Digital Call Vega and the Spot Threshold for Positive Volatility Sensitivity

Summary

The note derives when a cash-or-nothing digital call's price rises with volatility under Black–Scholes assumptions. It starts from the option's vega, whose sign is determined by the negative of d1 because the remaining density and discounting factors are positive. Thus positive sensitivity occurs when d1 is negative.

Rewriting that condition gives a threshold for the underlying spot relative to the strike, interest rate, volatility, and time to expiry. The argument is a sign check followed by algebra, rather than a numerical example or empirical result. It applies to the stated Black–Scholes digital call setup; it does not establish the same condition for other payoff structures or models. The question's initial expression has a typographical mismatch in its exponent, while the answer's derivation supplies the threshold it intends.

Key ideas

• The digital call's vega formula has a sign opposite to d1.
• The standard normal density and discount factor do not change the vega's sign.
• Positive volatility sensitivity occurs when d1 is below zero.
• The condition translates into a spot-price threshold involving strike, rate, volatility, and time to maturity.

Tags

# Answer by Magic is in the chain (score 9, accepted)


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

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

Can anyone give me some hints to proceed?

## Answer by Magic is in the chain (score 9, accepted)

https://quant.stackexchange.com/a/48819

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!

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.