# [Final] Practice problems solutions (1) ```Solution to Final Exam
423-F17
1. In the following question, the market model is NOT necessarily the Black-Scholes model.
(a) A European call option with strike price K = \$20 and exercise date in six months is traded at
C E (0) = \$2. The current price of the underlying stock, which pays no dividends, is S(0) = \$25 and
the continuously compounded annual interest rate is r = 3%. Find an arbitrage opportunity.
Solution: Notice that S(0)−Ke−r/2 = 25−20e−0.015 = 5.30 &gt; 2 = C E (0). An arbitrage opportunity
is as follows:
At time t = 0:
1) short-sell 1 share of the stock (collect S(0));
2) deposit Ke−r/2 in the bank;
By doing so, we make a profit of S(0) − Ke−r/2 − C E (0) &gt; 0. In six months we can exit the short
position without investing any money, so we have an arbitrage.
At time t = 1/2:
1) withdraw K dollars from the bank;
2) exercise the European call and buy the stock at price K;
3) return the stock to the original owner, exiting the short position.
(b) Prove that an American call option on a stock which pays no dividends should never be exercised
before maturity.
Solution: Since the stock pays no dividends, we have that C A (t) = C E (t) for every t ∈ [0, T ]. If we
exercise the option at time t the payoff is (S(t) − K)+ . It is clear then that the option should not be
exercised when S(t) ≤ K. When S(t) &gt; K, we can do better by trading the option in the market,
since
C E (t) ≥ S(t) − Ke−r(T −t) ≥ S(t) − K.
2. In this problem, we consider a binomial tree model and r denotes the discrete time interest rate per
period.
(a) Prove that in the N -step binomial model, for every t ∈ {0, . . . , N − 1} the stock price satisfies
&quot;
#
∗ S(t + 1)
S(t) = E
S(t) .
1+r
Solution: Since p∗ =
&quot;
E∗
r−d
u−d ,
we have
#
S(t + 1)
(1 + u)S(t)
(1 + d)S(t)
S(t) ∗
S(t) = p∗
+(1−p∗ )
=
[p (1 + u) + (1 − p∗ )(1 + d)] = S(t).
1+r
1+r
1+r
1+r
Consider a two-step Binomial model. You are given the following incomplete tree, corresponding to a
European put option with strike price K = 65.
0
2.6545
?
14.6
17.09
35.6
Figure 1: European put with K = 65.
(b) Compute the discrete time interest rate per period r and the risk-neutral probability p∗ .
Hint: Write an appropriate system of equations and solve it.
Solution: The pricing formulas in the binomial tree give us
2.6545 =
14.6(1 − p∗ )
14.6p∗ + 35.6(1 − p∗ )
, 17.09 =
,
1+r
1+r
whose solution is p∗ = 0.8 and r = 0.1.
(c) Find the price of the put option at t = 0 and draw the complete binomial tree for the stock price.
Solution: The price of the European put at time 0 is given by
P E (0) =
2.6545p∗ + 17.09(1 − p∗ )
= 5.04.
1+r
The terminal values of the stock when there is at least one jump down can be deduced from the
payoff of the European put. We get that S ud = 50.4 and S dd = 29.4. Then, we can compute
Sd =
50.4p∗ + 29.4(1 − p∗ )
= 42.
1+r
Now, since S ud = (1 + u)S d and S dd = (1 + d)S d , we have u = 50.4
42 − 1 = 0.2 and d =
We can then complete the binomial tree for the stock price, which is the following
86.4
72
60
50.4
42
29.4
Figure 2: Stock price.
29.4
42
− 1 = −0.3.
3. A chooser option is a contract that allows its holder to decide at a later (fixed) time t1 whether the
option will be a call or a put. The value V (t1 ) of the chooser option at time t1 is the maximum between
the values of the call and put options, that is V (t1 ) = max{C E (t1 ), P E (t1 )}.
In this problem we assume that the choices for the put and call have the same maturity T and the same
strike price K.
(a) Prove that
V (t1 ) = C E (t1 ) + max{0, Ke−r(T −t1 ) − S(t1 )}.
Solution: From the put-call parity we have
V (t1 ) = max{C E (t1 ), C E (t1 ) + Ke−r(T −t1 ) − S(t1 )} = C E (t1 ) + max{0, Ke−r(T −t1 ) − S(t1 )}.
(b) Prove that the price of chooser option at time 0 must be given by V (0) = C E (0) + PbE (0), where
C E denotes a European call option with maturity T and strike price K and PbE denotes a European
put option with maturity t1 and strike price Ke−r(T −t1 ) .
Solution: The term max{0, Ke−r(T −t1 ) − S(t1 )} = (Ke−r(T −t1 ) − S(t1 ))+ corresponds to the payoff
of a European put option with maturity t1 and strike price Ke−r(T −t1 ) . Thus, we can replicate the
chooser option buying one share of this put and the European call option with maturity T and strike
price K. By the non-arbitrage principle, V (0) must be equal to C E (0) + PbE (0).
(c) Consider a Black-Scholes model with r = 0.1, σ = 0.3, S(0) = 80, and no dividends. Compute the
price of the chooser option, for which the holder can choose at time t1 = 0.25 years whether to hold
the call or put option with maturity T = 0.5 years and strike price K = 85.
Solution: First, we find the price of the European call with maturity T and strike price K. We
have
√
S(0)
1
1 d1 = √
ln
+ (r + σ 2 )T = 0.06, d2 = d1 − σ T = −0.16,
K
2
σ T
and therefore C E (0) = S(0)Φ(d1 ) − Ke−r T Φ(d2 ) = 6.62.
For the European put with maturity t1 and strike price Ke−r(T −t1 ) we have
√
1 2 1 S(0)
b
d1 = √
ln
+
(r
+
σ )t1 = 0, db2 = db1 − σ t1 = −0.15,
−r(T
−t
)
1
2
σ t1
Ke
and PbE (0) = Ke−r(T −t1 ) Φ(−db2 )e−rt1 − S(0)Φ(−db1 ) = 5.25.
We have finally that V (0) = 6.62 + 5.25 = 11.87.
4. Consider a stock in the Black-Scholes model with volatility σ = 20%, initial price S(0) = \$100 and
suppose that the continuously compounded interest rate is r = 10%. You sell 300 six-months European
call options on the stock with strike price K = \$102.
(a) Compute the delta and the price of the call option.
Solution: We can compute d1 (0) = 0.28 and d2 (0) = 0.14. Then we have from the normal table
∆C E = Φ(d1 (0)) = 0.6103,
C E (0) = S(0)Φ(d1 (0)) − Ke−rT Φ(d2 (0)) = 7.1120.
(b) Find the zero-valued delta-neutral portfolio.
Solution: In this case the delta-neutral portfolio contains z = −300 calls.
Since 0 = ∆V = x + z∆C E , we have x = −z∆C E = 300 &times; (0.6103) = 183.09 shares of the stock.
Finally, since 0 = V (0) = xS(0)+y+zC E (0), the risk-free position is given by y = −xS(0)−zC E (0) =
−16174.23.
(c) The price of the stock after 5 days is \$98. Compare the values of the zero-valued delta-neutral and
the zero-valued unhedged portfolios. Assume you don’t make changes to both portfolios in these 5
days.
Solution: The unhedged portfolio U contains x = 0 shares of the stock, z = −300 calls and a
risk-free position of y = 300 &times; 7.1120 = 2133.60.
Let t = 5/356. We can compute
d1
5
365
= 0.13, d2
5
365
= −0.01,
and from the Black-Scholes formula we get CE (t) = 5.8561.
We compute the values of the delta-neutral and the unhedged portfolios
0.1&times;5
5
5
5
E
365
= −16174.23e
+ 183.09S
− 300C
= −11.56,
V
365
365
365
0.1&times;5
5
5
E
U
= 2133.60e 365 − 300C
= 379.69.
365
365
(d) Find the zero-valued delta-vega neutral portfolio by using an additional European put with expiration
in six months and strike price \$105. Show that the portfolio is also delta-gamma neutral.
Solution: For this European call option, we have calculated in part (a) that ∆C E = 0.6103, C E (0) =
7.1120. The gamma and vega of the call option are ΓC E = 0.0271 and VC E = 27.1251.
b = 105 is different and T = 0.5. We can compute then
For the European put option PbE , note that K
b
b
d1 (0) = 0.08 and d2 (0) = −0.06 which leads to PbE (0) = 5.5170, ∆PbE = −0.4681, ΓPbE = 0.0281 and
VPbE = 28.1194.
The delta-vega neutral portfolio contains z1 = −300 calls.
V
Since 0 = VV = z1 VC E + z2 VPbE , it contains z2 = −z1 VCbE = 289.39 put options.
PE
Since 0 = ∆V = x + z1 ∆C E + z2 ∆PbE , it contains x = −z1 ∆C E − z2 ∆PbE = 318.55 shares of the
stocks.
Since 0 = V (0) = xS(0) + y + z1 C E (0) + z2 PbE (0), the risk-free position is given by y = −xS(0) −
z1 C E (0) − z2 PbE (0) = −31317.79.
Finally, we compute gamma of the portfolio
ΓV = z1 ΓC E + z2 ΓPbE = −300 &times; 0.0271 + 289.39 &times; 0.0281 = 0.
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