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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:

S

1

:

S

2

:

When

S

2 is normalized, it becomes:

S

2

:

100

105 105

100

110 95

100/100

110/105 95/105

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Copyright © 2006 by Krzysztof Ostaszewski

We solve for the risk-neutral probability of an “up” move, q, by discounting the normalized cash flows of

S

2.

100

100 = q ⋅ 110

105 + ( 1 − q ) 105 95

  

, thus

q = 2 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

S

2 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:

6 c 105

100

0

where c is the call price. We discount its cash flows with interest and the

105

risk-neutral probabilities to get its price:

c

100 = 2 3

  

 6 105

  

 + 1 − 2 3

  

 ( ) 0

   

 

and

c = 3.81

.

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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:

S

1

( ) 1, S

2

( ) 1, S

1

( ) 2, S

2

( 2, )

1 105 110 110 120

21 105 110 100 100

3 100 95 95 95

4 100 95 100 90

We have the following evolution of

S

1: 110 105

100 100

95 100

100

Let us normalize

S

2 by dividing its cash flows by those of

S

1 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

105 = q 1,1 ( ) 120 110 + ( 1 q 1,1 ( ) ) 100 100 q 1,1 ( ) = 21 11

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Copyright © 2006 by Krzysztof Ostaszewski

For the lower node at time 1, q(1,0):

95

100 = q 1,0 ( ) 95 95 + ( 1 − q 1,0 ( ) ) 100 90 q 1,0 ( ) = 1 2

For the node at time 0, q(0,0):

100

100 = q 0,0 ( ) 110 105 + ( 1 − q 0,0 ( ) ) 100 95 q 0,0 ( ) = 21 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 ( ) = 21 41 11 21 = 11 41

The remaining probabilities are:

Q ( )

2

= q 0,0 ( ) ( ( 1 q ) ( ) 1,1 ) = 21 41 1 11 21  = 10 41

Q ( )

3

= ( 1 q 0,0 ( ) ) q 1,0 ( ) = 1 − 21 41

  

 ⋅ 1 2 = 10

41

Q ( )

4

= ( 1 q 0,0 ( ) ) ( 1 q 1,1 ( ) ) = 1 21 41  ⋅ 1 1 2  = 10 41

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

2,

1

( )

0/95

0/100

0/100

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Therefore:

2,

1

( )

100 = Q ( )

1

110 1 ( 2,

1

) = 11 41 100 110  = 10 41

.

The remaining Arrow-Debreu security values are:

2,

2

( )

100 = Q ( )

2

100 1 ( 2,

2

) = 10 41 100 100  = 10 41 ,

2,

3

( )

100 = Q ( )

3

95 1 ( 2,

3

) = 10 41 100 95  = 200 779 ,

2,

4

( )

100 = Q ( )

4

100 1 ( 2,

4

) = 10 41 100 100  = 10 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, (

1

) = 20 10 41  = 200 41 = 4.878049

Another example in the book prices a derivative with a payoff equal to

S

2

( ) 2

( )

2. Squaring the time 2 values for

S

2, we obtain the following payoffs for this derivative:

1 14,400

2 10,000

3 9,025

4 8,100

The price of this derivative is:

14,400 ( 2,

1

) + 10,000 2, (

2

) + 9,025 2, (

3

) + 8,100 2, (

4

)

=

= 10,243.90

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Copyright © 2006 by Krzysztof Ostaszewski

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 r. To make this model arbitrage-free, we must assume:

u > e

rh

> d

. The probability of an “up” move in the asset price is

q = e

rh

d

ud

(this is exactly the same formula as

q = 1 + id

ud

proved previously, except for the use of 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

m >

ln K d

N

S 0 ( )

 

  ln u

d

  

European call/put option prices:

c 0 ( ) = S 0 ( ) Φ ( m; N,q * ) Ke

rNh

Φ ( m; N,q )

, and

p 0 ( ) = Ke

rNh

( 1 − Φ ( m; N,q ) ) S 0 ( ) ( 1 − Φ ( m; N,q * ) )

,

where

Φ ( m; N,q ) = N!

j! N ( − j ) !

j=m

N

q

j

( 1 q )

Nj

and

q* = que

rh.

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The continuous-time limit of the binomial model is the Black-Scholes model. If we select

u = e

h, then the distribution of stock price changes is lognormal with mean

( r

2

/ 2 ) T

and variance 2

T

. 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 risk- neutral 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

1

( ) i, j = e

r i( )+1h

uV i ( + 1, j ) dV i ( + 1, j + 1 )

ud

2

( ) i, j = V i ( + 1, J + 1 ) V i ( + 1, j )

ud

( ) S i, j ( )

2, 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.

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Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 249 -

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 ) u

1

+ ( 1 − q ) S i ( 1, j ) d

1

)

,

where:

e

h

= 1 + ( ) h

,

( ) h = D i, j ( )

S i, j ( )

,

u

1

= 1 + u , d

1

= 1 + d ,

q = e

rh

d

ud

. This is intuitively similar to assuming that the stock price 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 put- call 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.

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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:

( S

AVG

X )

+

Average price put:

( XS

AVG

)

+

The exercise price can also be the average (called “floating strike options”):

Average strike call:

( S T ( ) S

AVG

)

+

Average strike put:

( S

AVG

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 ( ) S

min

)

+, and a

lookback put pays

( S

max

S T ( ) )

+. These options are path dependent, and thus quite difficult to price binomially. A variation is a high-water mark call

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Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 251 -

option, which pays

( S

max

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.

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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.

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Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 253 -

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:

KS* = 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:

S* = K 1 − e

rh

( 1 − q )

1 − e

rh

( 1 − q ) d

. The investor should exercise the option if 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).

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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

q

d

) m + q

d

d = e

rh

q

u

u

2

+ ( 1 − q

u

q

d

) m

2

+ q

d

d

2

= e

2rh+ 2h

q

u

u

3

+ ( 1 − q

u

q

d

) m

3

+ q

d

d

3

= e

3rh+3 2h

Here the parameters are

u = e

h

,m = 1, d = 1

u

and 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.

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Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 255 -

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 )

ud

( ) 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 d ( )

1 .

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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 )

ud

( ) S i, j ( )

In the Black-Scholes model for a call option,

Γ = Φ ( ) d

1

S

0

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. Path-

dependent 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.

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Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 257 -

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%.

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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 one- month period):

u = e

h

= e

0.10

1

12

= 1.029288 d = 1

u = 0.971545 q = e

rh

d

ud = e

0.0025

− 0.971545

1.029288 − 0.971545 = 0.5361

e

rh

= e

0.0025

=1.002503127

Using these parameters for the underlying we get this model:

$22.64 ⋅ 1.029288 = $23.31

$22 ⋅ 1.029288 = $22.64

$22 $22

$22 ⋅ 0.971545 = $21.37

$21.37 ⋅ 0.971545 = $20.76

The payoffs of the derivative are (note that 1 – 0.5361 = 0.4639):

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Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 259 -

$1.31 1

1.00253127 ( $1.31 ⋅ 0.5361 + $1.00 ⋅ 0.4639 ) = $1.16918

$1.08 $1.00

1

1.00253127 ( $1.00 ⋅ 0.5361 + $1.00 ⋅ 0.4639 ) = $0.99750 $1.00

The price of this derivative is

$1.08 =

1

1.00253127 ( $1.16918 ⋅ 0.5361 + $0.99750 ⋅ 0.4639 )

.

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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 r1d = r0

1+γ with probability 0.50,

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

r1d = e = 1+

(

γ

)

2.

We are given σ = 25%. This means that 1+γ = eσ = 1.28402542.

Given this, the short-term rate will be in two years:

4%⋅ 1.28402542

( )

2 = 6.59488508% with probability 0.25, 4% with probability 0.50, and

4%⋅ 1.28402542

( )

−2 = 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:

(20)

1.07387736 1.04⋅ 1+ 0.04

1.28402542

⎛⎝⎜ ⎞

⎠⎟

⋅ 0.25 ≈ 0.25034485.

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

1.04 ⋅ 0.50 ≈ 0.18499849.

If the floor were a two-year floor inclusive of both years, its total value would be

0.25034485 + 0.1849849 = 0.43534334.

(21)

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:

F = Se

r T( −t).

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

36060r

,e

36060r

= 984

975 = 328 325 .

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:

1 − P = 24.7282442 − 325 328 30, P = 1− 24.7282442 + 325

328 30 =

= − 23.7282442 + 29.725609756 =

= 5.997365556.

(22)

Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 264 -

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.

(23)

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 Black- Scholes 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

d

1

=

ln S PV X ( )

  

  + 1 2

2

t

t =

ln 24.7282442 325

328 30

 

 

 

  + 1 2

2

1 6 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.

(24)

Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 266 -

Exercise 6.

Using the same data as in the previous problem, demonstrate that a one- period 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

325

no matter what 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:

S 0 ( ) = [ 1 1 1 ] , S 1 ( ) =

1.009231 0.849231 1.5006589 1.009231 1.213188 0 1.009231 1.455825 0

 

 

 .

S(1) is an invertible matrix with determinant 0.4217175494. For the market to be arbitrage-free, all you need is a state-price vector

[

1 2 3

]

T,

i.e., solution to the system of equations:

1.009231

1

+1.009231

2

+1.009231

3

= 1 0.849231

1

+1.213188

2

+1.455825

3

= 1 1.5006589

1

+ 0⋅

2

+ 0⋅

3

= 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

1.5006589 = 0.66637395,

and we get a new, simpler 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

(25)

or

1.220201

2

+1.220201

3

= 0.39365477 1.220201

2

+1.460201

3

= 0.43409436

2

= 0.16666667,

3

= 0.15781303.

The market is arbitrage-free.

(26)

Spring 2003 Notes for SoA Course 6 exam, Copyright © 2003 by Krzysztof Ostaszewski - 268 -

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

p

1

=

1

( 1 + i ) = 0.66637395 328

325 = 0.67252509 p

2

=

2

( 1 + i ) = 0.16666667 328

325 = 0.16820514 p

3

=

3

( 1 + i ) = 0.15781303 328

325 = 0.15926977

References

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