Solutions
Math 441 Spring 2015
Second Midterm
(Practice)
Problem 1. Describe the Black-Scholes formula: explain to what type of financial derivative(s) it applies, and
what it says about their initial value.
Solution. The Black-Scholes formula determines the price of a European put or call, assuming
the underlying stock is given by the log-normal model. For a European call, the price is
cE = S0 (d1 )
In the above formula
i.e.
Ke
rT
(d2 )
(z) denotes the cumulative distribution function for the standard normal,
(z) =
Z
z
1
1
p e
2⇡
x2 /2
dx
and
d1 =
ln(S0 /K) + (r +
p
T
1 2
)T
2
,
d2 =
ln(S0 /K) + (r
p
T
1 2
)T
2
and S0 = current price of the stock, = volatility, r = zero risk rate, T = time to maturity of the
option and K = strike price. The corresponding formula for a European put is
pE = K ( d2 )e
rT
S0 ( d1 ).
Problem 2. The price of a stock today is $45. Assume that every 6 months the stock goes up by 10% or down
by 10%. Assume the zero risk rate is 5% (continuous compounding). Compute the price of the
following derivative:
8
American put T = 1 year , K = $48
< 1 (long)
5 (long)
Stock
:
1 (long)
European put T = 1 year , K = $45
Solution. First, let us compute the values of the stock at all nodes.
Su = 45(1.1) = 49.5
Sd = 45(0.9) = 40.5
Suu = 45(1.1)2 = 54.45
Sud = 45(1.1)(0.9) = 44.55
Sdd = 45(0.9)2 = 36.45
Moreover,
p=
er T d
e0.025 0.9
=
= 0.6265756
u d
1.1 0.9
Use this compute the pay o↵ functions at maturity. Let h denote the European put and H the
American put. Then, for the American put,
Huu = max{48
54.45, 0} = 0
Hud = max{48
44.55, 0} = 3.45
Hdd = max{48
36.45, 0} = 11.55,
huu = max{45
54.45, 0} = 0
hud = max{45
44.55, 0} = 0.45
hdd = max{45
36.45, 0} = 8.55.
and the European put
Then, using the binomial formula for the European put,
h0 = e
(0.05)
(0.6265756)2 (0) + 2(0.6265756)(1
=e
(0.05)
2(0.6265756)(0.3734244)(0.45) + (0.3734244)2 (8.55)
0.6265756)(0.45) + (1
0.6265756)2 (8.55)
⇡ (0.9512)(1.4028)
⇡ 1.3343
Now, we go node by node for the values of the American option, noting that the time step is 1/2
when using the 1-step formula
Hu = max{e
(0.05)/2
(pHuu + (1
p)Hud ), 48
Su }
= max{(0.9753)((0.6265756)(0) + (0.3734244)(3.45)), 1.5}
= max{1.4028, 1.5} = 1.4028
Hd = max{e
(0.05)/2
(pHud + (1
p)Hdd ), 48
Sd }
= max{(0.9753)((0.6265756)(3.45) + (0.3734244)(11.55)), 7.5}
= max{6.4747, 7.5} = 7.5.
H0 = max{e
(0.05)/2
(pHu + (1
p)Hd ), 48
45}
= max{(0.9753)((0.6265756)(1.4028) + (0.3734244)(7.5)), 3}
= max{3.6578, 3} = 3.6578.
Then, the prizes are H0 = 3.6578$, h0 = 1.3343$, and the price of the original portfolio is
3.6578 + 5S0 + 0.6407 = 3.6578 + 5(45) + 1.3343
= 3.6578 + 225 + 1.3343
= 229.99$.
Problem 3. Consider a stock which goes up by 5% and down by 5% every 4 months. Assuming the zero
risk rate is 6%, determine the current price of a butterfly spread comprised of European options
expering in 1 year, where
K1 = 10$ , K2 = 15$ , K3 = 20$.
Problem 4. A stock is trading today at $30. Suppose that in two months, the stock will be either $36 or $24.
Using a one-step binomial tree, value the following options which expire in two months:
(a) A European call option with K = 25, with a risk-free rate of r = 4%.
(b) A European call option with K = 20, with a risk-free rat of r = 4%.
(d) Repeat the above but now with European puts with the same K and same risk-free rates.
Problem 5. Determine the current price of a portfolio comprised of one (long) straddle with K = 20$ and
one (short) straddle with K = 40$, both with expiration T = 9 months. Assume that the zero
risk rate is 8% and that the underlying stock for both straddles is modelled by a log-normal with
annual volatility of 20% and S0 = 30$.
Solution. Let h denote the straddle at K = 20 and h0 denote the straddle at K = 40.
h00
⇧ = h0
So all we need to do is to find the value of each of the straddles. Since a straddle is just a long
position on a put and a long position on a call, the price of a straddle is given by adding up the
Black-Scholes formula for a put and a call, so
h0 = cE + pE = S0 (d1 )
Ke
rT
(d2 ) + K ( d2 )e
rT
S0 ( d 1 )
Simplifying (will save us time later when substituting)
h0 = S0 ( (d1 )
( d1 ))
Ke
rT
( (d2 )
( d2 ))
Substituting the parameters for the first straddle, we get
d1 =
d2 =
ln(30/20) + (0.08 + 12 (0.2)2 )(9/12)
p
⇡ 4.8046
(0.2) 9/12
ln(30/20) + (0.08 12 (0.2)2 )(9/12)
p
⇡ 4.5046
(0.2) 9/12
Substituting further (and using a calculator/Z-table to estimate
h0 = 30 [ (d1 )
⇡ 30(0.9999
⇡ 10.3969
( d1 )]
0)
20e
(0.08)(9/12)
20(0.9801)(0.9999
[ (d2 )
(·)),
( d2 )]
0)
Now we do the same for h0 , adding up again the Black-Scholes formulas for a call and a put
h00 = S0 ( (d1 )
( d1 ))
Ke
rT
( (d2 )
( d2 ))
where this time K = 40, so
d1 =
d2 =
and, just as before
ln(30/40) + (0.08 + 12 (0.2)2 )(9/12)
p
⇡
(0.2) 9/12
ln(30/40) + (0.08 12 (0.2)2 )(9/12)
p
⇡
(0.2) 9/12
h00 = 30 [ (d1 )
In conclusion,
2.1268
( d1 )]
40e
(0.08)(9/12)
[ (d2 )
⇡ 30(0.0167 0.9832) 40(0.9801)(0.007
⇡ 28.99 + 38.62 = 9.63
⇧ = 10.3969
2.4268
9.63 = 0.7669$.
( d2 )]
0.9923)
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