Quick Review
From the derivation of Last Article - Fokker Planck Equation, below is the final conclusion.
\[\frac{\partial p(s,t)}{\partial t} - \frac{1}{2}*\frac{\partial^2 [\sigma^2(s,t) p(s,t)]}{\partial s^2} + \frac{\partial [\mu(s,t) p(s,t)]}{\partial s} = 0\]Which describe the an euqation that the pdf of future underlying asset price should satisfy. And if the $\mu(s,t)$ and $\sigma(s,t)$ is the risk-neutral parameters for $dS$. Then the $p(s,t)$ is the risk-neutral probability for asset price.
In last article, we generally assume that the SDE for s is as below:
\[dS = \mu(S,t)dt + \sigma(S,t)dw\]In this article, we will derive the Dupire Formula.
By using it, we can calculate $\sigma^2(s,t)$ from market quotes.
Dupire Formula
To simpilify the derivation, we uses call option payoff as an example:
\[f(s) = (s-K)^+\]Other payoff functions just follow the same derivation procedures. Suppose $C(K,t)$ is the observable call option price from market, $K$ is strike price and $t$ is time to maturity.
Under risk-neutral measure, the $C(K,t)$ can be calculated below:
\[C(K,t) = P(0,t)\tilde{E}[(s-K)^+] \\ = P(0,t)\int_K^{+\infty}(s-K)p(s,t)ds\]Where $P(0,t)$ is the current price of zero coupon bond which pays $1 at time t. And it’s observable from market.
\[P(0,t) = e^{-\int_0^t r(u)du}\]Note that, $r(u)$ is deterministic here.
Three Derivatives
To obtain final result, we need to calculate 3 derivative: $\frac{\partial C}{\partial K}$, $\frac{\partial^2 C}{\partial K^2}$ and $\frac{\partial C}{\partial t}$:
- $\frac{\partial C}{\partial K}$
- $\frac{\partial^2 C}{\partial K^2}$
- $\frac{\partial C}{\partial t}$
\[\frac{\partial P(0,t)}{\partial t} = \frac{\partial (e^{\int_0^t r(u)du})}{\partial t} = -r(t)P(0,t)\]
Main Equation
$\frac{\partial C}{\partial t}$ would be our starting point, rearrange it:
\[\frac{\partial C}{\partial t} + r(t)C = P(0,t)\int_K^{+\infty}(s-K)\frac{\partial p(s,t)}{\partial t}ds\]Plug in Fokker Planck Equation to replace $\frac{\partial p(s,t)}{\partial t}$:
\[\frac{\partial C}{\partial t} + r(t)C = P(0,t)\int_K^{+\infty}(s-K)[\frac{1}{2}*\frac{\partial^2 [\sigma^2(s,t) p(s,t)]}{\partial s^2} - \frac{\partial [\mu(s,t) p(s,t)]}{\partial s}]ds\]Two Intergrals
Next step would be solve the 2 intergrals in the right hand side:
- $I_1 = \int_K^{+\infty}(s-K)\frac{\partial^2 [\sigma^2(s,t) p(s,t)]}{\partial s^2}ds$
Applying intergral by part, which
\[u = s - K, v = \frac{\partial [\sigma^2(s,t) p(s,t)]}{\partial s}\]So the intergral would be: \(I_1 = (s-K)\frac{\partial [\sigma^2(s,t) p(s,t)]}{\partial s}|^{+\infty}_{s=K} - \int_K^{+\infty}\frac{\partial [\sigma^2(s,t) p(s,t)]}{\partial s}ds \\ = 0 - 0 - \int_K^{+\infty}\frac{\partial [\sigma^2(s,t) p(s,t)]}{\partial s}ds\)
Where
\[\lim_{s \to +\infty}\frac{\partial [\sigma^2(s,t) p(s,t)]}{\partial s} = 0\]
One can intuitively think pdf $p(s,t)$ remains 0 when s is arbitrarily large.
Apply intergral by part again, which
\[u = s, v = \frac{\partial [\sigma^2(s,t) p(s,t)]}{\partial s}\]then
\[I_1 = -[s\frac{\partial [\sigma^2(s,t) p(s,t)]}{\partial s}|^{+\infty}_{s=K} - \int_K^{+\infty}\frac{\partial [\sigma^2(s,t) p(s,t)]}{\partial s}ds]\\ = -\sigma^2(s,t) p(s,t)|^{+\infty}_{s=K} \\ = \sigma^2(K,t) p(K,t)\]Also from $\frac{\partial^2 C}{\partial K^2}$, we know that:
\[\frac{\partial^2 C}{\partial K^2} = P(0,t)p(K,t)\]Plug it into $I_1$
\[I_1 = \frac{\sigma^2(K,t)}{P(0,t)}\frac{\partial^2 C}{\partial K^2}\]- $I_2 = \int_K^{+\infty}(s-K)\frac{\partial [\mu(s,t) p(s,t)]}{\partial s}]ds$
Intergral by part, which
\[u = s-K, v = \mu(s,t) p(s,t)\]Then,
\[I_2 = (s-K)\mu(s,t) p(s,t)|^{+\infty}_{s=K} - \int_K^{+\infty}\mu(s,t) p(s,t)ds \\ = - \int_K^{+\infty}\mu(s,t) p(s,t)ds\]To make this solvable, we need to further asseme that:
\[\mu(s,t) = [r(t)-q(t)]s\]Then
\[I_2 = -[r(t)-q(t)]\int_K^{+\infty}sp(s,t)ds\]From previous derivation, we already know:
\[\frac{-1}{P(0,t)}\frac{\partial C}{\partial K} = \int_K^{+\infty}p(s,t)ds\]and
\[\frac{C(K,t)}{P(0,t)} = \int_K^{+\infty}(s-K)p(s,t)ds\]So
\[I_2 = -[r(t)-q(t)][\frac{C(K,t)}{P(0,t)}-\frac{-K}{P(0,t)}\frac{\partial C}{\partial K}]\]Back to Main Equation
\[\frac{\partial C}{\partial t} + r(t)C = P(0,t)[\frac{1}{2}*I_1 - I_2]\]Pulg in $I_1$ and $I_2$:
\[\frac{\partial C}{\partial t} + r(t)C = P(0,t)[\frac{1}{2}*\frac{\sigma^2(K,t)}{P(0,t)}\frac{\partial^2 C}{\partial K^2} + [r(t)-q(t)][\frac{C(K,t)}{P(0,t)}-\frac{-K}{P(0,t)}\frac{\partial C}{\partial K}]]\\ = \frac{\sigma^2(K,t)}{2}\frac{\partial^2 C}{\partial K^2} + [r(t)-q(t)]C+[r(t)-q(t)]K\frac{\partial C}{\partial K}\]rearrange it:
\[\sigma^2(K,t) = \frac{\frac{\partial C}{\partial t}+q(t)C+[r(t)-q(t)]K\frac{\partial C}{\partial K}}{\frac{1}{2}\frac{\partial^2 C}{\partial K^2}}\]Note
This solution is different from some answers in the references. It’s becuase my assumtion for S is:
\[dS = [r(t)-q(t)]Sdt + \sigma(S,t)dw\]But their assumtion is lognormal distribution:
\[dS = [r(t)-q(t)]Sdt + \sigma(S,t)Sdw\]Below is their final answer:
\[\sigma^2(K,t) = \frac{\frac{\partial C}{\partial t}+q(t)C+[r(t)-q(t)]K\frac{\partial C}{\partial K}}{\frac{1}{2}K^2\frac{\partial^2 C}{\partial K^2}}\]References
Fokker Planck Equation Derivation: Local Volatility, Ornstein Uhlenbeck, and Geometric Brownian
The Dupire Formula
Derivation of Local Volatility
DerivationofFokker–Planck EquationfortheLocal VolatilityModel
