Hey in the COS method we use characteristic function of $\ln{S_T}$ to price european options (by recovering density from characteristic function). But how do we know that density exists? For example I would like to use this method to price options in Kou, NIG or CGMY model, do these Levy processes has density and we can use this method? I know the theorem which states that if characteristic function is integrable, i.e $\int_{\mathbb{R}}|\phi (t)|dt<\infty $ then there exists continuous density with respect to Lebesque measure. Is characteristic function integrable for processes which I mentioned and in general for all Levy processes?


1 Answer 1


Example: Kou Model

Well, let's have a look at the Kou model $$\phi_{\ln(S_t)}(u)=\exp\left(\underbrace{\ln\left(S_0e^{(r-q+\omega)t}\right)iu}_{=(\star)}\;\;\underbrace{-\frac{1}{2}\sigma^2tu^2}_{=(\star\star)}+\underbrace{\lambda t\left(\frac{p\eta_+}{\eta_+-iu}+\frac{p'\eta_-}{\eta_-+iu}-1\right)}_{=(\star\star\star)}\right),$$ where

  • $r$ and $q$ are interest rate and dividend yield
  • $\sigma$ is the volatility
  • $p$ and $p'$ are the probabilities of up and down jumps ($p+p'=1$)
  • $\eta_+>1$ and $\eta_->0$ relate to the (inverse) mean jump size
  • $\omega$ is the jump compensation ensuring that $S_te^{-(r-q)t}$ is a martingale.

I claim we can easily prove that \begin{align*} \int_\mathbb{R}|\phi(u)|\text{d}u<\infty. \end{align*}

  • Clearly, $|e^{(\star)}|\leq 1$ by Euler's formula. All these points lie on the unit circle.
  • Next, $|e^{(\star\star)}|$ decays super quickly, it's like a Gaussian bell curve. So $\int_\mathbb{R} |e^{(\star\star)}|<\infty$.
  • The $(\star\star\star)$ case is not much harder. Firstly, the $-\lambda t$ can be taken out of the integral and doesn't bother us. Next, rationalising the denominator, we get \begin{align*} \left|\exp\left(\frac{\lambda tp\eta_+}{\eta_+-iu}\right)\right| &= \left|\exp\left(\frac{\lambda tp\eta_+(\eta_++iu)}{\eta_+^2+u^2}\right)\right| \\ &= \left|\exp\left(\frac{\lambda tp\eta_+\eta_+}{\eta_+^2+u^2}\right)\exp\left(\frac{\lambda tp\eta_+ui}{\eta_+^2+u^2}\right)\right| \\ &\leq \exp\left(\frac{\lambda tp\eta_+\eta_+}{\eta_+^2+u^2}\right). \end{align*} This generates further exponential decay as $u\to\pm\infty$ and remains finite at $u=0$. In fact, we can estimate the term by a constant. The same applies to the $\eta_-$ fraction.

Thus, all in all, $(\star)$ is bounded by one, $(\star\star)$ looks like a bell curve and $(\star\star\star)$ is also bounded. Taken together, $\phi$ decays exponentially fast and is in $L^1(\mathbb{R})$.

A few notes

  • The Riemann-Lebesgue Lemma tells us that characteristic functions (Fourier transforms) converge to zero for $u\to\pm\infty$. All we need to show is that they converge quickly enough to zero.
  • You can do similar bounds for the NIG model and other characteristic functions
  • For these exponential Lévy processes, the probability density function is rarely known in closed-form. The characteristic function is very simple though. You can always obtain the PDF by (numerically) Fourier inverting the characteristic function.

Sidenote: Uncertainty Principle

One thing to keep in mind is the uncertainty principle: If $f$ spreads out, then its Fourier transform $\hat{f}$ is very `compact' and vice versa. The density of the log stock price is very spread out if the time horizon $t$ is large: there is a wide range of potential values for a stock price in 5 years time. However, the density for tomorrow's stock price is very peaked around today's value. Accordingly, the characteristic function is very spread out for short maturities but decays super quickly for long maturities. Example for the imaginary part of the characteristic function of the normal distribution (real part looks similar): enter image description here

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    $\begingroup$ Nice ..and this is why we have different methods for transform based option pricing for short and not-so-short time horizons, I'd argue. And why you always have to be extra careful when calculating those integrals with rigid methods like simple (as in: non-adaptive) FFT transform methods. $\endgroup$ Mar 4, 2021 at 7:19
  • $\begingroup$ @Kermittfrog Thank you! I always had the impression the Fourier literature emphasises the moneyness dimension way more: this method is very inaccurate for OTM options (e.g. FFT, COS or Lewis' approach) while this method is good for OTM options (e.g. Saddlepoint method). The time to maturity (in my reading) seems to get less attention. Do you have a method in mind that's particularly suited/awful when it comes to very short/long lived options? And I have to confess I always use adaptive (integral) methods to implement Fourier methods, I'm save here! :D $\endgroup$
    – Kevin
    Mar 4, 2021 at 9:45
  • $\begingroup$ I first used the FrFT method 'only' and simply pumped the number of nodes to some large value as my options' expiries where commonly >1M or so, so I was never truly in the problematic range in my applications. Lately, I brute-force-ishly use some adaptive quadrature schemes as I simply do not need a zillion option prices in one go anymore. So, sorry, I have no examples at hand. $\endgroup$ Mar 4, 2021 at 11:49
  • $\begingroup$ Thanks for your answer. As you wrote: ,,For these exponential Lévy processes, the probability density function is rarely known in closed-form.'' and what if the PDF doesnt exists (are there Levy processes for which density doesnt exists?,If density is not known in closed form, how do we know that it exists)? For example Poisson distribution has characteristic function but it doesnt have density, so what do we get from COS method in this case? How do we know that the model which we use has density and we can use COS method? $\endgroup$ Mar 4, 2021 at 13:06
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    $\begingroup$ @JohhnWhite Characteristic functions normally decay exponentially fast. So their integral over $\mathbb{R}$ is finite and you can use your theorem to deduce that a PDF exists. I showed you how to do the calculations for the Kou model, you can do similar calculations for other models. The Poisson distribution different from the other models because it's a discrete distribution. The PMF is the Radon Nikodym derivative wrt the counting measure, not the Lebesgue measure. Note that a characteristic function exists even if a PDF does not. $\endgroup$
    – Kevin
    Mar 4, 2021 at 18:50

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