# 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)

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My first stop is checking $Call(\cdot, \lambda \sigma^2_1 + (1 - \lambda) \sigma^2_2) \leq \lambda Call(\cdot, \sigma^2_1) + (1 - \lambda)Call(\cdot, \sigma^2_2)$. –  Richard Herron Nov 18 '11 at 11:25

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.

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What you have shown is that for any $\sigma$ there's a strike $K_{max}$ at which BS Formula is not convex (treating at the same time Call and Put case). That's nice and smart I accept this answer. Thx. –  TheBridge Nov 18 '11 at 15:27