In (ii)(c) of the question we are asked to calculate vega of the option. Could we still be asked to do this? Thank you
Can you explain how the vega of the call option is St*pdfN(d1)*sqrt(T-t) as when I try to apply the chain rule I end up with more terms than this? I have used the chain rule to say dct/dsigmat= St*pdfN(d1)*dd1/dsigma-Kexp(-r(T-t))pdfN(d2)*dd2/dsigma and since dd2/dsigma=dd1/dsigma-sqrt(T-t) we get dct/dsigmat= St*pdfN(d1)*dd1/dsigma-Kexp(-r(T-t))pdfN(d1-sigma*sqrt(T-t))*(dd1/dsigma-sqrt(T)) since we are at the money K=St and pdfN(d1)=1/sqrt(2pi)*exp(-0.5*d1^2) and pdfN(d2)=1/sqrt(2pi)*exp(-0.5*(d1-sigma*sqrt(T-t))^2)=1/sqrt(2pi)*exp(-0.5*(d1^2-2sigma*d1*sqrt(T-t)+sigma^2*(T-t))) I cannot see how we can simplify this down to St*pdfN(d1)*sqrt(T-t) Would the vega of the put option be St*pdfN(-d1)*sqrt(T) which by symmetry of the normal distribution will actually be the same as the vega of the call option? Could we be asked to calculate for the anything but the at the money option? I cannot see how we could do this for an option that was not at the money because presumably there is some neat cancelling which I have not spotted. If we could be asked can you explain how to get the vega of a non-ATM call option? Thank you
We know from part (ii)(a) that d1 = 0.15, and this will massively simplify pdfN(d1) to pdfN(0.15). Once you've made the simplification you'll see that there's no strict dependence on the option being at-the-money. Yes, the vega for a European put is the same as for the European call. There's a great list here.
