FFT

Still WIP

We will be looking at estimating the characteristic function of the end distribution, i.e. the underlying process at maturity.

For a random variable \(X\) the characteristic function \(\varphi_X(t) : \mathbb{R} \rightarrow \mathbb{C}\) is defined as the expected value of the function \(e^{itX}\)

\[ \varphi_X(t) = E[e^{itX}] = \int_{-\infty}^{\infty} e^{itx} dF_X(x) \]

As opposed to the density function the characteristic function will always exist. If the probability density function \(f_X\) exists then the characteristic function is defined as

\[ \psi_X(t) = \int_{-\infty}^{\infty} e^{itx} f_X(x) dx \]

which is equal to the Fourier transform of the density function \(f_X\).

For a stochastic process \(X_t\) and for any function \(t(s)\) such that \(\int_{-\infty}^{\infty} t(s) X_s ds\) converges for almost all realisations of \(X\), i.e. with probability 1, the characteristic function \(\psi\) is given as

\[ \psi_X(t) = E[e^{i\int_{-\infty}^{\infty} t(x) X(x) dx}] \]

CHECK THE ABOVE!! if this is true then we can use this by splitting the integral by the term structure. We then end of with a sum of integrals, which splits into a product of exponentials. If we assume that the product of exponentials are independent(which I think we can since the Brownian(even Levy?) increments are independent?) then the expected value is multiplicative and we can compute each one separately. \(t(x)\) must be a measure

Given a Levy Process the Levy-Khintchine gives us the expression of the characteristic function as

\[ L-K \]

where ... are the coefficients of the Levy-Ito decomposition

\[ L-I \]

We need to use the below but for that we need to rewrite the above somewhat Since FT is linear we can treat each element of the stream(each term of the expansion) as a separate transform??.

The QUESTION IS: can we do something about that expected value? We should be able to calculate \(\int t(s) X(s) ds\) since we have an approximation of \(X\) but what about the the expected value of the exponential?? Are we sure about that formula?? There is some trick here that we need to figure out....

IDEAS:

Jensen's inequality: \(f(E[X]) \leq E[f(X)]\)

If \(\{X_1 , ... , X_n\}\) are idenpendent variables with characteristic functions \(\phi_{X_i}\) then \(\phi_(X_1 + \cdots X_n)(t) = \prod_i \phi_{X_i} (t)\).

Central limit theorem says that \(W_{t_k} = lim_{N \rightarrow \infty} S_N(t_k)\), i.e. random approimation, where \(S_N(t_k) = (X_1 + \cdots + X_N) \sqrt{\Delta t}\) for a partition \(0 = t_0 \leq \cdots \leq t_N = 1\). So \(\phi_{W_t} = \phi_{lim_{N \rightarrow \infty} S(t_k)}\) is approximated by \(\phi_{\sum X_i \sqrt{\Delta t}} = \prod \phi_{X_i \sqrt{\Delta t}}\)