option prices:
where
c( 0) = S (0 )Φ(m;N,q *) − Ke− rNh Φ(m;N,q) , and
p( 0) = KerNh (1 − Φ (m; N,q )) − S( 0)(1 − Φ( m;N,q *)),
Φ( m;N,q) = ∑
N
j =m
and q* = que .
N!
N− j
q j (1 − q )
j! ( N − j )!
rh
Copyright © 2006 by Krzysztof Ostaszewski
The continuous-time limit of the binomial model is the Black-Scholes
h
model. If we select u = e , then the distribution of stock price changes is
2
lognormal with mean (r −
/ 2)T and variance 2T . As N gets large,
under these parameter selections, the binomial model’s valuation will
approach the Black-Scholes and can be taken to be the same for sufficiently
large N.
Binomial model parameters
In practical applications of the binomial model, it is worth noting that a
recombining tree dramatically reduces the number of computations required.
Also, one has to be careful in practice, as if parameter values are chosen
inappropriately, the model will no longer be arbitrage-free.
Recursive valuation
One can work the valuation tree backwards (right-to-left) to establish riskneutral probabilities and values of all securities. In general:
Value at current node =
= (Value in up node ⋅ Pr(up) +
+Value in down node⋅ Pr ( down))
or
V (i, j ) = e −rh (qV (i +1, j +1) + (1 − q)V (i + 1, j )) .
In order to value any security, one uses trading strategies that replicate those
securities with those with available prices. We generally work recursively,
solving for the replicating portfolio at each node
uV (i + 1, j ) − dV (i + 1, j + 1)
u−d
V (i +1, J +1) − V (i + 1, j )
i, j ) =
2(
( u − d )S (i, j)
1
(i, j) = e
−r ( i+1) h
, a solution of the above, is often referred to as the delta of the derivative;
you will see why this is when you get to the section on delta hedging.
2
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Dividends and other income
Remember that the price of a stock is reduced by the amount of the dividend
at the moment the dividend is paid. To incorporate the dividend payments
into the recursive model, the following formulas are used:
For dollar dividends
S(i −1, j ) = e − rh (q (S(i, j +1) + D(i, j + 1)) + (1 − q)(S(i, j ) + D(i, j )) )
S(i, j +1) + D(i, j + 1) = S(i −1, j )u
S(i, j ) + D(i, j ) = S(i −1, j )d
Thus, when working recursively, you must add the dividend to the stock
price before discounting. For dividends, which are proportions of stock
price,
S(i −1, j ) = e −( r− ) h (qS(i −1, j )u1 + (1 − q) S(i −1, j ) d1 ),
where: e = 1 +
h
(h ) , ( h) =
D(i, j)
u
d
, u1 =
, d1 =
,
S(i, j )
1+
1+
e rh − d
. This is intuitively similar to assuming that the stock price
q=
u−d
grows at a rate of r − . Note that the stock price tree will not recombine
when a dividend is paid as a constant dollar amount; it will, however, it will
recombine when the dividend is a constant proportion of the stock value.
Put-call parity for dividend-paying assets
We already know that for non-dividend paying stocks
c( 0) − p( 0) = S( 0)e − t − Ke − rT
The modifications to Black-Scholes model in order to accommodate a
dividend-paying stock are exactly the same as the modifications to the putcall parity.
Exotic derivatives
These are derivatives that go beyond the standard European put and call.
Many exotic options are path dependent and therefore difficult to value with
a binomial method, especially if they are American.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Main types of exotic derivatives:
- Digital (a.k.a. Binary) Option
Its payoff depends on whether the underlying asset value is above or below a
fixed amount on the expiration date. Cash-or-nothing call (put) options pay a
fixed amount, X, if the underlying asset value is above (below, respectively)
the exercise price K, and nothing otherwise. All-or-nothing call (put) options
pay the asset value if the terminal asset price is above (below) the exercise
price at expiration, and nothing otherwise. One-touch all-or-nothing options
pay off if the underlying asset goes above (call) or below (put) the exercise
price during the life of the option.
- Gap option
Payoff depends on whether the underlying asset value on the expiration date
is above or below a fixed amount that is different from that used for the
payoff. The payoff for a call option is S(T) – K if S(T) > H or zero otherwise.
- Asian option
Payoff depends on the average price of the underlying asset during the life of
the option. Commonly used in equity-indexed products (e.g., equity-linked
annuities), foreign currency options, and interest rate options for hedging
purposes.Asian options pay the difference between the average price and the
exercise price provided this is a positive quantity. The exercise price can be
fixed such that the payoffs are:
+
Average price call: (SAVG − X )
Average price put:
(X − S )
+
AVG
The exercise price can also be the average (called “floating strike options”):
Average strike call: (S(T ) − SAVG )
Average strike put: (SAVG − S( T ))
+
+
- Lookback option
For such an option, payoff depends on the underlying asset price at
expiration and also the minimum or maximum asset value attained during
+
the life of the option. Therefore, a lookback call pays (S(T ) − Smin ) , and a
lookback put pays (Smax − S(T )) . These options are path dependent, and
thus quite difficult to price binomially. A variation is a high-water mark call
+
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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option, which pays (Smax − K ) . Such an option is commonly found in
equity-indexed annuities.
+
- Barrier options
For such an option, its payoff depends on whether or not the value of the
underlying asset reaches a certain price level (a barrier) before expiration.
Knockout options become worthless or pay a fixed rebate if the asset value
reaches a barrier but otherwise have payoffs identical to a standard option.
- Down-and-out options become worthless or pay a fixed amount if the
asset value falls below a barrier.
- Down-and-in options only come into existence if the asset price falls
below a barrier.
- Up-and-out options become worthless or pay a fixed acount if the asset
value reaches a barrier.
- Up-and-in options only come into existence if the asset price reaches a
barrier.
- Double knockout options become worthless or pay a fixed amount if the
asset price reaches either of the lower or upper barrier.
- A standard call option is the sum of a down-and-out call option and a
down-and-in call option with the same exercise price and barrier.
Options on the minimum or maximum (a.k.a. rainbow options)
- Their payoffs depend on the values of several assets.
- Options pay the maximum or minimum of the several assets.
- A European option on the maximum of two assets is equivalent to
holding one of the assets plus a European option to exchange this asset
for the other one.
Cliquets
A cliquet option is a series of standard call options that pay the annual
increase in the underlying asset.
Quantos
Short for “quantity adjusted option”. Guaranteed exchange-rate contracts in
which the payoff on a foreign currency derivative is converted to the
domestic currency at a fixed exchange rate.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Other exotics
- Compound options = options on other options.
- Chooser options are options, which permit the choice between buying a
call or put option so that the option holder can select the option they
require at expiration depending on which is more valuable.
- Spread options = payoffs reflect the difference between the values of two
assets.
American options
Permit the holder to exercise at any time. This early exercise privilege
allows us to establish some bounds for the price of an American option.
+
Consider a call option. We have S( t ) ≥ C(t ) ≥ (S (t ) − K ) . If the option
cost were greater than the asset price, then you could sell the option, buy the
asset, and keep the difference, using the asset to fulfill the obligation on the
option you sold. If the option cost were less than the intrinsic value, then you
could buy the option and immediately exercise it for the intrinsic value,
making a riskless profit. In fact, the call price will be strictly greater than the
intrinsic value at all times, except possibly maturity or just immediately
prior to a dividend payment. Thus a rational investor will not exercise an
American call option between dividend payment dates, since such an option
can always be sold for more money than could be obtained by exercising it.
Therefore if there are no dividend payments then the price of an American
call option on a single asset is equal to the price of a European call option on
the same asset. It is rational to exercise an American option when the
intrinsic value immediately prior to the dividend payment is greater than the
value of the option after the dividend payment; so exercise when
S( t ) > S * (t − ), where S * (t − ) − K = D( t *) (the underlying price at which
the intrinsic value of the option immediately prior to the dividend payment is
equal to the option value immediately after the dividend payment). For
American put options, there is always a critical asset price below which it is
optimal to exercise the American put option early.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Numerical Valuation of American options using the Binomial Method
For an American put option, the value at each node is
(
P = Max K − S,e rh (q ( K − Su ) + (1 − q)( K − Sd )
+
+
))
This means that you should calculate the value recursively from right-to-left,
but replace the option value at any node with the intrinsic value at that node
if the intrinsic value at that node is greater than the recursive value
calculated. Note the three possibilities for the put option at each node that
affect the decision to exercise early:
- If the underlying price in both “up” and “down” states is less than the
strike price of the put option (i.e., the put is in the money in both
scenarios) then it is optimal to exercise early.
- If the underlying price in both “up” and “down” states is greater than
the strike price of the put option (i.e., the put is out of the money in
both scenarios) then it is not optimal to exercise early.
- If the option is in the money in the “down” state but not in the “up”
state, then you have to compute the critical price, S*; this is the price
at which the investor is indifferent between exercising early and
retaining the option. It can be found by solving the equation:
K − S* = e − rh q ⋅ 0 + (1 − q ) ⋅ ( K − S * d ) , i.e., the current intrinsic
value has to equal the current value if unexercised. This solves to:
(
)
1 − e −rh (1 − q)
. The investor should exercise the option if
S* = K
1 − e − rh (1 − q) d
the actual asset price, S, is below the critical price, S*.
The procedure for an American call option is similar, except the comparison
between the option’s value and it’s intrinsic value for the purposes of
substituting the greater of the two is made only at nodes at which a dividend
is paid.
Numerical Methods
They have become very important in finance. Reasons for that are:
- Security models have become more complex,
- Types of securities and derivatives are also complex,
- New developments in risk management technology, and
- Regulation and practice mandate more analysis (e.g., VaR).
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Some comments on numerical methods:
- Even closed form or analytic solutions still require numerical
techniques,
- Prices of some derivatives are governed by partial differential
equations, which require numerical techniques to solve,
- Lattice methods are numerical in nature,
- Monte Carlo, i.e., simulation, is numerical in nature.
Lattice models
- Lattice models approximate the distribution of the underlying with a
discrete one,
- Trinomial models have been shown to be more stable and more
accurate with half as many time intervals as a standard binomial
lattice. But they require more calculation.
The trinomial model can be described as follows:
“Up” movement by a factor u with probability q u
“Middle” movement by a factor m with probability 1 − q u − q d
“Down” movement by a factor d with probability q d
The probabilities are computed by equating the first three moments of the
lognormal distribution:
q u u + (1 − q u − qd )m + q d d = e rh
q u u 2 + (1 − q u − qd )m 2 + q d d 2 = e 2rh +
q u u 3 + (1 − q u − qd )m 3 + q d d 3 = e 3rh + 3
Here the parameters are u = e
h
,m = 1, d =
2
h
2
h
1
and
u
is chosen to make
the model arbitrage-free. This produces a recombining tree.
Monte Carlo simulation
Steps:
- Generate N values from the distribution of the ending asset value,
- Compute the option value for each underlying value,
- Discount this value at the risk-free rate,
- Compute the average option value for the sample.
As N gets large, the answer will converge to the true value.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Variance reduction techniques
They reduce the standard error more efficiently than running thousands of
simulations:
- Antithetic variable technique: Average a pair of unbiased estimators
to produce a new unbiased estimator whose variance is less than either
of the individual estimators assuming that the two individual
estimators are negatively correlated.
- Control variate method: Replace the problem under consideration with
a similar but simpler problem that has an analytic solution. The two
must be highly correlated.
- Stratified sampling: Break the sampling region into pieces (called
strata), then sample from the strata. You may also sample from only
significant strata: For example, if pricing an option, which is deeply
out of the money, don’t simulate asset prices that will only result in a
zero value for the option.
- Low-discrepancy methods: Use deterministic points that are as
uniformly distributed as possible rather than random simulations.
- Path dependency: One can combine simulated paths into “bundles” to
approximate the value of American-style derivatives.
Hedging
Dealers who write derivative contracts must manage the risk. To manage the
risk, one generally uses hedging, i.e., assuming positions that balance the
risk by offsetting it. The risk is measured with various sensitivity statistics.
They can be estimated by the use of a binomial model:
- Delta is the sensitivity of an option’s price to changes in the price of the
underlying:
∆(i, j ) =
V (i + 1, j +1) − V (i +1, j )
( u − d )S(i, j)
The hedge of the derivative uses the number of units of the underlying asset
in the replicating portfolio equal to the delta. If a portfolio of assets is
constructed so that the overall portfolio has a delta of zero, then the portfolio
is said to be delta neutral. In the Black-Scholes model of a call option, Delta
= N ( d1 ) .
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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- Gamma is the sensitivity of delta to changes in the price of the
underlying asset
Γ=
∆(i + 1, j +1) − ∆(i + 1, j )
( u − d )S(i, j)
In the Black-Scholes model for a call option, Γ =
Φ( d1 )
.
S0 T
- Theta is the sensitivity of the value of a derivative with respect to time,
time-decay measure:
Θ=
V (i + 2, j + 1) − V (i, j )
2h
- Rho is the sensitivity of the value of the derivative to interest rates.
- Vega is the sensitivity of the value of the derivative to volatility of the
underlying.
The sensitivities can be estimated using simulation. The hedge portfolio’s
sensitivities will change over time, and one must rebalance the hedge
repeatedly. There is a tradeoff between the safety of frequently rebalancing
the hedge portfolio and the transaction costs required to do so. Pathdependent derivatives present the biggest challenges in hedging, since the
hedging requirement can often change dramatically, especially when the
option is nearing expiration and the option is close to the money or some
other meaningful boundary (such as the barrier in a knockout option). A
static hedge is one that is established on the initial date and not rebalanced; it
is constructed in an attempt to replicate the path-dependent option’s payoffs
at critical price levels (such as near the boundary).
Economic roles of derivatives:
- Provide efficient method to achieve payoffs that are not readily available
otherwise,
- Path-dependent options can hedge stochastic volatility,
- Can help manage tax and regulatory considerations,
- Lower transaction costs than purchasing replicating securities
individually.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Insurance products
Life insurance products contain a variety of guarantees, including maturity
guarantees and return of premium guarantees upon death or withdrawal, of
which minimum interest rate guarantees are most significant. These
embedded options are quite a challenge to price. They require input
regarding the possible values of the asset portfolio backing the products, and
regarding the behavior of the policyholders. Equity-indexed products are
also common in some insurance companies (equity-linked annuities are a
new product since 1995).
Pension Plans
Embedded option: participant is entitled to the maximum of two alternative
benefit amounts (such as a defined benefit formula or the accumulated value
of a set of contributions). A trinomial valuation of this benefit based on two
variables (salary growth rate and implicit crediting rate) found that the
standard deterministic valuation typically employed by actuaries may
undervalue this option as much as 35%.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Exercise 1.
A European derivative pays the excess of the underlying asset price over
$23, plus a dollar (the dollar is paid even if the excess is zero), in two
months’ time. The current price of the underlying is $22, and the volatility of
the underlying asset’s continuously compounded rate of return is 10% per
year. The continuously compounded risk free interest rate in the risk-neutral
world is 0.25% per month. Us a two-period binomial model to estimate the
price of this European derivative.
Solution. The basic formulae for the binomial model are (for each onemonth period):
0.10
1
12
u=e =e
= 1.029288
1
d = = 0.971545
u
e rh − d
e 0.0025 − 0.971545
q=
=
= 0.5361
u − d 1.029288 − 0.971545
e rh = e 0.0025 =1.002503127
h
Using these parameters for the underlying we get this model:
$22.64 ⋅1.029288 = $23.31
$22 ⋅1.029288 = $22.64
$22
$22 ⋅ 0.971545 = $21.37
$22
$21.37 ⋅ 0.971545 = $20.76
The payoffs of the derivative are (note that 1 – 0.5361 = 0.4639):
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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$1.31
1
( $1.31⋅0.5361 + $1.00 ⋅ 0.4639) = $1.16918
1.00253127
$1.08
$1.00
1
( $1.00 ⋅0.5361 + $1.00 ⋅ 0.4639) = $0.99750
1.00253127
$1.00
The price of this derivative is
$1.08 =
1
( $1.16918 ⋅0.5361 + $0.99750 ⋅ 0.4639).
1.00253127
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Exercise 2.
You are given the following binomial model for the value of the short-term
interest rate, risk-free over a one-year period:
• One year from now this short-term interest rate is either:
r1u = r0 ⋅ (1 + γ ) with probability 0.50, or
r
r1d = 0 with probability 0.50,
1+ γ
where r0 in the initial short-term rate, and γ is a parameter. The
probabilities given are risk-neutral probabilities.
• Annualized volatility of this short-term interest rate is σ = 25%.
• The current value of the short-term rate is 4%.
A 2-year European (meaning that it pays only if the short-rate breaches the
floor at the end of two years) interest rate floor with a 3.5% strike level and a
notional amount of 100 is issued. This derivative security will pay the
difference between 3.5% and then current short-term interest rate as
calculated for the 100 notional amount, if such difference is positive.
Calculate the value of this interest rate floor.
Solution.
Recall that in the binomial model u = eσ t and d = e− σ t . When t = 1, this
becomes u = eσ and d = e− σ . In terms of the notation of this problem, you
can conclude that:
r1u
2
2σ
=
e
=
1
+
γ
.
(
)
r1d
We are given σ = 25%. This means that
1 + γ = eσ = 1.28402542.
Given this, the short-term rate will be in two years:
2
4% ⋅ (1.28402542 ) = 6.59488508% with probability 0.25,
4% with probability 0.50, and
−2
4% ⋅ (1.28402542 ) = 2.42612264% with probability 0.25.
Only the third outcome produces positive cash flows from the floor, which is
then worth
100 ⋅ ( 3.50% − 2.42612264%) = 100 ⋅1.07387736% = 1.07387736.
This cash flow is paid with probability 0.25, and its expected present value is
the price of the floor. We use the risk-free rate over the next year as the rate
for discounting for that year. Therefore the value of the floor is:
1.07387736
⋅ 0.25 ≈ 0.25034485.
0.04
⎛
⎞
1.04 ⋅ ⎜ 1 +
⎝ 1.28402542 ⎟⎠
Exercise 3.
Now assume in the previous problem that the floor only pays the difference
between 3.5% and the current short-term rate at the end of the first year, if
such difference is positive. What is the value of such one-year floor?
Solution.
At the end of the first year, the short rate is
4% ⋅ e0.25 ≈ 5.136102% with probability 0.50, and
4% ⋅ e−0.25 ≈ 3.115203% with probability 0.50.
Only the second outcome gives rise to a payment by the floor, with such
payment being 0.38479687 on the 100 notional amount. Its expected present
value is
0.38479687
⋅ 0.50 ≈ 0.18499849.
1.04
If the floor were a two-year floor inclusive of both years, its total value
would be
0.25034485 + 0.1849849 = 0.43534334.
Exercise 4.
You are given the following securities:
• 60-day Treasury Bill with face amount 1000.
• 150-day Treasury Bill with face amount 1000 and current price 975.
• Stock of ABCZ Corporation with current price of 24.7282442.
• European call option on the ABCZ stock, with current price of 1, time to
exercise of 60 days exactly, strike price of 30, for which the d1 parameter in
the Black-Scholes equation equals 0.70. The price of this option is equal
exactly to the price given by the Black-Scholes equation.
• European put option on the ABCZ stock, with time to exercise of 60 days,
and strike price of 30. The price of this option is equal exactly to the price
given by the Black-Scholes equation. You can assume that put-call parity
holds perfectly.
• Futures contract on a 90-day Treasury Bill, time to delivery of 60 days,
face amount of 1000, and current price of 984. You can assume that the
fundamental relation between
futures and the underlying holds.
You are also given the cumulative normal distribution values
z
–1.4
–0.7
0
0.7
1.4
N(z) 0.0808
0.242
0.5 0.758
0.9192
Calculate the current market price of the put option.
Solution. The fundamental relation between futures and underlying is:
r ( T −t )
.
F = Se
Here, the 90-day Treasury Bill futures go for 984, and the underlying, which
is the 150-day Treasury Bill, in 60 days deliverable as a 90-day Bill, goes for
975. Therefore, the continuously-compounded risk-free rate, assuming, for
simplicity, 360 days in a year, must satisfy:
984 = 975e
60
r
360
,e
60
r
360
=
984 328
=
.
975 325
This means that to discount anything by 60 days, you multiply by 325/328,
and to accumulate over 60 days, you multiply by 328/325.
From put-call parity, we get:
325
30,
328
325
P = 1− 24.7282442 +
30 =
328
= −23.7282442 + 29.725609756 =
= 5.997365556.
1− P = 24.7282442 −
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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While the $30 call sells for $1, $30 put sells for $5.997365556. This should
not be in any way a surprise, the put has $5 intrinsic value.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Exercise 5.
Using the same data as in problem No. 13, find the annualized volatility
(standard deviation) of the rate of return of ABCZ stock. You may assume
that 1 year has 360 days, and the price of the call is determined by the BlackScholes equation exactly, except that S, the price of the stock, is replaced by
the difference S – PV(Dividends paid on the stock during the life of the
option).
Solution.
We have


 24.7282442  1
+
ln 325
 S  1 2

ln 
+
t
30  2
PV
(
X
)
2


 328

d1 =
=
1
t
6
2
1
6
= 0.70
Therefore:
0.0833333
2
− 0.2857738 - 0.1840629 = 0
0.2857738 ± 0.2857738 2 − 4(0.0833333)(-0.1840629 )
=
2 ⋅ 0.0833333
= 3.98372953, or
= −0.5544439 (can' t be).
Therefore, = 398.37%. Don’t think this answer is unreasonable, as you
have a call option deeply out of the money, and the said call still has
significant value, $1. This can only happen if there is a reasonable chance of
stock exceeding $30 in price within sixty days, and for that to happen, the
stock must move up by roughly 24% within 1/6 of a year.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Exercise 6.
Using the same data as in the previous problem, demonstrate that a oneperiod market consisting of the 60-day T-bill, the put option, and the stock is
arbitrage free over one period. You are additionally given that the period is
60 days, and the stock price at the end of 60 days will be either 21, or 30, or
36.
Solution.
$1 invested in the Treasury Bill will end up being $
328
no matter what
325
happens.
$24.7282442 invested in stock of ABCZ ends up being $21, $30, or $36.
$5.997365556 invested in the put ends up being $9, $0, or $0, as the exercise
price is $30.
The market matrices are:
1.009231 0.849231 1.5006589 


S (0 ) = [1 1 1], S(1) = 1.009231 1.213188
0
.

1.009231 1.455825
0
S(1) is an invertible matrix with determinant 0.4217175494. For the market
T
to be arbitrage-free, all you need is a state-price vector [ 1
,
2
3]
i.e., solution to the system of equations:
1.009231
1
+1.009231
2
+1.009231
0.849231 1 +1.213188 2 +1.455825
1.5006589 1 + 0⋅ 2 + 0⋅ 3 = 1
3
3
=1
=1
Because S(1) is invertible, not only does a solution exist, it is unique, so that
this market is complete as well, as long as the solution is positive. It is pretty
obvious that
1
=
1
= 0.66637395, and we get a new, simpler
1.5006589
system of equations by putting that value in the first two equations.
1.009231
2
+1.009231
3
= 0.32747491
1.220201
2
+1.460201
3
= 0.43409436
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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or
1.220201
1.220201
2
= 0.16666667,
3
2
2
+1.220201
+1.460201
3
3
= 0.39365477
= 0.43409436
= 0.15781303. The market is arbitrage-free.
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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Exercise 7.
Determine whether the market described in the previous problem is
complete, and find its risk-neutral probabilities, if it is arbitrage-free.
Solution.
The market is complete because it is arbitrage-free and the matrix S(1) is
invertible. Because we already solved for the state-price vector, the second
part is almost a freebie.
1
= 0.66637395,
2
= 0.16666667,
3
= 0.15781303
and therefore
328
= 0.67252509
325
328
p2 = 2 (1+ i) = 0.16666667
= 0.16820514
325
328
p3 = 3 (1 + i) = 0.15781303
= 0.15926977
325
p1 =
(1+ i ) = 0.66637395
1
Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski
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