# Strike Arbitrage

In Stochastic Volatility Modelling, Chapter 2, the author derived the Dupire equation $$\mathbb{E}[\sigma_T^2|S_T = K] = 2\frac{\frac{dC}{dT} + qC +(r-q)K\frac{dC}{dK}}{K^2 \frac{d^2C}{dK^2}}.$$

The author discussed its denominator: he linked the denominator to a butterfly strategy. Then I cannot understand the following parts:

1. Options’ markets are arbitraged well enough that butterfly spreads do not have negative prices:3 the denominator in the Dupire formula is positive.

2. In a model, $$\frac{d^2C(K,T)}{dK^2} = e^{-rT}\mathbb{E}[\delta(S_T - K)]$$, where $$\delta(\cdot)$$ denotes the Dirac delta function. The condition $$\frac{d^2C(K,T)}{dK^2} > 0$$ is equivalent to requiring that the market implied density (what's implied density?) be positive.

For the second question:

The implied density is the density function we integrate call payoffs against to match market call prices, denoted $$f$$ here.

So, ignoring discount factors, the answer comes from Dirac delta function's properties:

$$\mathbf{E}[\delta(S-K)] = \int_{-\infty}^\infty \delta(S-K)f(S) dS = f(K)$$

Alternatively:

$$C = C(K) = \int_K^\infty (S-K)f(S) dS$$

$$\frac{\partial C}{\partial K} = \frac{\partial }{\partial K} \left(\int_K^\infty (S-K)f(S) dS \right)$$

$$= \int_K^\infty \frac{\partial }{\partial K}\left((S-K)f(S)\right)dS - (K-K)f(K)$$

$$= - \int_K^\infty f(S)dS$$

Then:

$$\frac{\partial^2 C}{\partial K^2} = \frac{\partial C}{\partial K} \left( - \int_K^\infty f(S)dS\right) = \left[-f(S)\right]\bigl\vert_{S=K}^{S=\infty} = f(K)$$

• Thanks, ir7. I see that it is equivalent to the implied density function f(K) > 0. But I have another question: it is called implied density. Like implied vol comes from the option price, what is this density implied from? – JuniorQuant Jul 31 at 2:07
• In equation $C(K) = \int_K^\infty (S-K)f(S) dS$, the market gives us $C(K)$. The $f(S)$ that satisfies the equation is called the market implied density function. That's its definition. But now we just proved that if the market gives as a continuum of call prices $C(K)$ for all strikes, then we simply compute it's second derivative wrt to $K$ (perhaps using finite difference approximations if the call price is not an analytic function of $K$) to get the density function. – ir7 Jul 31 at 2:15
• Now, the call price surface needs to be free of arbitrage (and this gives the connection with RN, risk neutral world). See Carr and Madan's A note on sufficient conditions for no arbitrage which shows that a grid of European call quotes is free of static arbitrage if all adjacent vertical spreads, butterfly spreads, and calendar spreads are nonnegatively priced. Ropper's Arbitrage Free Implied Volatility Surfaces includes a comprehensive framework for no-static arbitrage in a volatility surface/call price surface and provides sufficient and necessary conditions – ir7 Jul 31 at 2:26
• you're a lifesaver! I want to find a book and systematically learn this staff. Do you have any recommendations? Thank you for your time and help! – JuniorQuant Jul 31 at 2:36

If $${\frac {\partial^2 C} {\partial K ^2}}$$ was zero, then the price-strike curve would just be a straight sloping-downwards line, and it would cost the same to buy either two call options at strike $$K$$ (portfolio A), or one option each at strike $$K-1$$ and strike $$K+1$$ (portfolio B).

If you think about the payoffs at expiry where spot=$$S_t$$ of these two portfolios, you'll see that they are the same for $$S_t < K-1$$ (ie. both payoffs are 0) and the same for $$S_t > K+1$$ (ie. both payoffs are $$2(S_t - K)$$). BUT, between $$K-1 < S_t < K+1$$, portfolio B always pays more, because the option with strike $$K-1$$ is in the money earliest.

I've shown some graphs of this below. What this means is that portfolio B must cost more than portfolio A in a fairly priced market, and if you think about the shape of the price vs. strike curve, it means it must be concave (ie. $${\frac {\partial^2 C} {\partial K ^2}} > 0$$). • Oh, I see. So if we ignore the price of options, the butterfly strat basically locks a range to gain the profit. Thanks! – JuniorQuant Jul 30 at 4:07
• Hi StackG, I think it should be convex, right? But how can this be related to the implied density? Actually, what's the so-called implied density? There is no definition in the book. – JuniorQuant Jul 30 at 19:12
• Great answer below by ir7. It's worth mentioning that this is actually only the implied distribution in the RN measure, not the real-world measure, although I have seen many practitioners who are happy to use it as a real-world approximation too... – StackG Jul 31 at 0:52