Convexity of BS Equation for Call and Put

Question

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 1

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 2

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.