dstefan
07-08-2008, 02:12 PM
Chapter 5, Problem 7
Based on the linear Taylor expansion e^{-x} \approx 1 - x, new approximation formulas for the price of ATM options can be obtained by replacing r T and q T by 1 - e^{- r T} and 1 - e^{- q T}, respectively. The resulting formulas are
\label{approx_atm_call_nonzero_r_new}
C \approx \sigma S \sqrt{\frac{T}{2 \pi}} ~\frac{e^{-q T}+e^{-r T}}{2} ~+~ \frac{S ~(e^{- q T} - e^{- r T})}{2}
\label{approx_atm_put_nonzero_r_new}
P \approx \sigma S \sqrt{\frac{T}{2 \pi}} ~\frac{e^{-q T}+e^{-r T}}{2} ~-~ \frac{S ~(e^{- q T} - e^{- r T})}{2}
(i) Show that the Put--Call parity is satisfied by the approximations (\ref{approx_atm_call_nonzero_r_new}) and (\ref{approx_atm_put_nonzero_r_new}).
(ii) Estimate how good the new approximation (\ref{approx_atm_put_nonzero_r_new}) is, for an ATM put with S = 60, q=0.03, \sigma = 0.25, and T=1, if r=0.06, by computing the corresponding relative approximate error. Compare this error with the relative approximate error found in the previous exercise.
Solution:
(i) From (\ref{approx_atm_call_nonzero_r_new}) and (\ref{approx_atm_put_nonzero_r_new}), it is easy to see that
P - C + S e^{- q T} ~=~ - 2~\frac{S ~(e^{- q T} - e^{- r T})}{2} + S e^{- q T} ~=~ S e^{- r T} ~=~ K e^{- r T},
since K = S for ATM options.
(ii) Using the new approximation formula (\ref{approx_atm_put_nonzero_r_new}), we obtain that P_{approx new,r=0.06,q=0.03} ~=~ 4.861031. The Black-Scholes value of the put option is P_{BS,r=0.06,q=0.03} = 4.886985, and therefore
\frac{|P_{BS,r=0.06,q=0.03}-P_{approx new,r=0.06,q=0.03}|}{P_{BS,r=0.06,q=0.03}} ~=~ 0.005311 ~=~ 0.5311\%
Recall, from the previous exercise, that the approximation error corresponding to the original approximation formula is 1.4761\%. We conclude that, for this particular example, the new approximation formula is more accurate.
Based on the linear Taylor expansion e^{-x} \approx 1 - x, new approximation formulas for the price of ATM options can be obtained by replacing r T and q T by 1 - e^{- r T} and 1 - e^{- q T}, respectively. The resulting formulas are
\label{approx_atm_call_nonzero_r_new}
C \approx \sigma S \sqrt{\frac{T}{2 \pi}} ~\frac{e^{-q T}+e^{-r T}}{2} ~+~ \frac{S ~(e^{- q T} - e^{- r T})}{2}
\label{approx_atm_put_nonzero_r_new}
P \approx \sigma S \sqrt{\frac{T}{2 \pi}} ~\frac{e^{-q T}+e^{-r T}}{2} ~-~ \frac{S ~(e^{- q T} - e^{- r T})}{2}
(i) Show that the Put--Call parity is satisfied by the approximations (\ref{approx_atm_call_nonzero_r_new}) and (\ref{approx_atm_put_nonzero_r_new}).
(ii) Estimate how good the new approximation (\ref{approx_atm_put_nonzero_r_new}) is, for an ATM put with S = 60, q=0.03, \sigma = 0.25, and T=1, if r=0.06, by computing the corresponding relative approximate error. Compare this error with the relative approximate error found in the previous exercise.
Solution:
(i) From (\ref{approx_atm_call_nonzero_r_new}) and (\ref{approx_atm_put_nonzero_r_new}), it is easy to see that
P - C + S e^{- q T} ~=~ - 2~\frac{S ~(e^{- q T} - e^{- r T})}{2} + S e^{- q T} ~=~ S e^{- r T} ~=~ K e^{- r T},
since K = S for ATM options.
(ii) Using the new approximation formula (\ref{approx_atm_put_nonzero_r_new}), we obtain that P_{approx new,r=0.06,q=0.03} ~=~ 4.861031. The Black-Scholes value of the put option is P_{BS,r=0.06,q=0.03} = 4.886985, and therefore
\frac{|P_{BS,r=0.06,q=0.03}-P_{approx new,r=0.06,q=0.03}|}{P_{BS,r=0.06,q=0.03}} ~=~ 0.005311 ~=~ 0.5311\%
Recall, from the previous exercise, that the approximation error corresponding to the original approximation formula is 1.4761\%. We conclude that, for this particular example, the new approximation formula is more accurate.