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I have trouble understanding the use/implication of the Lévy-Khintchine theorem. One possible way to state it is the following:

The characteristic function $\varphi$ is infinitely divisible if and only if $\varphi = \exp{\psi}$ and

$$\psi(\lambda) = i \lambda a - \frac{\sigma^2 \lambda^2}{2} + \int_{\mathbb R} \left(\exp(i \lambda x) - 1 - \frac{i\lambda x}{1+x ^2}\right)\frac{1+x^2}{x^2} \nu(dx)$$

where $\nu$ is a finite measure on $\mathbb R$ with $\nu\{0\}=0$, $(a,\sigma^2,\nu)$ is the Lévy-Khintchine triplet.

Now, this formula looks a little intimidating. What would be its value/implications in words?

Somehow, one can use the theorem to decide the following question:

Is there a time-homogenous, stochastically continuous process $X(t)$ with independent increments and with $X(1)\sim Exp(1)$?

I unfortunately do not see how the theorem might help answer this question...

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Let $\nu$ a measure on $\mathbb{R}$ such that $\nu(\{0\})=0$ and $a,\sigma^2 \in \mathbb{R}$. Define $\psi$ by $$\psi(\lambda) = i \lambda a - \frac{\sigma^2 \lambda^2}{2} + \int_{\mathbb R} \left(\exp(i \lambda x) - 1 - \frac{i\lambda x}{1+x ^2}\right)\frac{1+x^2}{x^2} \nu(dx) \tag{1}$$

Then there exists a time-homogenous stochastically continuous process $(X_t)_t$ with independent increments such that $$\mathbb{E}e^{\imath \, \lambda \cdot X_t} = \exp(t \cdot \psi(\lambda))$$ in particular $\mathbb{E}e^{\imath \, \lambda \cdot X_1} = \exp(\psi(\lambda))$.

Thus, it suffices to find a finite measure $\nu$ and $b, \sigma^2 \in \mathbb{R}$ such that $$\exp(\psi(\lambda)) = \mathbb{E}e^{\imath \, \lambda \cdot X_1} = \mathbb{E}e^{\imath \, \lambda \cdot \text{Exp}(1)} = \frac{1}{1-\imath \, \lambda} = \exp(-\log(1-\imath \lambda))$$ where $\log$ denotes the principal value (so the imaginary part is in $(-\pi,\pi]$).

To find a suitable measure $\nu$ we start rewriting the expression $-\log(1-\imath \, t)$: $$\begin{align} -\log(1-\imath \, \lambda) &= \imath \int_0^{\lambda} \frac{1}{1-\imath \, t} \, dt = \imath \int_0^{\lambda} \int_0^{\infty} e^{-x \cdot (1-\imath \, t)} \, dx \, dt \\ &= \imath \, \int_0^{\infty} \int_0^{\lambda} e^{-x \cdot (1-\imath \, t)} \, dt \, dx = \int_0^{\infty} \frac{e^{\imath \, \lambda \cdot x}-1}{x} \cdot e^{-x} \, dx \tag{2} \end{align}$$

This already looks a bit similar to $(1)$. Indeed, $$\begin{align} - \log(1-\imath \, t) &\stackrel{(2)}{=} \int_0^{\infty} \frac{e^{\imath \, \lambda \cdot x}-1}{x} \cdot e^{-x} \, dx = \int_0^{\infty} \left(e^{\imath \, \lambda \cdot x}-1 - \frac{\imath \, \lambda \cdot x}{1+x^2} \right) \cdot \frac{e^{-x}}{x} \, dx + \imath \, \lambda \cdot a \end{align}$$ where $$a := \int_0^{\infty} \frac{e^{-x}}{1+x^2} \, dx$$

So we arrive at $- \log(1-\imath \, \lambda) = \psi(\lambda)$ where $$\begin{align} a &= \int_0^{\infty} \frac{e^{-x}}{1+x^2} \, dx \\ \sigma^2 &= 0 \\ \nu(dx) &:= \frac{x}{1+x^2} \cdot e^{-x} \cdot 1_{(0,\infty)}(x) \, dx \end{align}$$ Obviously, $\nu$ is a finite measure, $\nu(\{0\})=0$. By applying the theorem, we conclude that there exists a time-homogenous stochastically continuous process $(X_t)_t$ with independent increments such that $X_1 \sim \text{Exp}(1)$.

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  • $\begingroup$ Thanks! What is the intuitive meaning of the Lévy-Khintchine triplet $(a,\sigma^2,\nu)$? I especially can't imagine what $\nu$ really measures... $\endgroup$ – madison54 Mar 13 '13 at 12:28
  • $\begingroup$ Let $G \subseteq \mathbb{R} \backslash \{0\}$ such that $\nu(G)=0$. Then one can actually show that $(X_t)$ has (almost surely) no jumps with magnitude $G$, i.e $X_t - X_{t-} \notin G$ almost surely (where $X_{t-} := \lim_{s \uparrow t} X_s$). In the given example, this means that $(X_t)_t$ has only jumps with positive magnitude (since $\nu((-\infty,0))=0$). So $\nu$ somehow characterizes the jumps of the process. $\endgroup$ – saz Mar 13 '13 at 13:00

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