Heston Call Probabilities and Gil-Pelaez Fourier Inversion

# A clarification on the Heston option pricing formula


# A clarification on the Heston option pricing formula












I have carefully reconstructed all the computations that lead to the Heston option pricing formula for a call. I end up with this formula for the "adjusted" probabilities

$$ P_j\left(x,v,T;\ln K\right) = \int_{-\infty}^{\infty}\frac{d\phi}{2\,\pi}\,\frac{e^{-i\,\phi\,\ln\left(K\right)}\,f_j\left(x,v,T;\phi\right)}{i\,\phi}\,d\phi $$

while in the original paper and all textbooks the final formula is expressed as

$$ P_j\left(x,v,T;\ln K\right) = \frac{1}{2}+\int_{0}^{\infty}\frac{d\phi}{\pi}\,\text{Re}\left(\frac{e^{-i\,\phi\,\ln\left(K\right)}\,f_j\left(x,v,T;\phi\right)}{i\,\phi}\,\right)d\phi. $$

I understand that, if $f_j\left(x,v,T;\phi\right)$ is such that

$$ f_j\left(x,v,T;-\phi\right) = \bar{f_j}\left(x,v,T;\phi\right)\quad(1) $$

(where a bar indicates the complex conjugate) then we have immediately that

$$ \int_{-\infty}^{\infty}\frac{d\phi}{2\,\pi}\,\frac{e^{-i\,\phi\,\ln\left(K\right)}\,f_j\left(x,v,T;\phi\right)}{i\,\phi}\,d\phi = \int_0^\infty\frac{d\phi}{\pi}\,\text{Re}\left(\frac{e^{-i\,\phi\,\ln\left(K\right)}\,f_j\left(x,v,T;\phi\right)}{i\,\phi}\,\right)d\phi. $$

My problem is thus twofold. First I cannot see from the definition of $f_j\left(x,v,T;\phi\right)$ that the property (1) is satisfied (neither I have found any kind of discussion in other textbooks) and, second, even if (1) is proved I miss the $\frac{1}{2}$ that appears in the Heston formula.

Any help would be greatly appreciated.

## Answer by Quantuple (score 2, accepted)

https://quant.stackexchange.com/a/24886

I do not know what are the exact steps you followed, but here's my 2 cents:

\begin{align*} C(S_t,t) &= P(t,T) E^Q [ (S_T - K)^+ ] \\ & = P(t,T) E^Q [ S_T 1_{S_T \geq K} ] - K P(t,T) E^Q [ 1_{S_T \geq K} ] \end{align*}

Which can be re-written (using a change of numéraire)

\begin{align*} C(S_t,t) &= S_t E^{Q^S} [ 1_{S_T \geq K} ] - K P(t,T) E^Q [ 1_{S_T \geq K} ] \\ &= S_t P_1 - K P(t,T) P_2 \end{align*}

where $P_1 = Q^S( S_T \geq K)$ (resp. $P_2 = Q(S_T \geq K)$) are directly related to the cumulative distribution function of the stock price $S_T$ under the measure $Q^S$ (resp. $Q$) where the stock price (resp. the riskless money account) is chosen as numéraire.

Because $P_1$ and $P_2$ share very similar interpretations (up to a change of measure), I simply focus on $P_2$ in what follows.

If we knew the probability density function $q(s_t,v_t,t;S_T=s)$ in closed-form, we could trivially infer the desired cdf by integrating it and hence deduce $P_2$. This is not the case here.

However, we know the characteristic function $f_2$ of the log-stock price $s_T = \ln(S_T)$ under $Q$ (remember that Heston's model is an affine diffusion framework), which by definition writes $f_2(s_t,v_t,t;\phi) = E^Q[e^{i\phi s_T}]$. From this definition comes the equality you mention in your original post.

At this point, there exists a general result, known as the Gil-Pelaez inversion theorem, which tells you how to 'invert' $f_2(\phi)$ (which is basically a Fourier transform) to obtain the desired CDF. Here, the theorem writes

$$Q(s_T < \ln(K) ) = \frac{1}{2} - \frac{1}{\pi} \int_0^\infty \Re \left[ \frac{e^{-i \phi \ln(K)}f_2(\phi))}{i\phi} \right] d\phi$$

Making the missing 1/2 term appear. Noting that $P_2 = 1 - Q(s_T < \ln(K) )$ concludes the demonstration. Similar developments can be applied to obtain $P_1$. You can probably find demonstrations of the Gil-Pelaez inversion theorem in the literature.