Skip to content
All library documents

Black–Scholes Option Convexity in Implied Volatility

Article Quant Q&A · Author: TheBridge

Summary

The document asks whether Black–Scholes call and put prices are convex as functions of implied volatility. The answers explain that convexity can vary with moneyness rather than holding uniformly. Vega, the derivative of option value with respect to volatility, is shaped by the normal density evaluated at the option’s d1 value. Since that density peaks at zero, vega can have a maximum as volatility varies, which implies concavity of option value over the relevant interval for a strike chosen to place d1 at zero.

A separate answer gives intuition for out-of-the-money options: at very low volatility their value and vega are small, while increasing volatility can first increase vega, producing convexity in that region. Another observation is that option values increase with volatility but remain bounded, so they cannot be globally convex over the full range. Together, the responses caution against assigning one curvature property to every strike and volatility range; they provide arguments and intuition, not a complete classification for all parameter combinations.

Key ideas

  • Black–Scholes option value does not have one universal convexity pattern across strikes and volatility levels.
  • Vega depends on the normal density at d1, which reaches its maximum when d1 is zero.
  • For a strike associated with that vega maximum, the response uses the shape of vega to argue for concavity in volatility.
  • Out-of-the-money options can show convexity when vega rises from a very small value as volatility increases.
  • Because option values are increasing yet bounded in volatility, they cannot be globally convex over the entire range.

Tags

Full text
# Convexity of BS Equation for Call and Put


# Convexity of BS Equation for Call and Put












I have a simple question.

Is the Black-Scholes Formula convex with respect to Implied volatility parameter $\sigma$ (for calls or put) ?

When I say Black-Scholes I mean for a call the following one (on Forward price $F_t$):

$$Call (F_t,T-t, K, \sigma^2) = F_t.N(d_1) - K.e^{-r.(T-t)}.N(d_2)$$

$$d_1=\frac{Ln(F_t/K)+1/2.\sigma^2.(T-t)}{\sigma.\sqrt{T-t}}$$ $$d_2=d_1 - \sigma.\sqrt{T-t}$$

and for a put

$$Put (F_t,T-t, K, \sigma^2) = K.e^{-r.(T-t)}.N(-d_2)-F_t.N(-d_1) $$

PS: I know the answer is no but is there a fancy way to prove this (i.e. no brutal force differentiation of the vega)

## Answer by Brian B (score 8, accepted)

https://quant.stackexchange.com/a/2403

Sure. The formula for vega (you probably recall) is

$$ v(\sigma) = S n( d_1(\sigma) )\sqrt{T-t} $$

The gaussian PDF, $n(\cdot)$, is strictly non-convex, having a local maximum at zero. There is therefore a corresponding maximum of vega occurring where the strike $K_\text{max}$ solves $$ d_1(\sigma)=0 $$ which works out to $$ K_\text{max} = S \exp((r-q-\frac12\sigma^2)\sqrt{T-t}). $$

Therefore, for this strike we have for any $\sigma_1,\sigma_2$ such that $\sigma_1<\sigma<\sigma_2$, that $$ d_1(\sigma_1)<0=d_1(\sigma)<d_1(\sigma_2) $$ and since $0$ is the argmax of $n(\cdot)$ $$ n(d_1(\sigma_{1,2}))<n(d_1(\sigma))=1. $$

It follows that for any $\lambda \in [0,1]$ $$ \lambda v(\sigma_{1})+(1-\lambda) v(\sigma_{2})<v(\sigma) $$ proving concavity of vega.

## Answer by joelhoro (score 1)

https://quant.stackexchange.com/a/3712

The vega is quite linear for ATM options. It's convex mostly for OTM and ITM.

An intuitive explanation is that an OTM option with zero volatility will be worth zero. If you increase the volatility by 1% then most likely the price is still close to zero. Therefore the vega is zero (or tiny).

Now if you increase the volatility sufficiently, clearly at some point the option is going to have a reasonable value, and a positive vega. Therefore as you increased volatility, vega increased (from zero to something), which shows the convexity.

## Answer by Hans (score 1)

https://quant.stackexchange.com/a/10657

BS is increasing with respect to volatility, and bounded from above, i.e. the call by $F$ and put by $K$, as volatility goes to infinity. So it can not be convex with respect to volatility.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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