The One-Half Term in Fourier Inversion for Option Exercise Probabilities

# Where does 1/2 in Fourier Transform method of pricing options come from?


# Where does 1/2 in Fourier Transform method of pricing options come from?












I am reading Jianwe Zhu's Applications of Fourier Transform to Smile Modeling. On page 26, the author is describing how to use the Fourier tranform to price vanilla European call options. If $f_j$ is the Fourier transform of the density function of $x = \ln(S)$ (under measure $Q$), then the probability of exercise under $Q$ (that, is probability $x > \ln(K)$) is

$$F_j (x(T) > a) = \frac{1}{2\pi} \int _{\mathbb{R}} f_j(\phi) \Bigg(\int_a^{\infty} e^{-i\phi x} \mathrm{d}x\Bigg) \mathrm{d}\phi\text{.}\tag{1}$$

This equation makes sense to me. The author then says, equation (2.17),

> A further straightforward calculation yields $$F_j (x(T) > a) = \frac{1}{2} + \frac{1}{2\pi}\int_{\mathbb{R}}f_j(\phi)\frac{e^{-i\phi a}}{i\phi}\mathrm{d}\phi \text{.}\tag{2}$$

This equation does not make sense to me. Where did equation (2) come from? In particular, where did the $\frac{1}{2}$ come from?

## Answer by Pleb (score 5)

https://quant.stackexchange.com/a/69559

#### It comes from a direct application of the Fourier inversion theorem for a CDF:

For a general one-dimensional CDF $F_X(x)$, the Fourier inversion theorem can be described as:

\begin{align} F_X(x) &= \frac{1}{2} - \frac{1}{2\pi} \int_{-\infty}^\infty \frac{e^{-iux}\phi_X(u)}{iu} \: du\\ &=\frac{1}{2} - \frac{1}{\pi} \int_0^\infty \mathcal{R}\left[\frac{e^{-iux}\phi_X(u)}{iu}\right] \: du, \end{align}

where $\phi_X(u)$ is the characteristic function for $X$. See Schmelzle (2010) chapter 3.2 for full derivations.

With regards to the probability of exercise, $F_j(x(T)>a)$, first calculate the inner integral:

$$ F_j(x(T)>a) = \frac{1}{2\pi} \int_{\mathbb{R}}f_j(\phi)\frac{e^{-i\phi a}}{i\phi}\: d\phi. $$

Now, see that:

\begin{align} F_j(x(T)>a) & = 1-F_j(x(T)\leq a)\\ &= 1 - \left(\frac{1}{2} - \frac{1}{2\pi} \int_{\mathbb{R}} f_j(\phi)\frac{e^{-i\phi a}}{i\phi}\: d\phi\right)\\ &=\frac{1}{2} + \frac{1}{2\pi} \int_{\mathbb{R}} f_j(\phi)\frac{e^{-i\phi a}}{i\phi}\: d\phi,\\ \end{align} where we — in the second equality — have used that $F_j(x(T)\leq a)$ is a CDF and have inserted its corresponding Fourier inversion counterpart, as seen above. Furthermore, be aware that $f_j(\phi)$ is defined as the characteristic function per equation (2.13) in the book.