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Chapter 13
The Black-Scholes-Merton Model
March 3, 2009
13.1. The Black-Scholes option pricing model assumes that the probability distribution of the stock price in one year(or at any other future time) is lognormal. It assumes that the continuously compounded rate of return on the stock during the year is normal distributed.
13.2. The standard deviation of the percentage price change in time δt is σ
δt where σ is the volatility. In this problem σ = 0.3 and, assuming 252 trading days in one year, δt = 1/252 = 0.004 so that σ
δt = 0. 3
0 .004 = 0.019 or 1.9%.
13.3. The price of an option or other derivative when expressed in terms of the price of the underlying stock is independent of risk preference. Options therefore have the same value in a risk-neutral world as they do in the real world. We may therefore assume that the world is risk neutral for the purposes of valuing options. This samplifies the analysis. In a risk-neutral world, all securities have an expected return equal to risk-free interest rate. Also, in a risk-neutral world, the appropriate discount rate to use for expected future cash flows is the risk-free interest rate.
13.4. In this case S 0 = 50, K = 50, r = 0.1, σ = 0.3, T = 0.25, and
d 1 =
ln(50/50) + (0.1 + 0. 32 /2)0. 25
- 3
d 2 = d 1 − 0. 3
The European put price is
p = 50e−^0.^1 ×^0.^25 N (− 0 .0917) − 50 N (− 0 .2417) = 50 × 0. 4634 e−^0.^1 ×^0.^25 − 50 × 0 .4045 = 2. 37
or $2.37.
13.5. In this case we must subtract the present value of the dividend from the stock price before using Black-Scholes. Hence the appropriate value of S 0 is
S 0 = 50 − 1. 50 e−^0.^1667 ×^0.^1 = 48. 52
As before K = 50, r = 0.1, σ = 0.3 and T = 0.25. In this case
d 1 = ln(48. 52 /50) + (0.1 + 0. 09 /2)0. 25
- 3
d 2 = d 1 − 0. 3
The European put price is
50 N (0.1086)e−^0.^1 ×^0.^25 − 48. 52 N (− 0 .0414) = 50 × 0. 5432 e−^0.^1 ×^0.^25 − 48. 52 × 0 .4835 = 3. 03
or $3.03.
13.6. The implied volatility is the volatility that makes the Black-Scholes price of an option equal to its market price. It is calculated using an iterative procedure.
13.7. In this case, μ = 0.15 and σ = 0.25. From equation (13.7), the probability distribution for the rate of return over a 2-year period with continuous compounding is
φ
i.e., φ(0. 11875 , 0 .1768) The expected value of the ruturn is 11.875% per annum and the standard deviation is 17.68% per annum.
13.8. (a) The required probability is the probability of the stock price being above $40 in six months’ time. Suppose that the stock price in six months is ST
ln ST ∼ φ
[
ln 38 +
]
i.e., ln ST ∼ φ(3. 687 , 0 .247) Since ln 40 = 3.689, the required probability is
1 − N
= 1 − N (0.008)
From normal distribution tables N (0.008) = 0.5032, so that the required probability is 0.4968. In general the required probability is N (d 2 ). (b) In this case, the required probability is the probability of the stock price being less than $40 in six months’ time. It is
1 − 0 .4968 = 0. 5032
(b) If: f = e−r(T^ −t)
[
ln S +
r −
σ^2 2
(T − t)
]
then ∂f ∂t
= re−r(T^ −t)
[
ln S +
r − σ^2 2
(T − t)
]
− e−r(T^ −t)
r − σ^2 2
∂f ∂S
e−r(T^ −t) S ∂^2 f ∂S^2
e−r(T^ −t) S^2 The left-hand side of the Black-Scholes-Merton differential equation is
e−r(T^ −t)
[
r ln S + r
r − σ^2 2
(T − t) −
r − σ^2 2
]
=re−r(T^ −t)
[
ln S +
r − σ^2 2
(T − t)
]
=rf Hence equation (13.16) is satisfied.
13.12. This problem is related to Problem 12.10.
(a) If G(S, t) = h(t, T )Sn, then ∂G/∂t = htSn, ∂G/∂S = hnSn−^1 , and ∂^2 G/∂S^2 = hn(n − 1)Sn−^2 , where ht = ∂h/∂t. Substituting into the B-S-M differential equa- tion, we obtain ht + rhn +
σ^2 hn(n − 1) = rh
(b) The derivative is worth Sn^ when t = T. The boundary condition for this differential equation is therefore h(T, T ) = 1. (c) The equation h(t, T ) = e[0.^5 σ (^2) n(n−1)+r(n−1)](T −t)
satisfies the boundary condition, since it collapses to h = 1 when t = T. It can also be shown that it satisfies the differential equation in (a). Alternatively we can also solve the differential equation in (a) directly. The differential equation can be written ht h = −r(n − 1) −
σ^2 n(n − 1) The solution to this is
ln h = [−r(n − 1) −
σ^2 n(n − 1)]t + k
where k is a constant. Since ln h = 0 when t = T , it follows that
k = [r(n − 1) +
σ^2 n(n − 1)]T
So that ln h = [−r(n − 1) +
σ^2 n(n − 1)](T − t) or h(t, T ) = e[0.^5 σ (^2) n(n−1)+r(n−1)](T −t)
13.13. In this case S 0 = 52, K = 50, r = 0.12, σ = 0.3 and T = 0.25.
d 1 =
ln(52/50) + (0.12 + 0. 32 /2)0. 25
- 3
d 2 = d 1 − 0. 3
The price of the European call is
52 N (0.5365) − 50 e−^0.^12 ×^0.^25 N (0.3865) = 52 × 0. 7042 − 50 e−^0.^03 × 0 .6504 = 5. 06
or $5.06.
13.14. In this case S 0 = 69, K = 70, r = 0.05, σ = 0.35 and T = 0.5.
d 1 = ln(69/70) + (0.05 + 0. 352 /2) × 0. 5
- 35
d 2 = d 1 − 0. 35
The price of the European put is
70 e−^0.^05 ×^0.^5 N (0.0809) − 69 N (− 0 .1666) = 70e−^0.^025 × 0. 5323 − 69 × 0 .4338 = 6. 40
or $6.40.
13.15. Using the notation in Section 13.12, D 1 = D 2 = 1, and
K[1 − e−r(t^2 −t^1 )] = 65[1 − e−^0.^1 ×^0.^25 ] = 1. 60 K[1 − e−r(T^ −t^2 )] = 65[1 − e−^0.^1 ×^0.^1667 ] = 1. 07
Since D 1 < K[1 − e−r(t^2 −t^1 )] and D 2 < K[1 − e−r(T^ −t^2 )] So it is never optimal to exercise the call option early. DerivaGem shows that the value of the option is 10.94.
(d) c = SN (d 1 ) − Ke−r(T^ −t)N (d 2 ) ∂c ∂t = SN ′(d 1 )
∂d 1 ∂t − rKe−r(T^ −t)N (d 2 ) − Ke−r(T^ −t)N ′(d 2 )
∂d 2 ∂t From (b): SN ′(d 1 ) = Ke−r(T^ −t)N ′(d 2 ) Hence ∂c ∂t
= −rKe−r(T^ −t)N (d 2 ) + SN ′(d 1 )
∂d 1 ∂t
∂d 2 ∂t
Since d 1 − d 2 = σ
T − t
∂d 1 ∂t
∂d 2 ∂t
∂t (σ
T − t)
= − σ 2
T − t Hence ∂c ∂t = −rKe−r(T^ −t)N (d 2 ) − SN ′(d 1 )
σ 2
T − t (e) From differentiating the Black-Scholes formula for a call price we obtain ∂c ∂S = N (d 1 ) + SN ′(d 1 )
∂d 1 ∂S − Ke−r(T^ −t)N (d 2 )
∂d 2 ∂S From the results in (b) and (c), it follows that ∂c ∂S
= N (d 1 )
(f) Differentiating the result in (e) and using the result in (c), we obtain ∂^2 c ∂S^2 = N ′(d 1 )
∂d 1 ∂S = N ′(d 1 )
Sσ
T − t From the result in (d) and (e) ∂c ∂t
σ^2 S^2 ∂^2 c ∂S^2 = − rKe−r(T^ −t)N (d 2 ) − SN ′(d 1 ) σ 2
T − t
σ^2 S^2 N ′(d 1 )
Sσ
T − t =r[SN (d 1 ) − Ke−r(T^ −t)N (d 2 )] =rc
This shows that the Black-Scholes formula for a call option does indeed satisfy the Black-Scholes differencial equation.
(g) If S ≥ K, then d 1 , d 2 → +∞ as t → T , N (d 1 ) = N (d 2 ) = 1. From (d), c = S − K. If S < K, then d 1 , d 2 → −∞ as t → T , N (d 1 ) = N (d 2 ) = 0. From (d), c = 0. Therefore c satisfies the boundary condition for a European call option, i.e., that c = max(S − K, 0) as t → T.
13.18. From the Black-Scholes equations
p + S 0 = Ke−rT^ N (−d 2 ) − S 0 N (−d 1 ) + S 0
Because 1 − N (−d 1 ) = N (d 1 ), this is
Ke−rT^ N (−d 2 ) + S 0 N (d 1 )
Also: c + Ke−rT^ = S 0 N (d 1 ) − Ke−rT^ N (d 2 ) + Ke−rT Because 1 − N (d 2 ) = N (−d 2 ), this is also
Ke−rT^ N (−d 2 ) + S 0 N (d 1 )
The Black-Scholes equations are therefore consistent with put-call parity.
13.19. This problem naturally leads on to the material in Chapter 18 on volatility smiles. Using DerivaGem we obtain the following table of implied volatilities: Maturity (months) Strike price ($) 3 6 12 45 37.78 34.99 34. 50 34.15 32.78 32. 55 31.98 30.77 30. The option prices are not exactly consistent with Black-Scholes. If they were, the implied volatilities would be all the same. We usually find in practice that low strike price options on a stock have significantly higher implied volatilities than high strike price options on the same stock.
13.20. Black’s approach in effct assumes that the holder of option must decide at time zero whether it is a European option maturing at time tn (the final ex-dividend date) or a European option maturing at time T. In fact the holder of the option has more flexibility than this. The holder can choose to exercise at time tn if the stock price at that time is above some level but not otherwise. Furthermore, if the option is not exercised at time tn, it can still be exercised at time T. It appears that Black’s approach understates the ture option value. This is because the holder of the option has more alternative strategies for decideing when to exercise the option than the two alternatives implicitly assumed by the approach. These alternatives add value to the option.
13.23. If f = S−^2 r/σ 2 , then ∂f ∂t
∂f ∂S
2 r σ^2
S−^
2 r σ^2 −^1
∂^2 f ∂S^2
2 r σ^2
2 r σ^2
S−^
2 r σ^2 −^2
∂f ∂t
∂f ∂S
σ^2 S^2
∂^2 f ∂S^2 = rS−^2 r/σ
2 = rf
This show that the Black-Scholes equation is satisfied. S−^2 r/σ 2 could therefore be the price of a traded security.
13.24. No. This is mainly because when markets are efficient the impact of dilution from executive stock options or warrants is reflected in the stock price as soon as they are announced and does not need to be taken into account again when the options are valued.
13.25. In this case, S 0 = K = 50, r = 0.05, σ = 0.25, T = 5, N = 10, 000 , 000 and M = 3, 000 , 000. From Section 13.10, the value of each employee option is the value of N/(N + M ) regular call options on the company’s stock, and the total cost is M times this. Thus applying Black-Scholes Formula,
d 1 =
ln(50/50) + (0.05 + 0. 252 /2)
- 25
d 2 = d 1 − 0. 25
The regular European call price is
c = 50N (0.7267) − 50 e−^0.^05 ×^5 N (0.1677), = 16. 2517.
Therefore the cost to the company of the employee stock option issue is
M ·
N
N + M
· c
=3, 000 , 000 ×
× 16. 2517 ,
or $37,503,923.08.
13.26. In this case, S 0 = 50, μ = 0.18, σ = 0.3. The probability distribution of the stock price in two years, ST , is lognormal and is, from equation (13.3), given by:
ln ST ∼ φ
[
ln 50 +
]
i.e., ln ST ∼ φ(4. 18 , 0 .42) The mean stock price is from equation (13.4)
50 e^0.^18 ×^2 = 50e^0.^36 = 71. 67
and the standard deviation is, from equation (13.5)
502 e^2 ×^0.^18
e^0.^09 ×^2 − 1 = 31. 83
95% confidence intervals for ln ST are
- 18 − 1. 96 × 0. 42 and 4 .18 + 1. 960. 42
i.e.,
- 35 and 5. 01 These correspond to 95% confidence limits for ST of
e^3.^35 and e^5.^01
i.e.,
- 52 and 150. 44
13.27. The calculation are shown in the table below
Σui = 0. 09471 Σu^2 i = 0. 01145
and an estimate of standard deviation of weekly return is : √
- 01145 13
14 × 13
The volatility per annum of 0. 0288
52 = 0.2079, or 20.79%. The standard error of this estimate is
- 2079 √ 2 × 14
or 3.93% per annum.
then ∂f ∂t
= −S^2 (r + σ^2 )e(r+σ (^2) )(T −t)
∂f ∂S = 2Se(r+σ (^2) )(T −t)
∂^2 f ∂S^2 = 2e(r+σ
(^2) )(T −t)
The left-hand side of the Black-Scholses differential equation is:
− S^2 (r + σ^2 )e(r+σ (^2) )(T −t)
- 2rS^2 e(r+σ (^2) )(T −t)
- σ^2 S^2 e(r+σ (^2) )(T −t)
=rS^2 e(r+σ
(^2) )(T −t)
=rf
Hence the Black-Scholses equation is satisfied.
13.29. In this case S 0 = 30, K = 29, r = 0.05, σ = 0.25 and T = 1/3.
d 1 = ln(30/29) + (0.05 + 0. 252 /2)/ 3
- 25
d 2 = d 1 − 0. 25
N (0.4225) = 0. 6637 N (0.2782) = 0. 6096
N (− 0 .4225) = 0. 3363 N (− 0 .2782) = 0. 3904
(a) The European call price is
30 × 0. 6637 − 29 e−^0.^05 /^3 × 0 .6096 = 2. 52
or $2.52. (b) The American call price is the same as the European call price. It is $2.52. (c) The European put price is
29 e−^0.^05 /^3 × 0. 3904 − 30 × 0 .3363 = 1. 05
or $1.05. (d) Put-call parity states that:
c + Ke−rT^ = p + S 0
In this case c = 2.52, S 0 = 30, K = 29, p = 1.05 and e−rT^ = 0.9835 and it is easy to verify that the relationship is satisfied.
13.30. (a) The present value of the dividend must be subtracted from the stock price. This gives a new stock price of:
30 − 0. 5 e−^0.^125 ×^0.^05 = 29. 5031
and
d 1 = ln(29. 5031 /29) + (0.05 + 0. 252 /2)/ 3
- 25
d 2 = d 1 − 0. 25
N (d 1 ) = 0. 6205 , N (d 2 ) = 0. 5645 The price of the option is thherefore
- 5031 × 0. 6205 − 29 e−^0.^05 /^3 × 0 .5645 = 2. 21
or $2.21. (b) Since N (−d 1 ) = 0.3795; N (−d 2 ) = 0. 4355 the value of the option when it is a European put is
29 e−^0.^05 /^3 × 0. 4355 − 29. 5031 × 0 .3795 = 1. 22
or $1.22. (c) If t 1 denotes the time when the dividend is paid:
K[1 − e−r(T^ −t^1 )] = 29(1 − e−^0.^05 ×^0.^2083 ) = 0. 3005
This is less than the dividend. Hence the option should be exercised immediately before the ex-dividend date for a sufficiently high value of the stock price.
13.31. We first value the option assuming that it is not exercised early, we set the time to maturity equal to 0.5. There is a dividend of 0.4 in 2 months and 5 months. Other parameters are S 0 = 18, K = 20, r = 10%, σ = 30%. DerivaGem gives the price as 0.7947. We next value the option assuming that it is exercised at the five-month point just before the final dividend. DerivaGem gives the price as 0.7668. The price given by Black’s approximation is therefore 0.7947. DerivaGem also shows that the correct American option price calculate with 100 time steps is 0.8243. It is never optimal to exercise the option immediately before the first ex-dividend date when D 1 ≤ K[1 − e−r(t^2 −t^1 )] Where D 1 is the size of the frist dividend, and t 1 and t 2 are the times of the first and second dividend respectively. Hence we must have:
D 1 ≤ 20[1 − e−^0.^1 ×^0.^25 ]
