Assume for simplicity that the expiration time of an option is $1$ the initial stock price is $1$ and there is no dividend yield and the risk free return is $0$.

How is it possible to show that the following holds for the implied volatility : $$ \sigma_{\operatorname{imp}}^2 = \operatorname{O}(\log K)$$

Where $K$ is the strike price of the option.

  • 1
    $\begingroup$ Does the Roger Lee reference mentioned in the first sentences of this paper google.be/url?sa=t&source=web&rct=j&url=https://… help? $\endgroup$
    – Quantuple
    Commented Aug 12, 2016 at 15:34
  • $\begingroup$ You need a jump-free distribution for the underlying I guess? $\endgroup$
    – Mats Lind
    Commented Aug 12, 2016 at 16:07

1 Answer 1



  • $T,K$ are the maturity and the strike of a vanilla call of price $C(T, K)$.

  • $(S_t)_{t\in [0,T]}$ the price process of the underlying.

  • $\mathbb{Q}$ is the risk neutral measure.

  • $x(T,K) = \log(K/S_0) - rT$ is the log-moneyness.

Fix a maturiy $T^* > 0$ and define :

$$C(K) := C^{BS}(T^*, K, \sigma_{imp}(K))$$

S.t the R.H.S is the Black-Schiles price of a vanilla call with strike $K$ and maturity $T^*$

If $\mathbb{E}_{\mathbb{Q}}\left(S_{T^*}^2\right) < \infty$, one can easily show that :

$$\forall \ K > 0 : \quad C(K) \leq e^{-rT^*} \frac{\mathbb{E}_{\mathbb{Q}}\left(S_{T^*}^2\right)}{4K}$$

We will prove the following inequality : $$ \exists \ \kappa > 0, \ \forall \ x > \kappa : \sigma_{imp}^2(x(K)) \leq 2\frac{x(K)}{T^*}\quad \quad (*)$$

First, notice that by using the bound on the call price we have : $$\lim_{x\to +\infty} C(K(x)) = 0 \quad \quad (1)$$

Knowing that $\sigma \mapsto C^{BS}(T^*, K, \sigma)$ is continuous and increasing, it suffices to show that : $$C^{BS}(T^*, K, \sigma_{imp}(x)) \leq C^{BS}(T^*, K, \sqrt{2\frac{x(K)}{T^*}})$$

Now, using l’hôpital’s rule we can prove that :

$$ \lim_{x\to +\infty} C^{BS}(T^*, K, \sqrt{2\frac{x(K)}{T^*}}) = \frac{S_0}{2}$$ So : $$ \exists \ A > 0, \ \forall \ x>A: \quad C^{BS}(T^*, K, \sqrt{2\frac{x(K)}{T^*}}) - \frac{S_0}{2} \geq - \frac{S_0}{4}$$ So that : $$ \exists \ A > 0, \ \forall \ x>A: \quad C^{BS}(T^*, K, \sqrt{2\frac{x(K)}{T^*}}) \geq \frac{S_0}{4}$$

By (1), the definition of the limit gives: $$\exists \ B > 0, \ \forall \ x>B: \quad C(K(x)) \leq \frac{S_0}{4}$$ Puting $\kappa := \max(A,B)$ completes the proof.

Now expressing (*) in the strike forme rather than log-moneyness proves that : $$ \sigma_{imp}^2(K)= \underset{K\to +\infty}{\mathcal{O}}(\log(K))$$


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