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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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  • $\begingroup$ My first stop is checking $Call(\cdot, \lambda \sigma^2_1 + (1 - \lambda) \sigma^2_2) \leq \lambda Call(\cdot, \sigma^2_1) + (1 - \lambda)Call(\cdot, \sigma^2_2)$. $\endgroup$
    – Richard Herron
    Commented Nov 18, 2011 at 11:25

3 Answers 3

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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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  • $\begingroup$ 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. $\endgroup$
    – TheBridge
    Commented Nov 18, 2011 at 15:27
  • $\begingroup$ The conclusion from this derivation is a little weak, though it does refute the question as posed. It only shows for each $\sigma$ there is some $K$ at which vega is decreasing for values greater than $\sigma$. The fact is BS is not convex with respect to volatility for any arbitrary given parameters, including strikes. Also , it is not necessary to use $\sigma_1$ in the derivation. One only needs to show decrement of vega for a $\sigma$ pair. $\endgroup$
    – Hans
    Commented Mar 21, 2014 at 19:54
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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.

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  • $\begingroup$ This reasoning is really vague and the conclusion wrong. $\endgroup$
    – Hans
    Commented Mar 21, 2014 at 19:03
  • $\begingroup$ The conclusion is wrong, because for large enough volatility, BS is concave not convex. The original question asks whether BS is always convex. The answer should be in the negative. $\endgroup$
    – Hans
    Commented Mar 21, 2014 at 19:53
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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.

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