Hi All,
5.4. Practice question:
d1 = [LN(50/50) + (4% - 2% + 18%^2/2)*0.5] / [18%*SQRT(0.50)] = 0.1422; d2 = d1 - 18%*SQRT(0.50) = 0.0149 N(d1) = 0.5565 and N(d2) = 0.5060 c = S*exp(-qT)*N(d1) - K*exp(-rT)*N(d2) c = $49.502*0.5565 - $50*exp(-4%*6/12)*0.5060 = $2.753
I calculated this question similar to the example in Hull (Ch.14, call option with continuous div yield).
First I adjusted the stock price with the dividend yield: Sadj=S*exp(-0.02*0.25)=49.50
than recalculated the d1 with the adjusted S (using only r(f) as drift to avoid 'double counting'):
d1 = [LN(49.50/50) + (4% + 18%^2/2)*0.5] / [18%*SQRT(0.50)]
c=49.50*N(d1)- K*exp(-rT)*N(d2)
resulting only a small difference: C=2.419
My question is theoretical:
do the two methods have the same meaning (mathematical or rational) thus can I use the above calculation technique whenever exam tests BSM with dividend yield?
Thanks for your clarification!
5.4. Practice question:
d1 = [LN(50/50) + (4% - 2% + 18%^2/2)*0.5] / [18%*SQRT(0.50)] = 0.1422; d2 = d1 - 18%*SQRT(0.50) = 0.0149 N(d1) = 0.5565 and N(d2) = 0.5060 c = S*exp(-qT)*N(d1) - K*exp(-rT)*N(d2) c = $49.502*0.5565 - $50*exp(-4%*6/12)*0.5060 = $2.753
I calculated this question similar to the example in Hull (Ch.14, call option with continuous div yield).
First I adjusted the stock price with the dividend yield: Sadj=S*exp(-0.02*0.25)=49.50
than recalculated the d1 with the adjusted S (using only r(f) as drift to avoid 'double counting'):
d1 = [LN(49.50/50) + (4% + 18%^2/2)*0.5] / [18%*SQRT(0.50)]
c=49.50*N(d1)- K*exp(-rT)*N(d2)
resulting only a small difference: C=2.419
My question is theoretical:
do the two methods have the same meaning (mathematical or rational) thus can I use the above calculation technique whenever exam tests BSM with dividend yield?
Thanks for your clarification!
