Question

Suppose that we buy an asymmetric butterfly on a stock using
calls with strike prices 40, 47.50, and 60. We denote the price of the call option with strike
price K by C(K). We assume that we can trade options for a number of shares that can be
any positive integer (and not necessarily a multiple of 100). Suppose that C(40) = 11.45,
C(47.50) = 7.00, and C(60) = 2.08.
Let the number of options used be N(K) for the strike price K. We assume that each option
is for one share; we completely ignore the comcept of a contract in this problem. We take
N(K) to be a positive integer no matter whether the options are bought or sold.
(a) Are the options for strike price 40 bought or sold ?
(b) Are the options for strike price 47.50 bought or sold ?
(c) Are the options for strike price 60 bought or sold ?
(d) Suppose that each N(K) is a positive integer and that we choose the numbers
N(K) to be as small as possible. Determine N(40), N(47.50), and N(60). There is only one
correct answer.
(e) What is the total (net) amount that we pay for this butterfly ?
(f) If T = 1/2 (6 months), if the risk-free interest rate r = 0.05, and if the stock
price ST = 58.50 at expiration, what is our profit for this transaction (expressed as a real
number, which is negative if, and only if, we have a net loss) ? We use the number of shares
given in part (d) above.

Answer 1

K1 = 40, K2 = 47.50, and K3 = 60

(a) The options for strike price 40 will be bought.

(b) The options for strike price 47.50 will be sold.

(c) The options for strike price 60 will be bought.

(d) N(40) = (K3 - K2) / (K3 - K1) = (60 - 47.50) / (60 - 40) =  0.6250 ; since we want it to be an integer, we multiply this by 8. Hence, N(40) = 8 x 0.6250 = 5

N(47.50) = 8

N(60) = 8 x (1 - 0.6250) = 3

(e) The total (net) amount that we pay for this butterfly = N(40) x C(40) - N(47.50) x C(47.50) + N(60) x C(60) = 5 x 11.45 - 8 x 7.00 + 3 x 2.08 = 7.49

(f) Payoff from the position = N(40) x max (S - K1, 0) - N(47.50) x max (S - K2, 0) + N(60) x max (S - K3, 0) = 5 x max (58.50 - 40, 0) - 8 x max (58.50 - 47.50, 0) + 3 x max (58.50 - 60, 0) =  4.50

Our profit = 4.50 - 7.49 x (1 + r x t) = 4.50 - 7.49 x (1 + 0.05 x 1/2) = - 3.18

(Negative sign means there is a loss of 3.18)