6
$\begingroup$

Statement: if $c(t)$ is the price of the digital cash-or-nothing call option, then direct calculation (under Black-Scholes assumptions) shows that $$\frac{\partial c(t))}{\partial \sigma^2 }>0 \quad\text{if and only if}\quad S(t)<Xe^{-(r+\frac{1}{2} \sigma^2 )(T-t)}.$$

I fail to prove this statement (I do not even know how to start).

Can anyone give me some hints to proceed?

$\endgroup$
7
$\begingroup$

Hints:

You know the vega of a digital call option formula:

$V=-\frac{e^{-r(T-t)}}{\sigma} d_1 n\left(d_2\right)$

Where n is the standard normal density, which is positive. Sigma and exponential are also positive, so the sign of V is down to the sign of $d_1$. Which is negative when:

$d_1 <0$

$\ln \frac{S}{X}+\left(r+0.5\sigma^2\right)(T-t)<0$

$S<X e^{-\left(r+0.5\sigma^2\right)(T-t)}$

Hope this helps!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.