Risk-Neutral Density, Option Gamma, and Model Assumptions

# Answer by Kevin (score 4, accepted)


# Does the risk neutral pdf that is derived using Litzenberger-Breeden Method correspond to gamma and it's integral correspond to delta?












I derived the pdf using the butterfly prices and the curve looks like gamma of an option at every strike. Is that the case or am I missing something to get the pricing of an option?

## Answer by Kevin (score 4, accepted)

https://quant.stackexchange.com/a/73580

In many option pricing models, the option price is homogeneous of degree one, $$C(\lambda S_t,\lambda K)=\lambda C(S_t,K).$$ This property applies to most stochastic volatility and exponential Lévy models. One big exception are local volatility models. Essentially, the condition holds if doubling $S_t$ also doubles $S_T$ (because, for example, $S_T=S_te^{X_{T-t}}$). Then, \begin{align} C(\lambda S_t,\lambda K)&=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t[\max\{\lambda S_T-\lambda K,0\}] \\ &=\lambda e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t[\max\{S_T-K,0\}]\\ &=\lambda C(S_t,K). \end{align}

Differentiating with respect to $\lambda$ implies \begin{align*} S_t\frac{\partial C}{\partial S_t}+K\frac{\partial C}{\partial K}=C. \end{align*} This resembles Euler's homogeneous function theorem. Due to monotonicity in strike, we know that $\frac{\partial C}{\partial K}<0$.

Further differentiation with respect to $S_t$ and $K$ implies \begin{align*} S_t^2\frac{\partial^2 C}{\partial S_t^2}=K^2\frac{\partial^2 C}{\partial K^2}. \end{align*} Due to convexity in strike, we know that $\frac{\partial^2 C}{\partial K^2}>0$.

From Breeden-Litzenberger (1978), \begin{align*} q(K) &= e^{rT}\frac{\partial^2 C}{\partial K^2}\\ &= e^{rT}\frac{S_t^2}{K^2}\frac{\partial^2 C}{\partial S^2} \\ &= e^{rT}\frac{S_t^2}{K^2}\Gamma. \end{align*} You can thus see that the risk-neutral density is indeed closely linked to gamma. You're right.

Note that the equations $S_t^2\frac{\partial^2 C}{\partial S_t^2}=K^2\frac{\partial^2 C}{\partial K^2}$ and $S_t\frac{\partial C}{\partial S_t}+K\frac{\partial C}{\partial K}=C$ can be used to calculate gamma and delta in an (almost) model-free fashion, namely \begin{align} \Delta &=\frac{C}{S_t}-\frac{K}{S_t}\frac{\partial C}{\partial K}, \\ \Gamma &=\frac{K^2}{S^2_t}\frac{\partial^2 C}{\partial K^2}. \end{align}

## Answer by Andrea (score 2)

https://quant.stackexchange.com/a/81790

Definitely not.

The 2 quantities are very very different.

In order to see it, check what is required to compute them.

For $C_{KK}$, one needs the ability to compute $C(S_0, K)$ for every strike $K$: these can be read from market data (implied vol, spot, rates) with a bit of interpolation. But everybody will ultimately agree what they are.

On the other hand, $C_{SS}$ requires the values of $C(S, K)$ for every spot price $S$ (btw, which $K$? Let's say some ATM strike $K_{ATM}$). Where do we get that? Only God knows what would happen if the spot were different. We only observe $S_0$, nothing else.

You might ask: how do people compute delta & gamma? They need big assumptions, which go under the names of: sticky delta, sticky moneyness, sticky strike, sticky model etc... And, no 2 traders would agree which one is correct.

So, no, they are not the same thing.