dstefan
06-10-2008, 04:18 PM
Chapter 3, Problem 10
Show that an ATM call on an underlying asset paying dividends continuously at rate q is worth more than an ATM put with the same maturity if and only if q \leq r, where r is the constant risk free rate. Use the Put-Call parity, and then use the Black-Scholes formula to prove this result.
Solution:
For at-the-money options, i.e., with S=K, the Put-Call parity can be written as
C - P = S e^{-q(T-t)} - K e^{-r(T-t)} ~=~ K e^{-q(T-t)} - K e^{-r(T-t)} = K e^{-r(T-t)} ~\left( e^{(r-q)(T-t)} - 1 \right)
Therefore, C \geq P if and only if e^{(r-q)(T-t)} \geq 1, which is equivalent to r \geq q.
Alternatively, the Black-Scholes formulas for at-the-money options can be written as
C = K e^{-q(T-t)} N( d_1) - K e^{-r(T-t)} N( d_2); P = K e^{-r(T-t)} N(-d_2) - K e^{-q(T-t)} N(-d_1)
where
d_1 = \left( \frac{r-q}{\sigma} + \frac{\sigma}{2} \right) \sqrt{T-t}; d_2 = \left( \frac{r-q}{\sigma} - \frac{\sigma}{2} \right) \sqrt{T-t}
Then
C \geq P \Longleftrightarrow e^{-q(T-t)} ( N( d_1) + N(-d_1) ) \geq e^{-r(T-t)} ( N( d_2) + N(-d_2) )
\Longleftrightarrow e^{-q(T-t)} \geq e^{-r(T-t)} ~\Longleftrightarrow~ r \geq q
since N( d_1) + N(-d_1) = N( d_2) + N(-d_2) = 1.
Show that an ATM call on an underlying asset paying dividends continuously at rate q is worth more than an ATM put with the same maturity if and only if q \leq r, where r is the constant risk free rate. Use the Put-Call parity, and then use the Black-Scholes formula to prove this result.
Solution:
For at-the-money options, i.e., with S=K, the Put-Call parity can be written as
C - P = S e^{-q(T-t)} - K e^{-r(T-t)} ~=~ K e^{-q(T-t)} - K e^{-r(T-t)} = K e^{-r(T-t)} ~\left( e^{(r-q)(T-t)} - 1 \right)
Therefore, C \geq P if and only if e^{(r-q)(T-t)} \geq 1, which is equivalent to r \geq q.
Alternatively, the Black-Scholes formulas for at-the-money options can be written as
C = K e^{-q(T-t)} N( d_1) - K e^{-r(T-t)} N( d_2); P = K e^{-r(T-t)} N(-d_2) - K e^{-q(T-t)} N(-d_1)
where
d_1 = \left( \frac{r-q}{\sigma} + \frac{\sigma}{2} \right) \sqrt{T-t}; d_2 = \left( \frac{r-q}{\sigma} - \frac{\sigma}{2} \right) \sqrt{T-t}
Then
C \geq P \Longleftrightarrow e^{-q(T-t)} ( N( d_1) + N(-d_1) ) \geq e^{-r(T-t)} ( N( d_2) + N(-d_2) )
\Longleftrightarrow e^{-q(T-t)} \geq e^{-r(T-t)} ~\Longleftrightarrow~ r \geq q
since N( d_1) + N(-d_1) = N( d_2) + N(-d_2) = 1.