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

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)

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

Can anyone give me some hints to proceed?

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

Hope this helps!