# 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.

Any comments and advice would be greatly appreciated! Thank you for your time and help!

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 '20 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 '20 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 '20 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 '20 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 '20 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 '20 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 '20 at 0:52