Show that $\frac{\partial c(t))}{\partial \sigma^2 }>0 \text{ if and only if } S(t)<Xe^{-r(r+\frac{1}{2} \sigma^2 )(T-t)}.$

Question

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?

Answer 1

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!