Time-independent local volatility

Suppose somebody provides us with a surface of European call prices $C(\tau,K)$ where $\tau$ stands for time-to-maturity and $K$ for the strike. By Dupire's results, there is a unique local volatility function $\sigma(\tau,K)$ that generates these prices, and it can be expressed from them as $$\sigma(\tau,K) = \frac{2C_\tau}{K^2C_{KK}},$$ here for simplicity I am assuming that interest rate is zero. Now, if we just have $C(T,K)$ for a single maturity $\tau = T$, is that true that there exists a unique time-independent local volatility $\sigma(K)$ that generates this price at that maturity? In case it does, is there an analytic formula for that function?

Yes, there is a unique time homogeneous local vol model. This is proven in http://www.sciencedirect.com/science/article/pii/S0304414912002487. There is a slight generalization required that if the option-implied density is zero somewhere, the corresponding local vol is infinite in that region, giving a "gap diffusion".

No, there is no nice formula for the local vol in this case.

• Thank you for the reference. Before I delve into the complicated details of the paper, would you be so kind as to point out whether there is and if so where it is that the time homogeneous local volatility is described? – Hans Jan 7 '16 at 4:15
• The Noble paper does not explicitly describe the local volatility function. In fact I would not suggest to try to go into the proofs of that paper unless your goal is to prove similar results yourself. – q.t.f. Jan 11 '16 at 16:13
• So you are saying that: the paper has proved the existence of time homogeneous local volatility function that produces a given marginal probability distribution at a given time, and that local volatility function is unique; however, the paper does not give the explicit form of the local volatility function. Is that right? – Hans Jan 12 '16 at 2:09

No.

In practice the local volatility model has a finite number of slices, so a single slice works as well. Now the problem is : how to compute the time derivative ? Well without adding any information you know that $$C(0,K) = (S_0-K)_+$$ so you could try $$C_\tau = \frac{C(\tau,K)-C(0,K)}{\tau}$$ but it is a very crude approximation. What you may want to do instead is first to use the Dupire formula w.r.t. implied volatilities and consider the whole surface flat and equal to you slice, thus taking $\Sigma_\tau=0$. It is equivalent to use the formula w.r.t. total implied variances $w(\tau,K):=\Sigma^2(\tau,K)\tau$ and compute the derivative as above : $$\lim_{\tau \rightarrow 0_+} w(\tau,K) = 0$$ $$w_\tau(\tau,K) \eqsim \frac{w(\tau,K)-w(0_+,K)}{\tau} = \Sigma^2(\tau,K)$$

Here is the formula w.r.t. total variances (a good reference is Gatheral's The Volatility Surface: A Practitioner's Guide) :

$$\sigma^2(\tau,K) = \frac{w_\tau(K)}{\displaystyle1-\frac{y}{w(\tau,K)}w_y(\tau,K)+\frac{1}{4}\left(-\frac{1}{4}-\frac{1}{w(\tau,K)}+\frac{y^2}{w^2(\tau,K)}\right)(w_y(\tau,K))^2+\frac{1}{2}w_{yy}(\tau,K)}$$ where $y=\ln(K/S_O)$ and $w_\tau(K)=\Sigma(K)^2$. You see here that the denominator still depends on $\tau$ through $w$. There is no way around this.

The vanilla price approach will yield the same conclusion since the equivalent time-derivative has to be computed using the Black-Scholes $\theta$ greek (derivative w.r.t. to time to maturity) with the implied volatility at the considered strike. The Black-Scholes theta depends on the time to maturity even without drift.

Also the local volatility generated by an implied volatility surface is unique obviously (satisfies Dupire equation).

• In my case I have the IVol $\Sigma(T,\cdot)$ available, and I would like to find time-independent LVol $\sigma$ that leads $\Sigma$. I am not sure your answer solves that issue at all. – Ulysses Feb 23 '15 at 13:03
• I wrote down the equation and it necessarily depends on time so my first intuition was wrong. I edited my answer. – vanna Feb 24 '15 at 20:57
• Regarding the formula, the dependence on $\tau$ in each term of the right hand side of the equation does not imply dependence of $\tau$ of the right hand side as a whole. The time dependence may well cancel for some form of $w$. So this answer fails to reach the conclusion. – Hans Jan 7 '16 at 3:59
• The question was about the existence of such surface given any input. My answer hints at counter examples : take any non trivial SVI – vanna Jan 7 '16 at 12:04