# Stability of Finite Difference method for Breeden-Litzenberger

I am trying to derive a risk-neutral density from European call option prices using a second order finite difference scheme. Let $$C(K,T)$$ be the price of a European call with strike $$K$$ and expiry $$T$$ then the risk-neutral density of the log price returns can be computed using Breeden-Litzenberger theorem.

I have given a set of strikes $$K_1, \ldots, K_n$$ (not equidistant and can lie far away) with corresponding implied volatilities $$\sigma(K_1,T), \ldots, \sigma(K_n,T)$$. Say I want to calculate the density for some interior strike $$k_p$$ with $$1 < p < n$$, I do: \begin{align} f(K) & \approx e^{rT} \frac{C(K_p+\Delta,T)-2C(K_p,T)+C(K_p-\Delta, T)}{(\Delta)^2} (*) \end{align}

My question is how to choose $$\Delta$$? Ideally I choose $$\Delta$$ as small as possible because the above approximation is valid for $$\Delta \to 0$$. However, when I choose $$\Delta$$ small (say $$\Delta=0.01$$), the density tends to be negative for more strike points than when I choose $$\Delta$$ larger. This is mainly the case for shorter expiries (1 month, 3 months) where the smile can be quite steep. I don't see this problem for longer maturities.

Questions

(1) Why do I see this behaviour? One important note is that to calculate $$C(K_p \pm \Delta, T)$$, I use a cubic spline to interpolate the volatility smile to obtain $$\sigma(K_p \pm \Delta,T)$$. The European call price is calculated using the Black model. It seems that when using the cubic spline it actually creates more arbitrages when choosing $$\Delta$$ small. I don't want to use an arbitrage free smoothing model because the objective is to test the raw vol smile for arbitrage ...

(2) Since the strikes can be quite far away and are non equidistant, I have tried using an non-equidistant version of $$(*)$$ where I basically set $$\Delta = K_i - K_j$$, but these $$\Delta's$$ can be quite large so I am not sure if the approximation is really valid in that case. I have to say that in this case there are very few arbitrages to detect.

• Wait, do I understand correctly that you are wondering why you see negative densities in an arbitrage situation? – Raskolnikov Jun 5 '20 at 6:55
• That is not my question. The question is (1) why the approximation becomes seems to become less stable when $\Delta$ is smaller for short expiries and (2) practical advice on how to do this without any arbitrage free smoothing technique. – user39039 Jun 5 '20 at 8:01